{"id":375,"date":"2022-05-18T16:39:02","date_gmt":"2022-05-18T16:39:02","guid":{"rendered":"https:\/\/pressbooks.ccconline.org\/accintrostats\/chapter\/additional-information-and-full-hypothesis-test-examples\/"},"modified":"2022-11-09T16:26:18","modified_gmt":"2022-11-09T16:26:18","slug":"additional-information-and-full-hypothesis-test-examples","status":"publish","type":"chapter","link":"https:\/\/pressbooks.ccconline.org\/accintrostats\/chapter\/additional-information-and-full-hypothesis-test-examples\/","title":{"raw":"Chapter 10.6: Additional Information and Full Hypothesis Test Examples","rendered":"Chapter 10.6: Additional Information and Full Hypothesis Test Examples"},"content":{"raw":"&nbsp;\r\n<ul>\r\n \t<li>In a <span data-type=\"term\">hypothesis test<\/span> problem, you may see words such as \"the level of significance is 1%.\" The \"1%\" is the preconceived or preset <em data-effect=\"italics\">\u03b1<\/em>.<\/li>\r\n \t<li>The statistician setting up the hypothesis test selects the value of <em data-effect=\"italics\">\u03b1<\/em> to use <strong>before<\/strong> collecting the sample data.<\/li>\r\n \t<li><strong>If no level of significance is given, a common standard to use is <em data-effect=\"italics\">\u03b1<\/em> = 0.05.<\/strong><\/li>\r\n \t<li>When you calculate the <em data-effect=\"italics\">p<\/em>-value and draw the picture, the <em data-effect=\"italics\">p<\/em>-value is the area in the left tail, the right tail, or split evenly between the two tails. For this reason, we call the hypothesis test left, right, or two tailed.<\/li>\r\n \t<li>The <strong>alternative hypothesis<\/strong>, \\({H}_{a}\\), tells you if the test is left, right, or two-tailed. It is the <strong>key<\/strong> to conducting the appropriate test.<\/li>\r\n \t<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em><strong>never<\/strong> has a symbol that contains an equal sign.<\/li>\r\n \t<li><strong>Thinking about the meaning of the<\/strong><span data-type=\"term\"><em data-effect=\"italics\">p<\/em>-value<\/span>: A data analyst (and anyone else) should have more confidence that he made the correct decision to reject the null hypothesis with a smaller <em data-effect=\"italics\">p<\/em>-value (for example, 0.001 as opposed to 0.04) even if using the 0.05 level for alpha. Similarly, for a large <em data-effect=\"italics\">p<\/em>-value such as 0.4, as opposed to a <em data-effect=\"italics\">p<\/em>-value of 0.056 (alpha = 0.05 is less than either number), a data analyst should have more confidence that she made the correct decision in not rejecting the null hypothesis. This makes the data analyst use judgment rather than mindlessly applying rules.<\/li>\r\n<\/ul>\r\n<p id=\"fs-idp42898080\">The following examples illustrate a left-, right-, and two-tailed test.<\/p>\r\n\r\n<div id=\"fs-idp41487520\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"fs-idp41487648\"><em data-effect=\"italics\">H<sub>o<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 5, <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &lt; 5<\/p>\r\nTest of a single population mean. <em data-effect=\"italics\">H<sub>a<\/sub><\/em> tells you the test is left-tailed. The picture of the <em data-effect=\"italics\">p<\/em>-value is as follows:\r\n<div id=\"fs-idm151356816\" class=\"bc-figure figure\"><span id=\"id14703034\" data-type=\"media\" data-alt=\"Normal distribution curve of a single population mean with a value of 5 on the x-axis and the p-value points to the area on the left tail of the curve.\"><img src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/05\/fig-ch09_09_01-02-1.jpg\" alt=\"Normal distribution curve of a single population mean with a value of 5 on the x-axis and the p-value points to the area on the left tail of the curve.\" width=\"380\" data-media-type=\"image\/jpg\" \/><\/span><\/div>\r\n<\/div>\r\n<div id=\"fs-idp42408528\" class=\"statistics try\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\r\n<div data-type=\"title\">Try It<\/div>\r\n<div id=\"fs-idp67352928\" data-type=\"exercise\">\r\n<div data-type=\"problem\">\r\n<p id=\"eip-idm132446832\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 10, <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &lt; 10<\/p>\r\n<p id=\"eip-idp3861728\">Assume the <em data-effect=\"italics\">p<\/em>-value is 0.0935. What type of test is this? Draw the picture of the <em data-effect=\"italics\">p<\/em>-value.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\" data-type=\"example\">\r\n\r\n<em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2264 0.2\u2003\u2003<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &gt; 0.2\r\n<p id=\"fs-idp117549824\">This is a test of a single population proportion. <em data-effect=\"italics\">H<sub>a<\/sub><\/em> tells you the test is <strong>right-tailed<\/strong>. The picture of the <em data-effect=\"italics\">p<\/em>-value is as follows:<\/p>\r\n\r\n<div id=\"fs-idm2304848\" class=\"bc-figure figure\"><span id=\"id14703170\" data-type=\"media\" data-alt=\"Normal distribution curve of a single population proportion with the value of 0.2 on the x-axis. The p-value points to the area on the right tail of the curve.\"><img src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/fig-ch09_09_02a-1.jpg\" alt=\"Normal distribution curve of a single population proportion with the value of 0.2 on the x-axis. The p-value points to the area on the right tail of the curve.\" width=\"380\" data-media-type=\"image\/jpg\" \/><\/span><\/div>\r\n<\/div>\r\n<div id=\"fs-idm126098448\" class=\"statistics try\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\r\n<div data-type=\"title\">Try It<\/div>\r\n<div id=\"eip-417\" data-type=\"exercise\">\r\n<div id=\"fs-idp131553552\" data-type=\"problem\">\r\n<p id=\"eip-idm89526848\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2264 1, <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &gt; 1<\/p>\r\n<p id=\"eip-idm24425280\">Assume the <em data-effect=\"italics\">p<\/em>-value is 0.1243. What type of test is this? Draw the picture of the <em data-effect=\"italics\">p<\/em>-value.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\" data-type=\"example\">\r\n<p id=\"element-38a\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 50\u2003\u2003<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 50<\/p>\r\nThis is a test of a single population mean. <em data-effect=\"italics\">H<sub>a<\/sub><\/em> tells you the test is <strong>two-tailed<\/strong>. The picture of the <em data-effect=\"italics\">p<\/em>-value is as follows.\r\n<div id=\"fs-idm117972816\" class=\"bc-figure figure\"><span id=\"id4829633\" data-type=\"media\" data-alt=\"Normal distribution curve of a single population mean with a value of 50 on the x-axis. The p-value formulas, 1\/2(p-value), for a two-tailed test is shown for the areas on the left and right tails of the curve.\"><img src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/fig-ch09_09_03N-1.jpg\" alt=\"Normal distribution curve of a single population mean with a value of 50 on the x-axis. The p-value formulas, 1\/2(p-value), for a two-tailed test is shown for the areas on the left and right tails of the curve.\" width=\"380\" data-media-type=\"image\/jpg\" \/><\/span><\/div>\r\n<\/div>\r\n<div id=\"fs-idm63250432\" class=\"statistics try\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\r\n<div data-type=\"title\">Try It<\/div>\r\n<div data-type=\"exercise\">\r\n<div data-type=\"problem\">\r\n<p id=\"eip-idp41687312\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.5, <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 0.5<\/p>\r\n<p id=\"eip-idm5668432\">Assume the <em data-effect=\"italics\">p<\/em>-value is 0.2564. What type of test is this? Draw the picture of the <em data-effect=\"italics\">p<\/em>-value.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm12892576\" class=\"bc-section section\" data-depth=\"1\">\r\n<h3 data-type=\"title\">Full Hypothesis Test Examples<\/h3>\r\n<div class=\"textbox textbox--examples\" data-type=\"example\">\r\n<div data-type=\"exercise\">\r\n<div id=\"id5300933\" data-type=\"problem\">\r\n\r\nJeffrey, as an eight-year old, <strong>established a mean time of 16.43 seconds<\/strong> for swimming the 25-yard freestyle, with a <strong>standard deviation of 0.8 seconds<\/strong>. His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for <strong>15 25-yard freestyle swims<\/strong>. For the 15 swims, <strong>Jeffrey's mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds.<\/strong> Conduct a hypothesis test using a preset <em data-effect=\"italics\">\u03b1<\/em> = 0.05. Assume that the swim times for the 25-yard freestyle are normal.\r\n\r\n<\/div>\r\n<div id=\"id5036614\" data-type=\"solution\">\r\n\r\nSet up the Hypothesis Test:\r\n\r\nSince the problem is about a mean, this is a <strong>test of a single population mean<\/strong>.\r\n\r\n<em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 16.43\u2003\u2003<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &lt; 16.43\r\n<p id=\"element-241\">For Jeffrey to swim faster, his time will be less than 16.43 seconds. The \"&lt;\" tells you this is left-tailed.<\/p>\r\nDetermine the distribution needed:\r\n\r\n<strong>Random variable: <\/strong>\\(\\overline{X}\\) = the mean time to swim the 25-yard freestyle.\r\n\r\n<strong>Distribution for the test: <\/strong>\\(\\overline{X}\\) is normal (population <span data-type=\"term\">standard deviation<\/span> is known: <em data-effect=\"italics\">\u03c3<\/em> = 0.8)\r\n\r\n\\(\\overline{X}~N\\left(\\mu ,\\frac{{\\sigma }_{X}}{\\sqrt{n}}\\right)\\) Therefore, \\(\\overline{X}~N\\left(16.43,\\frac{0.8}{\\sqrt{15}}\\right)\\)\r\n\r\n<em data-effect=\"italics\">\u03bc<\/em> = 16.43 comes from <em data-effect=\"italics\">H<sub>0<\/sub><\/em> and not the data. <em data-effect=\"italics\">\u03c3<\/em> = 0.8, and <em data-effect=\"italics\">n<\/em> = 15.\r\n\r\nCalculate the <em data-effect=\"italics\">p<\/em>-value using the normal distribution for a mean:\r\n<p id=\"element-706\"><em data-effect=\"italics\">p<\/em>-value = <em data-effect=\"italics\">P<\/em>(\\(\\overline{x}\\) &lt; 16) = 0.0187 where the sample mean in the problem is given as 16.<\/p>\r\n<em data-effect=\"italics\">p<\/em>-value = 0.0187 (This is called the <strong>actual level of significance<\/strong>.) The <em data-effect=\"italics\">p<\/em>-value is the area to the left of the sample mean is given as 16.\r\n\r\n<strong>Graph:<\/strong>\r\n<div id=\"hyptest11_ex1\" class=\"bc-figure figure\"><span id=\"id4282005\" data-type=\"media\" data-alt=\"Normal distribution curve for the average time to swim the 25-yard freestyle with values 16, as the sample mean, and 16.43 on the x-axis. A vertical upward line extends from 16 on the x-axis to the curve. An arrow points to the left tail of the curve.\"><img src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/fig-ch09_11_01N-1.jpg\" alt=\"Normal distribution curve for the average time to swim the 25-yard freestyle with values 16, as the sample mean, and 16.43 on the x-axis. A vertical upward line extends from 16 on the x-axis to the curve. An arrow points to the left tail of the curve.\" width=\"380\" data-media-type=\"image\/jpg\" \/><\/span><\/div>\r\n<em data-effect=\"italics\">\u03bc<\/em> = 16.43 comes from <em data-effect=\"italics\">H<sub>0<\/sub><\/em>. Our assumption is <em data-effect=\"italics\">\u03bc<\/em> = 16.43.\r\n\r\n<strong>Interpretation of the <em data-effect=\"italics\">p<\/em>-value: If <em data-effect=\"italics\">H<sub>0<\/sub><\/em> is true<\/strong>, there is a 0.0187 probability (1.87%)that Jeffrey's mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87% chance is small, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event.\r\n\r\nCompare <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value:\r\n<p id=\"element-890\"><em data-effect=\"italics\">\u03b1<\/em> = 0.05 <em data-effect=\"italics\">p<\/em>-value = 0.0187 <em data-effect=\"italics\">\u03b1<\/em> &gt; <em data-effect=\"italics\">p<\/em>-value<\/p>\r\n<p id=\"element-543\"><strong>Make a decision:<\/strong> Since <em data-effect=\"italics\">\u03b1<\/em> &gt; <em data-effect=\"italics\">p<\/em>-value, reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>.<\/p>\r\n&nbsp;\r\n\r\nThis means that you reject <em data-effect=\"italics\">\u03bc<\/em> = 16.43. In other words, you do not think Jeffrey swims the 25-yard freestyle in 16.43 seconds but faster with the new goggles.\r\n\r\n<strong>Conclusion:<\/strong> At the 5% significance level, we conclude that Jeffrey swims faster using the new goggles. The sample data show there is sufficient evidence that Jeffrey's mean time to swim the 25-yard freestyle is less than 16.43 seconds.\r\n<p id=\"element-252\">The <em data-effect=\"italics\">p<\/em>-value can easily be calculated.<\/p>\r\n\r\n<div id=\"fs-idm31641280\" class=\"statistics calculator\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\r\n<p id=\"element-362\">Press <code>STAT<\/code> and arrow over to <code>TESTS<\/code>. Press <code>1:Z-Test<\/code>. Arrow over to <code>Stats<\/code> and press <code>ENTER<\/code>. Arrow down and enter 16.43 for <em data-effect=\"italics\">\u03bc<sub>0<\/sub><\/em> (null hypothesis), .8 for <em data-effect=\"italics\">\u03c3<\/em>, 16 for the sample mean, and 15 for <em data-effect=\"italics\">n<\/em>. Arrow down to <em data-effect=\"italics\">\u03bc<\/em> : (alternate hypothesis) and arrow over to &lt; <em data-effect=\"italics\">\u03bc<sub>0<\/sub><\/em>. Press <code>ENTER<\/code>. Arrow down to <code>Calculate<\/code> and press <code>ENTER<\/code>. The calculator not only calculates the <em data-effect=\"italics\">p<\/em>-value (<em data-effect=\"italics\">p<\/em> = 0.0187) but it also calculates the test statistic (<em data-effect=\"italics\">z<\/em>-score) for the sample mean. <em data-effect=\"italics\">\u03bc<\/em> &lt; 16.43 is the alternative hypothesis. Do this set of instructions again except arrow to <code>Draw<\/code>(instead of <code>Calculate<\/code>). Press <code>ENTER<\/code>. A shaded graph appears with <em data-effect=\"italics\">z<\/em> = -2.08 (test statistic) and <em data-effect=\"italics\">p<\/em> = 0.0187 (<em data-effect=\"italics\">p<\/em>-value). Make sure when you use <code>Draw<\/code> that no other equations are highlighted in <em data-effect=\"italics\">Y<\/em> = and the plots are turned off.<\/p>\r\n\r\n<\/div>\r\nWhen the calculator does a <em data-effect=\"italics\">Z<\/em>-Test, the <code>Z-Test<\/code> function finds the <em data-effect=\"italics\">p<\/em>-value by doing a normal probability calculation using the <span data-type=\"term\">central limit theorem<\/span>:\r\n<p id=\"fs-idp159988784\">\\(P\\left(\\overline{x}&lt;16\\right)=\\)<code>2nd DISTR normcdf<\/code>\\(\\left(-10^99,16,16.43,0.8\/\\sqrt{15}\\right)\\) .<\/p>\r\nThe Type I and Type II errors for this problem are as follows:\r\n<p id=\"element-385\">The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.)<\/p>\r\nThe Type II error is that there is not evidence to conclude that Jeffrey swims the 25-yard free-style, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25-yard free-style, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.)\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm112426112\" class=\"statistics try\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\r\n<div data-type=\"title\">Try It<\/div>\r\n<div id=\"fs-idp78508304\" data-type=\"exercise\">\r\n<div id=\"fs-idp117367632\" data-type=\"problem\">\r\n\r\nThe mean throwing distance of a football for Marco, a high school freshman quarterback, is 40 yards, with a standard deviation of two yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, Marco\u2019s mean distance was 45 yards. The coach thought the different grip helped Marco throw farther than 40 yards. Conduct a hypothesis test using a preset <em data-effect=\"italics\">\u03b1<\/em> = 0.05. Assume the throw distances for footballs are normal.\r\n<p id=\"eip-idp30205504\">First, determine what type of test this is, set up the hypothesis test, find the <em data-effect=\"italics\">p<\/em>-value, sketch the graph, and state your conclusion.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm24767312\" class=\"statistics calculator\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\r\n<p id=\"eip-idm96892800\">Press STAT and arrow over to TESTS. Press 1:Z-Test. Arrow over to Stats and press ENTER. Arrow down and enter 40 for <em data-effect=\"italics\">\u03bc<\/em>0 (null hypothesis), 2 for <em data-effect=\"italics\">\u03c3<\/em>, 45 for the sample mean, and 20 for <em data-effect=\"italics\">n<\/em>. Arrow down to <em data-effect=\"italics\">\u03bc<\/em>: (alternative hypothesis) and set it either as &lt;, \u2260, or &gt;. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the <em data-effect=\"italics\">p<\/em>-value but it also calculates the test statistic (<em data-effect=\"italics\">z<\/em>-score) for the sample mean. Select &lt;, \u2260, or &gt; for the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with test statistic and <em data-effect=\"italics\">p<\/em>-value. Make sure when you use Draw that no other equations are highlighted in <em data-effect=\"italics\">Y<\/em> = and the plots are turned off.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"note\" data-has-label=\"true\" data-label=\"\">\r\n<div data-type=\"title\">Historical Note (<a class=\"autogenerated-content\" href=\"#fs-idp41487520\">(Figure)<\/a>)<\/div>\r\n<p id=\"eip-idp233123200\">The traditional way to compare the two probabilities, <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value, is to compare the critical value (<em data-effect=\"italics\">z<\/em>-score from <em data-effect=\"italics\">\u03b1<\/em>) to the test statistic (<em data-effect=\"italics\">z<\/em>-score from data). The calculated test statistic for the <em data-effect=\"italics\">p<\/em>-value is \u20132.08. (From the Central Limit Theorem, the test statistic formula is \\(z=\\frac{\\overline{x}-{\\mu }_{X}}{\\left(\\frac{{\\sigma }_{X}}{\\sqrt{n}}\\right)}\\). For this problem, \\(\\overline{x}\\) = 16, <em data-effect=\"italics\">\u03bc<sub>X<\/sub><\/em> = 16.43 from the null hypothes is, <em data-effect=\"italics\">\u03c3<sub>X<\/sub><\/em> = 0.8, and <em data-effect=\"italics\">n<\/em> = 15.) You can find the critical value for <em data-effect=\"italics\">\u03b1<\/em> = 0.05 in the normal table (see <strong>15.Tables<\/strong> in the Table of Contents). The <em data-effect=\"italics\">z<\/em>-score for an area to the left equal to 0.05 is midway between \u20131.65 and \u20131.64 (0.05 is midway between 0.0505 and 0.0495). The <em data-effect=\"italics\">z<\/em>-score is \u20131.645. Since \u20131.645 &gt; \u20132.08 (which demonstrates that \u03b1 &gt; <em data-effect=\"italics\">p<\/em>-value), reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>. Traditionally, the decision to reject or not reject was done in this way. Today, comparing the two probabilities <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value is very common. For this problem, the <em data-effect=\"italics\">p<\/em>-value, 0.0187 is considerably smaller than <em data-effect=\"italics\">\u03b1<\/em>, 0.05. You can be confident about your decision to reject. The graph shows <em data-effect=\"italics\">\u03b1<\/em>, the <em data-effect=\"italics\">p<\/em>-value, and the test statistic and the critical value.<\/p>\r\n\r\n<div id=\"eip-idm47467424\" class=\"bc-figure figure\"><span id=\"eip-idp126239824\" data-type=\"media\" data-alt=\"Distribution curve comparing the \u03b1 to the p-value. Values of -2.15 and -1.645 are on the x-axis. Vertical upward lines extend from both of these values to the curve. The p-value is equal to 0.0158 and points to the area to the left of -2.15. \u03b1 is equal to 0.05 and points to the area between the values of -2.15 and -1.645.\"><img src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/fig-ch09_11_02-1.png\" alt=\"Distribution curve comparing the \u03b1 to the p-value. Values of -2.15 and -1.645 are on the x-axis. Vertical upward lines extend from both of these values to the curve. The p-value is equal to 0.0158 and points to the area to the left of -2.15. \u03b1 is equal to 0.05 and points to the area between the values of -2.15 and -1.645.\" width=\"380\" data-media-type=\"image\/png\" \/><\/span><\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\" data-type=\"example\">\r\n<div data-type=\"exercise\">\r\n<div id=\"id5291254\" data-type=\"problem\">\r\n\r\nA college football coach records the mean weight that his players can bench press as <strong>275 pounds<\/strong>, with a <strong>standard deviation of 55 pounds<\/strong>. Three of his players thought that the mean weight was <strong>more than<\/strong> that amount. They asked <strong>30<\/strong> of their teammates for their estimated maximum lift on the bench press exercise. The data ranged from 205 pounds to 385 pounds. The actual different weights were (frequencies are in parentheses) <span id=\"set-1\" data-type=\"list\" data-list-type=\"labeled-item\" data-display=\"inline\"><span data-type=\"item\">205(3)\u00a0 \u00a0<\/span><span data-type=\"item\">215(3)\u00a0 \u00a0<\/span><span data-type=\"item\">225(1)\u00a0 \u00a0<\/span><span data-type=\"item\">241(2)\u00a0 \u00a0<\/span><span data-type=\"item\">252(2)\u00a0 \u00a0<\/span><span data-type=\"item\">265(2)\u00a0 \u00a0<\/span><span data-type=\"item\">275(2)\u00a0 \u00a0 <\/span><span data-type=\"item\">313(2)\u00a0 \u00a0<\/span><span data-type=\"item\">316(5)\u00a0 \u00a0<\/span><span data-type=\"item\">338(2)\u00a0 \u00a0<\/span><span data-type=\"item\">341(1)\u00a0 \u00a0<\/span><span data-type=\"item\">345(2)\u00a0 \u00a0<\/span><span data-type=\"item\">368(2)\u00a0 \u00a0<\/span><span data-type=\"item\">385(1)<\/span><\/span>.\r\n\r\nConduct a hypothesis test using a 2.5% level of significance to determine if the bench press mean is <strong>more than 275 pounds<\/strong>.\r\n\r\n<\/div>\r\n<div id=\"id5342777\" data-type=\"solution\">\r\n\r\nSet up the Hypothesis Test:\r\n\r\nSince the problem is about a mean weight, this is a <strong>test of a single population mean<\/strong>.\r\n\r\n<em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 275<span data-type=\"newline\">\r\n<\/span><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &gt; 275<span data-type=\"newline\">\r\n<\/span>This is a right-tailed test.\r\n\r\nCalculating the distribution needed:\r\n<p id=\"element-975\">Random variable: \\(\\overline{X}\\) = the mean weight, in pounds, lifted by the football players.<\/p>\r\n<strong>Distribution for the test:<\/strong> It is normal because <em data-effect=\"italics\">\u03c3<\/em> is known.\r\n\r\n\\(\\overline{X}~N\\left(275,\\frac{55}{\\sqrt{30}}\\right)\\)\r\n\r\n\\(\\overline{x}=286.2\\) pounds (from the data).\r\n\r\n<em data-effect=\"italics\">\u03c3<\/em> = 55 pounds <strong>(Always use <em data-effect=\"italics\">\u03c3<\/em> if you know it.)<\/strong> We assume <em data-effect=\"italics\">\u03bc<\/em> = 275 pounds unless our data shows us otherwise.\r\n\r\nCalculate the <em data-effect=\"italics\">p<\/em>-value using the normal distribution for a mean and using the sample mean as input (see <a class=\"autogenerated-content\" href=\"\/contents\/d0ba1833-f0d2-4195-8765-3c436745f0fb\">(Figure)<\/a> for using the data as input):\r\n\r\n\\(p\\text{-value}=P\\left(\\overline{x}&gt;286.2\\right)=0.1323\\).\r\n\r\n<strong>Interpretation of the <em data-effect=\"italics\">p<\/em>-value:<\/strong> If <em data-effect=\"italics\">H<sub>0<\/sub><\/em> is true, then there is a 0.1331 probability (13.23%) that the football players can lift a mean weight of 286.2 pounds or more. Because a 13.23% chance is large enough, a mean weight lift of 286.2 pounds or more is not a rare event.\r\n<div id=\"hyptest11_ex3\" class=\"bc-figure figure\"><span id=\"id4317824\" data-type=\"media\" data-alt=\"Normal distribution curve of the average weight lifted by football players with values of 275 and 286.2 on the x-axis. A vertical upward line extends from 286.2 to the curve. The p-value points to the area to the right of 286.2.\"><img src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/fig-ch09_11_03N-1.jpg\" alt=\"Normal distribution curve of the average weight lifted by football players with values of 275 and 286.2 on the x-axis. A vertical upward line extends from 286.2 to the curve. The p-value points to the area to the right of 286.2.\" width=\"380\" data-media-type=\"image\/jpg\" \/><\/span><\/div>\r\nCompare <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value:\r\n\r\n<em data-effect=\"italics\">\u03b1<\/em> = 0.025 <em data-effect=\"italics\">p<\/em>-value = 0.1323\r\n\r\n<strong>Make a decision:<\/strong> Since <em data-effect=\"italics\">\u03b1<\/em> &lt;<em data-effect=\"italics\">p<\/em>-value, do not reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>.\r\n\r\n<strong>Conclusion:<\/strong> At the 2.5% level of significance, from the sample data, there is not sufficient evidence to conclude that the true mean weight lifted is more than 275 pounds.\r\n\r\nThe <em data-effect=\"italics\">p<\/em>-value can easily be calculated.\r\n<div id=\"fs-idm79549936\" class=\"statistics calculator\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\r\n<p id=\"element-233\">Put the data and frequencies into lists. Press <code>STAT<\/code> and arrow over to <code>TESTS<\/code>. Press <code>1:Z-Test<\/code>. Arrow over to <code>Data<\/code> and press <code>ENTER<\/code>. Arrow down and enter 275 for <em data-effect=\"italics\">\u03bc<sub>0<\/sub><\/em>, 55 for <em data-effect=\"italics\">\u03c3<\/em>, the name of the list where you put the data, and the name of the list where you put the frequencies. Arrow down to <em data-effect=\"italics\">\u03bc:<\/em> and arrow over to &gt; <em data-effect=\"italics\">\u03bc<sub>0<\/sub><\/em>. Press <code>ENTER<\/code>. Arrow down to <code>Calculate<\/code> and press <code>ENTER<\/code>. The calculator not only calculates the <em data-effect=\"italics\">p<\/em>-value (<em data-effect=\"italics\">p<\/em> = 0.1331, a little different from the previous calculation - in it we used the sample mean rounded to one decimal place instead of the data) but it also calculates the test statistic (<em data-effect=\"italics\">z<\/em>-score) for the sample mean, the sample mean, and the sample standard deviation. <em data-effect=\"italics\">\u03bc<\/em> &gt; 275 is the alternative hypothesis. Do this set of instructions again except arrow to <code>Draw<\/code> (instead of <code>Calculate<\/code>). Press <code>ENTER<\/code>. A shaded graph appears with <em data-effect=\"italics\">z<\/em> = 1.112 (test statistic) and <em data-effect=\"italics\">p<\/em> = 0.1331 (<em data-effect=\"italics\">p<\/em>-value). Make sure when you use <code>Draw<\/code> that no other equations are highlighted in <em data-effect=\"italics\">Y<\/em> = and the plots are turned off.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"element-207\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<div data-type=\"exercise\">\r\n<div id=\"id4319892\" data-type=\"problem\">\r\n<p id=\"fs-idp90231232\">Statistics students believe that the mean score on the first statistics test is 65. A statistics instructor thinks the mean score is higher than 65. He samples ten statistics students and obtains the scores <span id=\"set-2\" data-type=\"list\" data-list-type=\"labeled-item\" data-display=\"inline\"><span data-type=\"item\">65\u00a0 \u00a0<\/span><span data-type=\"item\">65\u00a0 \u00a0<\/span><span data-type=\"item\">70\u00a0 \u00a0<\/span><span data-type=\"item\">67\u00a0 \u00a0<\/span><span data-type=\"item\">66\u00a0 \u00a0<\/span><span data-type=\"item\">63\u00a0 \u00a0<\/span><span data-type=\"item\">63\u00a0 \u00a0<\/span><span data-type=\"item\">68\u00a0 \u00a0<\/span><span data-type=\"item\">72\u00a0 \u00a0<\/span><span data-type=\"item\">71<\/span><\/span>. He performs a hypothesis test using a 5% level of significance. The data are assumed to be from a normal distribution.<\/p>\r\n\r\n<\/div>\r\n<div id=\"id5147242\" data-type=\"solution\">\r\n\r\nSet up the hypothesis test:\r\n\r\nA 5% level of significance means that <em data-effect=\"italics\">\u03b1<\/em> = 0.05. This is a test of a <strong>single population mean<\/strong>.\r\n\r\n<em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 65\u2003\u2003<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &gt; 65\r\n\r\nSince the instructor thinks the average score is higher, use a \"&gt;\". The \"&gt;\" means the test is right-tailed.\r\n<p id=\"fs-idp165624976\">Determine the distribution needed:<\/p>\r\n<strong>Random variable:<\/strong>\\(\\overline{X}\\) = average score on the first statistics test.\r\n\r\n<strong>Distribution for the test:<\/strong> If you read the problem carefully, you will notice that there is <strong>no population standard deviation given<\/strong>. You are only given <em data-effect=\"italics\">n<\/em> = 10 sample data values. Notice also that the data come from a normal distribution. This means that the distribution for the test is a student's <em data-effect=\"italics\">t<\/em>.\r\n<p id=\"element-240\">Use <em data-effect=\"italics\">t<\/em><sub>df<\/sub>. Therefore, the distribution for the test is <em data-effect=\"italics\">t<\/em><sub>9<\/sub> where <em data-effect=\"italics\">n<\/em> = 10 and <em data-effect=\"italics\">df<\/em> = 10 - 1 = 9.<\/p>\r\nCalculate the <em data-effect=\"italics\">p<\/em>-value using the Student's <em data-effect=\"italics\">t<\/em>-distribution:\r\n\r\n<em data-effect=\"italics\">p<\/em>-value = <em data-effect=\"italics\">P<\/em>(\\(\\overline{x}\\) &gt; 67) = 0.0396 where the sample mean and sample standard deviation are calculated as 67 and 3.1972 from the data.\r\n\r\n<strong>Interpretation of the <em data-effect=\"italics\">p<\/em>-value:<\/strong> If the null hypothesis is true, then there is a 0.0396 probability (3.96%) that the sample mean is 65 or more.\r\n<div id=\"hyptest11_ex4\" class=\"bc-figure figure\"><span id=\"id5241154\" data-type=\"media\" data-alt=\"Normal distribution curve of average scores on the first statistic tests with 65 and 67 values on the x-axis. A vertical upward line extends from 67 to the curve. The p-value points to the area to the right of 67.\"><img src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/fig-ch09_11_04N-1.jpg\" alt=\"Normal distribution curve of average scores on the first statistic tests with 65 and 67 values on the x-axis. A vertical upward line extends from 67 to the curve. The p-value points to the area to the right of 67.\" width=\"380\" data-media-type=\"image\/jpg\" \/><\/span><\/div>\r\nCompare <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value:\r\n\r\nSince <em data-effect=\"italics\">\u03b1<\/em> = 0.05 and <em data-effect=\"italics\">p<\/em>-value = 0.0396. <em data-effect=\"italics\">\u03b1<\/em> &gt; <em data-effect=\"italics\">p<\/em>-value.\r\n\r\n<strong>Make a decision:<\/strong> Since <em data-effect=\"italics\">\u03b1<\/em> &gt; <em data-effect=\"italics\">p<\/em>-value, reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>.\r\n<p id=\"fs-idm21057040\">This means you reject <em data-effect=\"italics\">\u03bc<\/em> = 65. In other words, you believe the average test score is more than 65.<\/p>\r\n<strong>Conclusion:<\/strong> At a 5% level of significance, the sample data show sufficient evidence that the mean (average) test score is more than 65, just as the math instructor thinks.\r\n\r\nThe <em data-effect=\"italics\">p<\/em>-value can easily be calculated.\r\n<div id=\"fs-idm14472336\" class=\"statistics calculator\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\r\n\r\nPut the data into a list. Press <code>STAT<\/code> and arrow over to <code>TESTS<\/code>. Press <code>2:T-Test<\/code>. Arrow over to <code>Data<\/code> and press <code>ENTER<\/code>. Arrow down and enter 65 for <em data-effect=\"italics\">\u03bc<\/em><sub>0<\/sub>, the name of the list where you put the data, and 1 for <code>Freq:<\/code>. Arrow down to <em data-effect=\"italics\">\u03bc<\/em>: and arrow over to &gt; <em data-effect=\"italics\">\u03bc<\/em><sub>0<\/sub>. Press <code>ENTER<\/code>. Arrow down to <code>Calculate<\/code> and press <code>ENTER<\/code>. The calculator not only calculates the <em data-effect=\"italics\">p<\/em>-value (<em data-effect=\"italics\">p<\/em> = 0.0396) but it also calculates the test statistic (<em data-effect=\"italics\">t<\/em>-score) for the sample mean, the sample mean, and the sample standard deviation. <em data-effect=\"italics\">\u03bc<\/em> &gt; 65 is the alternative hypothesis. Do this set of instructions again except arrow to <code>Draw<\/code> (instead of <code>Calculate<\/code>). Press <code>ENTER<\/code>. A shaded graph appears with <em data-effect=\"italics\">t<\/em> = 1.9781 (test statistic) and <em data-effect=\"italics\">p<\/em> = 0.0396 (<em data-effect=\"italics\">p<\/em>-value). Make sure when you use <code>Draw<\/code> that no other equations are highlighted in <em data-effect=\"italics\">Y<\/em> = and the plots are turned off.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm186895712\" class=\"statistics try\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\r\n<div data-type=\"title\">Try It<\/div>\r\n<div data-type=\"exercise\">\r\n<div data-type=\"problem\">\r\n<p id=\"eip-idm59977568\">It is believed that a stock price for a particular company will grow at a rate of \\$5 per week with a standard deviation of \\$1. An investor believes the stock won\u2019t grow as quickly. The changes in stock price is recorded for ten weeks and are as follows: \\$4, \\$3, \\$2, \\$3, \\$1, \\$7, \\$2, \\$1, \\$1, \\$2. Perform a hypothesis test using a 5% level of significance. State the null and alternative hypotheses, find the <em data-effect=\"italics\">p<\/em>-value, state your conclusion, and identify the Type I and Type II errors.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\" data-type=\"example\">\r\n<div data-type=\"exercise\">\r\n<div id=\"id5304801\" data-type=\"problem\">\r\n\r\nJoon believes that 50% of first-time brides in the United States are younger than their grooms. She performs a hypothesis test to determine if the percentage is <strong>the same or different from 50%<\/strong>. Joon samples <strong>100 first-time brides<\/strong> and <strong>53<\/strong> reply that they are younger than their grooms. For the hypothesis test, she uses a 1% level of significance.\r\n\r\n<\/div>\r\n<div id=\"id4150001\" data-type=\"solution\">\r\n<p id=\"element-399\">Set up the hypothesis test:<\/p>\r\nThe 1% level of significance means that <em data-effect=\"italics\">\u03b1<\/em> = 0.01. This is a <strong>test of a single population proportion<\/strong>.\r\n\r\n<em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.50\u2003\u2003<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 0.50\r\n\r\nThe words <strong>\"is the same or different from\"<\/strong> tell you this is a two-tailed test.\r\n\r\nCalculate the distribution needed:\r\n<p id=\"element-783\"><strong>Random variable:<\/strong><em data-effect=\"italics\">P\u2032<\/em> = the percent of of first-time brides who are younger than their grooms.<\/p>\r\n<p id=\"element-581\"><strong>Distribution for the test:<\/strong> The problem contains no mention of a mean. The information is given in terms of percentages. Use the distribution for <em data-effect=\"italics\">P\u2032<\/em>, the estimated proportion.<\/p>\r\n<p id=\"fs-idm81154768\">\\({P}^{\\prime }~N\\left(p,\\sqrt{\\frac{p\\cdot q}{n}}\\right)\\) Therefore, \\({P}^{\\prime }~N\\left(0.5,\\sqrt{\\frac{0.5\\cdot 0.5}{100}}\\right)\\)<\/p>\r\n<p id=\"fs-idm107085552\">where <em data-effect=\"italics\">p<\/em> = 0.50, <em data-effect=\"italics\">q<\/em> = 1\u2212<em data-effect=\"italics\">p<\/em> = 0.50, and <em data-effect=\"italics\">n<\/em> = 100<\/p>\r\n<p id=\"fs-idp14246672\">Calculate the <em data-effect=\"italics\">p<\/em>-value using the normal distribution for proportions:<\/p>\r\n<em data-effect=\"italics\">p<\/em>-value = <em data-effect=\"italics\">P<\/em> (<em data-effect=\"italics\">p\u2032<\/em> &lt; 0.47 or <em data-effect=\"italics\">p\u2032<\/em> &gt; 0.53) = 0.5485\r\n<p id=\"fs-idp133040416\">where <em data-effect=\"italics\">x<\/em> = 53, <em data-effect=\"italics\">p\u2032<\/em> = \\(\\frac{x}{n}\\text{\u00a0=\u00a0}\\frac{\\text{53}}{\\text{100}}\\) = 0.53.<\/p>\r\n<strong>Interpretation of the <em data-effect=\"italics\">p<\/em>-value:<\/strong> If the null hypothesis is true, there is 0.5485 probability (54.85%) that the sample (estimated) proportion \\(p\\text{'}\\) is 0.53 or more OR 0.47 or less (see the graph in <a class=\"autogenerated-content\" href=\"#hyptest11_ex5\">(Figure)<\/a>).\r\n<div id=\"hyptest11_ex5\" class=\"bc-figure figure\"><span id=\"id5140475\" data-type=\"media\" data-alt=\"Normal distribution curve of the percent of first time brides who are younger than the groom with values of 0.47, 0.50, and 0.53 on the x-axis. Vertical upward lines extend from 0.47 and 0.53 to the curve. 1\/2(p-values) are calculated for the areas on outsides of 0.47 and 0.53.\"><img src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/fig-ch09_11_05N-1.jpg\" alt=\"Normal distribution curve of the percent of first time brides who are younger than the groom with values of 0.47, 0.50, and 0.53 on the x-axis. Vertical upward lines extend from 0.47 and 0.53 to the curve. 1\/2(p-values) are calculated for the areas on outsides of 0.47 and 0.53.\" width=\"380\" data-media-type=\"image\/jpg\" \/><\/span><\/div>\r\n<p id=\"element-327\"><em data-effect=\"italics\">\u03bc<\/em> = <em data-effect=\"italics\">p<\/em> = 0.50 comes from <em data-effect=\"italics\">H<sub>0<\/sub><\/em>, the null hypothesis.<\/p>\r\n<em data-effect=\"italics\">p\u2032<\/em> = 0.53. Since the curve is symmetrical and the test is two-tailed, the <em data-effect=\"italics\">p\u2032<\/em> for the left tail is equal to 0.50 \u2013 0.03 = 0.47 where <em data-effect=\"italics\">\u03bc<\/em> = <em data-effect=\"italics\">p<\/em> = 0.50. (0.03 is the difference between 0.53 and 0.50.)\r\n\r\nCompare <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value:\r\n\r\nSince <em data-effect=\"italics\">\u03b1<\/em> = 0.01 and <em data-effect=\"italics\">p<\/em>-value = 0.5485. <em data-effect=\"italics\">\u03b1<\/em> &lt; <em data-effect=\"italics\">p<\/em>-value.\r\n\r\n<strong>Make a decision:<\/strong> Since <em data-effect=\"italics\">\u03b1<\/em> &lt; <em data-effect=\"italics\">p<\/em>-value, you cannot reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>.\r\n\r\n<strong>Conclusion:<\/strong> At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of first-time brides who are younger than their grooms is different from 50%.\r\n\r\nThe <em data-effect=\"italics\">p<\/em>-value can easily be calculated.\r\n<div id=\"fs-idm35225600\" class=\"statistics calculator\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\r\n\r\nPress <code>STAT<\/code> and arrow over to <code>TESTS<\/code>. Press <code>5:1-PropZTest<\/code>. Enter .5 for <em data-effect=\"italics\">p<\/em><sub>0<\/sub>, 53 for <em data-effect=\"italics\">x<\/em> and 100 for <em data-effect=\"italics\">n<\/em>. Arrow down to <code>Prop<\/code> and arrow to <code>not equals<\/code> <em data-effect=\"italics\">p<\/em><sub>0<\/sub>. Press <code>ENTER<\/code>. Arrow down to <code>Calculate<\/code> and press <code>ENTER<\/code>. The calculator calculates the <em data-effect=\"italics\">p<\/em>-value (<em data-effect=\"italics\">p<\/em> = 0.5485) and the test statistic (<em data-effect=\"italics\">z<\/em>-score). <code>Prop not equals<\/code> .5 is the alternate hypothesis. Do this set of instructions again except arrow to <code>Draw<\/code> (instead of <code>Calculate<\/code>). Press <code>ENTER<\/code>. A shaded graph appears with <em data-effect=\"italics\">z<\/em> = 0.6 (test statistic) and <em data-effect=\"italics\">p<\/em> = 0.5485 (<em data-effect=\"italics\">p<\/em>-value). Make sure when you use <code>Draw<\/code> that no other equations are highlighted in <em data-effect=\"italics\">Y<\/em> = and the plots are turned off.\r\n\r\n<\/div>\r\nThe Type I and Type II errors are as follows:\r\n\r\nThe Type I error is to conclude that the proportion of first-time brides who are younger than their grooms is different from 50% when, in fact, the proportion is actually 50%. (Reject the null hypothesis when the null hypothesis is true).\r\n<p id=\"fs-idp134008544\">The Type II error is there is not enough evidence to conclude that the proportion of first time brides who are younger than their grooms differs from 50% when, in fact, the proportion does differ from 50%. (Do not reject the null hypothesis when the null hypothesis is false.)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp8711616\" class=\"statistics try\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\r\n<div data-type=\"title\">Try It<\/div>\r\n<div data-type=\"exercise\">\r\n<div data-type=\"problem\">\r\n<p id=\"eip-idp103490704\">A teacher believes that 85% of students in the class will want to go on a field trip to the local zoo. She performs a hypothesis test to determine if the percentage is the same or different from 85%. The teacher samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1% level of significance.<\/p>\r\n<p id=\"eip-idp10599024\">First, determine what type of test this is, set up the hypothesis test, find the <em data-effect=\"italics\">p<\/em>-value, sketch the graph, and state your conclusion.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\" data-type=\"example\">\r\n<div id=\"element-312\" data-type=\"exercise\">\r\n<div id=\"id5318755\" data-type=\"problem\">\r\n\r\nSuppose a consumer group suspects that the proportion of households that have three cell phones is 30%. A cell phone company has reason to believe that the proportion is not 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three cell phones.\r\n\r\n<\/div>\r\n<div id=\"id4213559\" data-type=\"solution\">\r\n\r\nSet up the Hypothesis Test:\r\n\r\n<em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.30 <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 0.30\r\n\r\nDetermine the distribution needed:\r\n\r\nThe <strong>random variable<\/strong> is <em data-effect=\"italics\">P\u2032<\/em> = proportion of households that have three cell phones.\r\n\r\nThe <strong>distribution<\/strong> for the hypothesis test is \\(P\\text{'}~N\\left(0.30,\\sqrt{\\frac{\\left(0.30\\right)\\cdot \\left(0.70\\right)}{150}}\\right)\\)\r\n\r\n&nbsp;\r\n<div data-type=\"exercise\">\r\n<div id=\"id4146441\" data-type=\"problem\">\r\n\r\na. The value that helps determine the <em data-effect=\"italics\">p<\/em>-value is <em data-effect=\"italics\">p\u2032<\/em>. Calculate <em data-effect=\"italics\">p\u2032<\/em>.\r\n\r\n<\/div>\r\n<div id=\"id4274849\" data-type=\"solution\" data-print-placement=\"end\">\r\n\r\na. <em data-effect=\"italics\">p\u2032<\/em> = \\(\\frac{x}{n}\\) where <em data-effect=\"italics\">x<\/em> is the number of successes and <em data-effect=\"italics\">n<\/em> is the total number in the sample.\r\n\r\n<em data-effect=\"italics\">x<\/em> = 43, <em data-effect=\"italics\">n<\/em> = 150\r\n\r\n<em data-effect=\"italics\">p\u2032<\/em> = \\(\\frac{\\text{43}}{\\text{150}}\\)\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"exercise\">\r\n<div id=\"id5150803\" data-type=\"problem\">\r\n\r\nb. What is a <strong>success<\/strong> for this problem?\r\n\r\n<\/div>\r\n<div id=\"id5036464\" data-type=\"solution\" data-print-placement=\"end\">\r\n\r\nb. A success is having three cell phones in a household.\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"element-519\" data-type=\"exercise\">\r\n<div id=\"id4292096\" data-type=\"problem\">\r\n<p id=\"element-781\">c. What is the level of significance?<\/p>\r\n\r\n<\/div>\r\n<div id=\"id5242701\" data-type=\"solution\" data-print-placement=\"end\">\r\n\r\nc. The level of significance is the preset <em data-effect=\"italics\">\u03b1<\/em>. Since <em data-effect=\"italics\">\u03b1<\/em> is not given, assume that <em data-effect=\"italics\">\u03b1<\/em> = 0.05.\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"exercise\">\r\n<div id=\"id4673389\" data-type=\"problem\">\r\n\r\nd. Draw the graph for this problem. Draw the horizontal axis. Label and shade appropriately.<span data-type=\"newline\">\r\n<\/span>Calculate the <em data-effect=\"italics\">p<\/em>-value.\r\n\r\n<\/div>\r\n<div id=\"id5212164\" data-type=\"solution\" data-print-placement=\"end\">\r\n\r\nd. <em data-effect=\"italics\">p<\/em>-value = 0.7216\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"exercise\">\r\n<div id=\"id5104601\" data-type=\"problem\">\r\n\r\ne. Make a decision. _____________(Reject\/Do not reject) <em data-effect=\"italics\">H<sub>0<\/sub><\/em> because____________.\r\n\r\n<\/div>\r\n<div id=\"id5137008\" data-type=\"solution\" data-print-placement=\"end\">\r\n\r\ne. Assuming that <em data-effect=\"italics\">\u03b1<\/em> = 0.05, <em data-effect=\"italics\">\u03b1<\/em> &lt; <em data-effect=\"italics\">p<\/em>-value. The decision is do not reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em> because there is not sufficient evidence to conclude that the proportion of households that have three cell phones is not 30%.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idm14663600\" class=\"statistics try\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\r\n<div data-type=\"title\">Try It<\/div>\r\n<div data-type=\"exercise\">\r\n<div data-type=\"problem\">\r\n<p id=\"eip-idm17000432\">Marketers believe that 92% of adults in the United States own a cell phone. A cell phone manufacturer believes that number is actually lower. 200 American adults are surveyed, of which, 174 report having cell phones. Use a 5% level of significance. State the null and alternative hypothesis, find the <em data-effect=\"italics\">p<\/em>-value, state your conclusion, and identify the Type I and Type II errors.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-idp32935184\">The next example is a poem written by a statistics student named Nicole Hart. The solution to the problem follows the poem. Notice that the hypothesis test is for a single population proportion. This means that the null and alternate hypotheses use the parameter <em data-effect=\"italics\">p<\/em>. The distribution for the test is normal. The estimated proportion <em data-effect=\"italics\">p<\/em>\u2032 is the proportion of fleas killed to the total fleas found on Fido. This is sample information. The problem gives a preconceived <em data-effect=\"italics\">\u03b1<\/em> = 0.01, for comparison, and a 95% confidence interval computation. The poem is clever and humorous, so please enjoy it!<\/p>\r\n\r\n<div class=\"textbox textbox--examples\" data-type=\"example\">\r\n<div id=\"fs-idp65427488\" data-type=\"exercise\">\r\n<div id=\"id5257658\" data-type=\"problem\">\r\n<p id=\"eip-idm151173424\">My dog has so many fleas,<span data-type=\"newline\">\r\n<\/span> They do not come off with ease.<span data-type=\"newline\">\r\n<\/span> As for shampoo, I have tried many types<span data-type=\"newline\">\r\n<\/span> Even one called Bubble Hype,<span data-type=\"newline\">\r\n<\/span> Which only killed 25% of the fleas,<span data-type=\"newline\">\r\n<\/span> Unfortunately I was not pleased.<\/p>\r\nI've used all kinds of soap,<span data-type=\"newline\">\r\n<\/span> Until I had given up hope<span data-type=\"newline\">\r\n<\/span> Until one day I saw<span data-type=\"newline\">\r\n<\/span> An ad that put me in awe.\r\n\r\nA shampoo used for dogs<span data-type=\"newline\">\r\n<\/span> Called GOOD ENOUGH to Clean a Hog<span data-type=\"newline\">\r\n<\/span> Guaranteed to kill more fleas.\r\n\r\nI gave Fido a bath<span data-type=\"newline\">\r\n<\/span> And after doing the math<span data-type=\"newline\">\r\n<\/span> His number of fleas<span data-type=\"newline\">\r\n<\/span> Started dropping by 3's!\r\n\r\nBefore his shampoo<span data-type=\"newline\">\r\n<\/span> I counted 42.<span data-type=\"newline\">\r\n<\/span> At the end of his bath,<span data-type=\"newline\">\r\n<\/span> I redid the math<span data-type=\"newline\">\r\n<\/span> And the new shampoo had killed 17 fleas.<span data-type=\"newline\">\r\n<\/span> So now I was pleased.\r\n\r\nNow it is time for you to have some fun<span data-type=\"newline\">\r\n<\/span> With the level of significance being .01,<span data-type=\"newline\">\r\n<\/span> You must help me figure out<span data-type=\"newline\">\r\n<\/span> Use the new shampoo or go without?\r\n\r\n<\/div>\r\n<div id=\"id5100583\" data-type=\"solution\">\r\n\r\nSet up the hypothesis test:\r\n\r\n<em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2264 0.25 \u2003\u2003<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &gt; 0.25\r\n\r\nDetermine the distribution needed:\r\n\r\nIn words, CLEARLY state what your random variable \\(\\overline{X}\\) or <em data-effect=\"italics\">P\u2032<\/em> represents.\r\n<p id=\"fs-idp88897968\"><em data-effect=\"italics\">P\u2032<\/em> = The proportion of fleas that are killed by the new shampoo<\/p>\r\nState the distribution to use for the test.\r\n\r\n<strong>Normal:<\/strong>\\(N\\left(0.25,\\sqrt{\\frac{\\left(0.25\\right)\\left(1-0.25\\right)}{42}}\\right)\\)\r\n\r\n<strong>Test Statistic:<\/strong><em data-effect=\"italics\">z<\/em> = 2.3163\r\n\r\nCalculate the <em data-effect=\"italics\">p<\/em>-value using the normal distribution for proportions:\r\n\r\n<em data-effect=\"italics\">p<\/em>-value = 0.0103\r\n\r\nIn one to two complete sentences, explain what the <em data-effect=\"italics\">p<\/em>-value means for this problem.\r\n\r\nIf the null hypothesis is true (the proportion is 0.25), then there is a 0.0103 probability that the sample (estimated) proportion is 0.4048 \\(\\left(\\frac{17}{42}\\right)\\) or more.\r\n\r\nUse the previous information to sketch a picture of this situation. CLEARLY, label and scale the horizontal axis and shade the region(s) corresponding to the <em data-effect=\"italics\">p<\/em>-value.\r\n<div id=\"hyptest11_ex6\" class=\"bc-figure figure\"><span id=\"id4295819\" data-type=\"media\" data-alt=\"Normal distribution graph of the proportion of fleas killed by the new shampoo with values of 0.25 and 0.4048 on the x-axis. A vertical upward line extends from 0.4048 to the curve and the area to the left of this is shaded in. The test statistic of the sample proportion is listed.\"><img src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/fig-ch09_11_06-1.jpg\" alt=\"Normal distribution graph of the proportion of fleas killed by the new shampoo with values of 0.25 and 0.4048 on the x-axis. A vertical upward line extends from 0.4048 to the curve and the area to the left of this is shaded in. The test statistic of the sample proportion is listed.\" width=\"380\" data-media-type=\"image\/jpg\" \/><\/span><\/div>\r\n<p id=\"element-372\">Compare <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value:<\/p>\r\nIndicate the correct decision (\u201creject\u201d or \u201cdo not reject\u201d the null hypothesis), the reason for it, and write an appropriate conclusion, using complete sentences.\r\n<table summary=\"\">\r\n<thead valign=\"middle\">\r\n<tr>\r\n<th data-align=\"center\">alpha<\/th>\r\n<th data-align=\"center\">decision<\/th>\r\n<th data-align=\"center\">reason for decision<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td data-align=\"center\">0.01<\/td>\r\n<td data-align=\"center\">Do not reject \\({H}_{0}\\)<\/td>\r\n<td data-align=\"center\"><em data-effect=\"italics\">\u03b1<\/em> &lt; <em data-effect=\"italics\">p<\/em>-value<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<strong>Conclusion:<\/strong> At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of fleas that are killed by the new shampoo is more than 25%.\r\n\r\nConstruct a 95% confidence interval for the true mean or proportion. Include a sketch of the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval.\r\n<div id=\"hyptest11_ex7\" class=\"bc-figure figure\"><span id=\"id5244822\" data-type=\"media\" data-alt=\"Normal distribution graph of the proportion of fleas killed by the new shampoo with values of 0.26, 17\/42, and 0.55 on the x-axis. A vertical upward line extends from 0.26 and 0.55. The area between these two points is equal to 0.95.\"><img src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/fig-ch09_11_07-1.jpg\" alt=\"Normal distribution graph of the proportion of fleas killed by the new shampoo with values of 0.26, 17\/42, and 0.55 on the x-axis. A vertical upward line extends from 0.26 and 0.55. The area between these two points is equal to 0.95.\" width=\"380\" data-media-type=\"image\/jpg\" \/><\/span><\/div>\r\n<strong>Confidence Interval:<\/strong> (0.26,0.55) We are 95% confident that the true population proportion <em data-effect=\"italics\">p<\/em> of fleas that are killed by the new shampoo is between 26% and 55%.\r\n<div id=\"id5328435\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\r\n<div data-type=\"title\">Note<\/div>\r\n<p id=\"eip-idm81920704\">This test result is not very definitive since the <em data-effect=\"italics\">p<\/em>-value is very close to alpha. In reality, one would probably do more tests by giving the dog another bath after the fleas have had a chance to return.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\" data-type=\"example\">\r\n<div data-type=\"exercise\">\r\n<div data-type=\"problem\">\r\n<p id=\"eip-idm29000496\">The National Institute of Standards and Technology provides exact data on conductivity properties of materials. Following are conductivity measurements for 11 randomly selected pieces of a particular type of glass.<\/p>\r\n<p id=\"eip-idm46067920\">1.11;\u00a0 \u00a01.07;\u00a0 \u00a01.11;\u00a0 \u00a01.07;\u00a0 \u00a01.12;\u00a0 \u00a01.08;\u00a0 \u00a00.98;\u00a0 \u00a00.98;\u00a0 \u00a01.02;\u00a0 \u00a00.95;\u00a0 \u00a00.95 <span data-type=\"newline\">\r\n<\/span>Is there convincing evidence that the average conductivity of this type of glass is greater than one? Use a significance level of 0.05. Assume the population is normal.<\/p>\r\n\r\n<\/div>\r\n<div data-type=\"solution\">\r\n<p id=\"eip-idp69766080\">Let\u2019s follow a four-step process to answer this statistical question.<\/p>\r\n\r\n<ol id=\"eip-idp69766512\" type=\"1\">\r\n \t<li><strong>State the Question<\/strong>: We need to determine if, at a 0.05 significance level, the average conductivity of the selected glass is greater than one. Our hypotheses will be\r\n<ol id=\"eip-idp129554720\" type=\"a\">\r\n \t<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2264 1<\/li>\r\n \t<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &gt; 1<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li><strong>Plan<\/strong>: We are testing a sample mean without a known population standard deviation. Therefore, we need to use a Student's-t distribution. Assume the underlying population is normal.<\/li>\r\n \t<li><strong>Do the calculations<\/strong>: We will input the sample data into the TI-83 as follows.\r\n<div id=\"eip-idp122053552\" class=\"bc-figure figure\"><span id=\"eip-idp167220912\" data-type=\"media\" data-alt=\"\"><img src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/CNX_Stats_C09_M09_001a-1.jpg\" alt=\"\" width=\"250\" data-media-type=\"image\/jpeg\" \/><\/span><\/div>\r\n<div id=\"eip-idp29927184\" class=\"bc-figure figure\"><span id=\"eip-idp29927312\" data-type=\"media\" data-alt=\"\"><img src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/CNX_Stats_C09_M09_001b-1.jpg\" alt=\"\" width=\"250\" data-media-type=\"image\/jpeg\" \/><\/span><\/div>\r\n<div id=\"eip-idp126675984\" class=\"bc-figure figure\"><span id=\"eip-idp156983600\" data-type=\"media\" data-alt=\"\"><img src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/CNX_Stats_C09_M09_001c-1.jpg\" alt=\"\" width=\"250\" data-media-type=\"image\/jpeg\" \/><\/span><\/div>\r\n<div id=\"eip-idp154420944\" class=\"bc-figure figure\"><span id=\"eip-idp154421072\" data-type=\"media\" data-alt=\"\"><img src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/CNX_Stats_C09_M09_001d-1.jpg\" alt=\"\" width=\"250\" data-media-type=\"image\/jpeg\" \/><\/span><\/div><\/li>\r\n \t<li><strong>State the Conclusions<\/strong>: Since the <em data-effect=\"italics\">p<\/em>-value* (<em data-effect=\"italics\">p<\/em> = 0.036) is less than our alpha value, we will reject the null hypothesis. It is reasonable to state that the data supports the claim that the average conductivity level is greater than one.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"eip-767\" class=\"textbox textbox--examples\" data-type=\"example\">\r\n<div id=\"eip-178\" data-type=\"exercise\">\r\n<div data-type=\"problem\">\r\n<p id=\"eip-idp173526416\">In a study of 420,019 cell phone users, 172 of the subjects developed brain cancer. Test the claim that cell phone users developed brain cancer at a greater rate than that for non-cell phone users (the rate of brain cancer for non-cell phone users is 0.0340%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.<\/p>\r\n\r\n<\/div>\r\n<div data-type=\"solution\">\r\n<p id=\"eip-idp3953440\">We will follow the four-step process.<\/p>\r\n\r\n<ol id=\"eip-idp191218624\" type=\"1\">\r\n \t<li>We need to conduct a hypothesis test on the claimed cancer rate. Our hypotheses will be\r\n<ol id=\"eip-idp171226608\" type=\"a\">\r\n \t<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2264 0.00034<\/li>\r\n \t<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &gt; 0.00034<\/li>\r\n<\/ol>\r\n<p id=\"eip-idm39752176\" class=\"finger\">If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes cancer-causing environments, we want to minimize the chances of incorrectly identifying causes of cancer.<\/p>\r\n<\/li>\r\n \t<li>We will be testing a sample proportion with <em data-effect=\"italics\">x<\/em> = 172 and <em data-effect=\"italics\">n<\/em> = 420,019. The sample is sufficiently large because we have <em data-effect=\"italics\">np<\/em> = 420,019(0.00034) = 142.8, <em data-effect=\"italics\">nq<\/em> = 420,019(0.99966) = 419,876.2, two independent outcomes, and a fixed probability of success <em data-effect=\"italics\">p<\/em> = 0.00034. Thus we will be able to generalize our results to the population.<\/li>\r\n \t<li>The associated TI results are <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-idm80776672\" class=\"bc-figure figure\"><span id=\"eip-idp39356816\" data-type=\"media\" data-alt=\"\"><img src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/CNX_Stats_C09_M09_002a-1.jpg\" alt=\"\" width=\"250\" data-media-type=\"image\/jpeg\" \/><\/span><\/div>\r\n<div id=\"eip-idp38098208\" class=\"bc-figure figure\"><span id=\"eip-idp38098336\" data-type=\"media\" data-alt=\"\"><img src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/CNX_Stats_C09_M09_002b-1.jpg\" alt=\"\" width=\"250\" data-media-type=\"image\/jpeg\" \/><\/span><\/div><\/li>\r\n \t<li>Since the <em data-effect=\"italics\">p<\/em>-value = 0.0073 is greater than our alpha value = 0.005, we cannot reject the null. Therefore, we conclude that there is not enough evidence to support the claim of higher brain cancer rates for the cell phone users.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\" data-type=\"example\">\r\n<div data-type=\"exercise\">\r\n<div data-type=\"problem\">\r\n<p id=\"eip-idp47042256\">According to the US Census there are approximately 268,608,618 residents aged 12 and older. Statistics from the Rape, Abuse, and Incest National Network indicate that, on average, 207,754 rapes occur each year (male and female) for persons aged 12 and older. This translates into a percentage of sexual assaults of 0.078%. In Daviess County, KY, there were reported 11 rapes for a population of 37,937. Conduct an appropriate hypothesis test to determine if there is a statistically significant difference between the local sexual assault percentage and the national sexual assault percentage. Use a significance level of 0.01.<\/p>\r\n\r\n<\/div>\r\n<div data-type=\"solution\">\r\n\r\nWe will follow the four-step plan.\r\n<ol id=\"eip-idp113522624\" type=\"1\">\r\n \t<li>We need to test whether the proportion of sexual assaults in Daviess County, KY is significantly different from the national average.<\/li>\r\n \t<li>Since we are presented with proportions, we will use a one-proportion <em data-effect=\"italics\">z<\/em>-test. The hypotheses for the test will be\r\n<ol id=\"eip-idp17854992\" type=\"a\">\r\n \t<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.00078<\/li>\r\n \t<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 0.00078<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>The following screen shots display the summary statistics from the hypothesis test.\r\n<div id=\"fs-idm3544720\" class=\"bc-figure figure\"><span id=\"eip-idp26791072\" data-type=\"media\" data-alt=\"\"><img src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/CNX_Stats_C09_M09_003a-1.jpg\" alt=\"\" width=\"250\" data-media-type=\"image\/jpeg\" \/><\/span><\/div>\r\n<div id=\"eip-idp38471600\" class=\"bc-figure figure\"><span id=\"eip-idp38471728\" data-type=\"media\" data-alt=\"\"><img src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/CNX_Stats_C09_M09_003b-1.jpg\" alt=\"\" width=\"250\" data-media-type=\"image\/jpeg\" \/><\/span><\/div><\/li>\r\n \t<li>Since the <em data-effect=\"italics\">p<\/em>-value, <em data-effect=\"italics\">p<\/em> = 0.00063, is less than the alpha level of 0.01, the sample data indicates that we should reject the null hypothesis. In conclusion, the sample data support the claim that the proportion of sexual assaults in Daviess County, Kentucky is different from the national average proportion.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp69608272\" class=\"summary\" data-depth=\"1\">\r\n<h3 data-type=\"title\">Chapter Review<\/h3>\r\n<p id=\"fs-idp168110512\">The <span data-type=\"term\">hypothesis test<\/span> itself has an established process. This can be summarized as follows:<\/p>\r\n\r\n<ol>\r\n \t<li>Determine <em data-effect=\"italics\">H<sub>0<\/sub><\/em> and <em data-effect=\"italics\">H<sub>a<\/sub><\/em>. Remember, they are contradictory.<\/li>\r\n \t<li>Determine the random variable.<\/li>\r\n \t<li>Determine the distribution for the test.<\/li>\r\n \t<li>Draw a graph, calculate the test statistic, and use the test statistic to calculate the <em data-effect=\"italics\">p<\/em>-value. (A <em data-effect=\"italics\">z<\/em>-score and a <em data-effect=\"italics\">t<\/em>-score are examples of test statistics.)<\/li>\r\n \t<li>Compare the preconceived <em data-effect=\"italics\">\u03b1<\/em> with the <em data-effect=\"italics\">p<\/em>-value, make a decision (reject or do not reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>), and write a clear conclusion using English sentences.<\/li>\r\n<\/ol>\r\nNotice that in performing the hypothesis test, you use <em data-effect=\"italics\">\u03b1<\/em> and not <em data-effect=\"italics\">\u03b2<\/em>. <em data-effect=\"italics\">\u03b2<\/em> is needed to help determine the sample size of the data that is used in calculating the <em data-effect=\"italics\">p<\/em>-value. Remember that the quantity 1 \u2013 <em data-effect=\"italics\">\u03b2<\/em> is called the <strong>Power of the Test<\/strong>. A high power is desirable. If the power is too low, statisticians typically increase the sample size while keeping <em data-effect=\"italics\">\u03b1<\/em> the same.If the power is low, the null hypothesis might not be rejected when it should be.\r\n\r\n<\/div>\r\n<div class=\"practice\" data-depth=\"1\">\r\n<div data-type=\"exercise\">\r\n<div id=\"eip-135\" data-type=\"problem\">\r\n\r\nAssume <em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 9 and <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &lt; 9. Is this a left-tailed, right-tailed, or two-tailed test?\r\n\r\n<\/div>\r\n<div data-type=\"solution\">\r\n\r\nThis is a left-tailed test.\r\n\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"exercise\">\r\n<div id=\"eip-286\" data-type=\"problem\">\r\n\r\nAssume <em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2264 6 and <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &gt; 6. Is this a left-tailed, right-tailed, or two-tailed test?\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp45725152\" data-type=\"exercise\">\r\n<div id=\"eip-116\" data-type=\"problem\">\r\n\r\nAssume <em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.25 and <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 0.25. Is this a left-tailed, right-tailed, or two-tailed test?\r\n\r\n<\/div>\r\n<div data-type=\"solution\">\r\n\r\nThis is a two-tailed test.\r\n\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"exercise\">\r\n<div data-type=\"problem\">\r\n<p id=\"eip-195\">Draw the general graph of a left-tailed test.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"exercise\">\r\n<div data-type=\"problem\">\r\n\r\nDraw the graph of a two-tailed test.\r\n\r\n<\/div>\r\n<div data-type=\"solution\">\r\n<div id=\"fs-idm149132800\" class=\"bc-figure figure\"><span id=\"eip-idp62459136\" data-type=\"media\" data-alt=\"\"><img src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/CNX_Stats_C09_M09_item002annoN-1.jpg\" alt=\"\" width=\"380\" data-media-type=\"image\/jpeg\" \/><\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"exercise\">\r\n<div id=\"eip-648\" data-type=\"problem\">\r\n\r\nA bottle of water is labeled as containing 16 fluid ounces of water. You believe it is less than that. What type of test would you use?\r\n\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"exercise\">\r\n<div data-type=\"problem\">\r\n\r\nYour friend claims that his mean golf score is 63. You want to show that it is higher than that. What type of test would you use?\r\n\r\n<\/div>\r\n<div data-type=\"solution\">\r\n<p id=\"eip-68\">a right-tailed test<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"eip-729\" data-type=\"exercise\">\r\n<div id=\"eip-491\" data-type=\"problem\">\r\n\r\nA bathroom scale claims to be able to identify correctly any weight within a pound. You think that it cannot be that accurate. What type of test would you use?\r\n\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"exercise\">\r\n<div data-type=\"problem\">\r\n<p id=\"eip-458\">You flip a coin and record whether it shows heads or tails. You know the probability of getting heads is 50%, but you think it is less for this particular coin. What type of test would you use?<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-idp66689312\" data-type=\"solution\">\r\n<p id=\"eip-139\">a left-tailed test<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"exercise\">\r\n<div id=\"eip-519\" data-type=\"problem\">\r\n\r\nIf the alternative hypothesis has a not equals ( \u2260 ) symbol, you know to use which type of test?\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"eip-504\" data-type=\"exercise\">\r\n<div id=\"eip-634\" data-type=\"problem\">\r\n\r\nAssume the null hypothesis states that the mean is at least 18. Is this a left-tailed, right-tailed, or two-tailed test?\r\n\r\n<\/div>\r\n<div data-type=\"solution\">\r\n\r\nThis is a left-tailed test.\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"eip-320\" data-type=\"exercise\">\r\n<div data-type=\"problem\">\r\n\r\nAssume the null hypothesis states that the mean is at most 12. Is this a left-tailed, right-tailed, or two-tailed test?\r\n\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"exercise\">\r\n<div data-type=\"problem\">\r\n\r\nAssume the null hypothesis states that the mean is equal to 88. The alternative hypothesis states that the mean is not equal to 88. Is this a left-tailed, right-tailed, or two-tailed test?\r\n\r\n<\/div>\r\n<div data-type=\"solution\">\r\n\r\nThis is a two-tailed test.\r\n\r\n<\/div>\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div id=\"fs-idp139727497473120\" class=\"free-response\" data-depth=\"1\">\r\n<h3 data-type=\"title\">Homework<\/h3>\r\n<p id=\"eip-idm30094416\"><em data-effect=\"italics\">For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in <a class=\"autogenerated-content\" href=\"\/contents\/c0449c55-aa47-4f1c-bd5f-0521652f0e82\">(Figure)<\/a>. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.<\/em><\/p>\r\n\r\n<div id=\"id8395739\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\r\n<div data-type=\"title\">Note<\/div>\r\n<p id=\"eip-idm120632832\">If you are using a Student's-<em data-effect=\"italics\">t<\/em> distribution for one of the following homework problems, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, however.)<\/p>\r\n\r\n<\/div>\r\n<div data-type=\"exercise\">\r\n<div id=\"id8395759\" data-type=\"problem\">\r\n<p id=\"fs-idp44498592\">1) A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using alpha = 0.05, is the data highly inconsistent with the claim?<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-idm164410224\" data-type=\"solution\">\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"exercise\">\r\n<div id=\"id8395793\" data-type=\"problem\">\r\n\r\n2) From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant of around 2.1 years. A survey of 40 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level?\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"exercise\">\r\n<div id=\"id9424298\" data-type=\"problem\">\r\n\r\n3) The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of 20\u00a2. A study was done to test the claim that the mean cost of a daily newspaper is \\$1.00. Twelve costs yield a mean cost of 95\u00a2 with a standard deviation of 18\u00a2. Do the data support the claim at the 1% level?\r\n\r\n<\/div>\r\n<div id=\"fs-idm17421856\" data-type=\"solution\">\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"exercise\">\r\n<div id=\"id9424338\" data-type=\"problem\">\r\n<p id=\"element-977\">4) An article in the <em data-effect=\"italics\">San Jose Mercury News<\/em> stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 49 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1% level?<\/p>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"element-932a\" data-type=\"exercise\">\r\n<div id=\"id9424572\" data-type=\"problem\">\r\n\r\n5) The mean number of sick days an employee takes per year is believed to be about ten. Members of a personnel department do not believe this figure. They randomly survey eight employees. The number of sick days they took for the past year are as follows: 12;\u00a0 4;\u00a0 15;\u00a0 3;\u00a0 11;\u00a0 8;\u00a0 6;\u00a0 8. Let <em data-effect=\"italics\">x<\/em> = the number of sick days they took for the past year. Should the personnel team believe that the mean number is ten?\r\n\r\n<\/div>\r\n<div id=\"fs-idm143535568\" data-type=\"solution\">\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"exercise\">\r\n<div id=\"id9424624\" data-type=\"problem\">\r\n\r\n6) In 1955, <em data-effect=\"italics\">Life Magazine<\/em> reported that the 25 year-old mother of three worked, on average, an 80 hour week. Recently, many groups have been studying whether or not the women's movement has, in fact, resulted in an increase in the average work week for women (combining employment and at-home work). Suppose a study was done to determine if the mean work week has increased. 81 women were surveyed with the following results. The sample mean was 83; the sample standard deviation was ten. Does it appear that the mean work week has increased for women at the 5% level?\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"exercise\">\r\n<div id=\"id9400577\" data-type=\"problem\">\r\n\r\n7) Your statistics instructor claims that 60 percent of the students who take her Elementary Statistics class go through life feeling more enriched. For some reason that she can't quite figure out, most people don't believe her. You decide to check this out on your own. You randomly survey 64 of her past Elementary Statistics students and find that 34 feel more enriched as a result of her class. Now, what do you think?\r\n\r\n<\/div>\r\n<div id=\"fs-idm68403536\" data-type=\"solution\">\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"exercise\">\r\n<div id=\"id9400611\" data-type=\"problem\">\r\n\r\n8) A Nissan Motor Corporation advertisement read, \u201cThe average man\u2019s I.Q. is 107. The average brown trout\u2019s I.Q. is 4. So why can\u2019t man catch brown trout?\u201d Suppose you believe that the brown trout\u2019s mean I.Q. is greater than four. You catch 12 brown trout. A fish psychologist determines the I.Q.s as follows: 5;\u00a0 \u00a04;\u00a0 \u00a07;\u00a0 \u00a03;\u00a0 \u00a06;\u00a0 \u00a04;\u00a0 \u00a05;\u00a0 \u00a03;\u00a0 \u00a06;\u00a0 \u00a03;\u00a0 \u00a08;\u00a0 \u00a05. Conduct a hypothesis test of your belief.\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"exercise\">\r\n<div id=\"id9400954\" data-type=\"problem\">\r\n\r\n9) Refer to <a href=\"#element-410\">Exercise 9.119<\/a>. Conduct a hypothesis test to see if your decision and conclusion would change if your belief were that the brown trout\u2019s mean I.Q. is <strong>not<\/strong> four.\r\n\r\n<\/div>\r\n<div id=\"fs-idm153124944\" data-type=\"solution\">\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"exercise\">\r\n<div id=\"id9400994\" data-type=\"problem\">\r\n\r\n10) According to an article in <em data-effect=\"italics\">Newsweek<\/em>, the natural ratio of girls to boys is 100:105. In China, the birth ratio is 100: 114 (46.7% girls). Suppose you don\u2019t believe the reported figures of the percent of girls born in China. You conduct a study. In this study, you count the number of girls and boys born in 150 randomly chosen recent births. There are 60 girls and 90 boys born of the 150. Based on your study, do you believe that the percent of girls born in China is 46.7?\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"element-658\" data-type=\"exercise\">\r\n<div id=\"id9401185\" data-type=\"problem\">\r\n\r\n11) A poll done for <em data-effect=\"italics\">Newsweek<\/em> found that 13% of Americans have seen or sensed the presence of an angel. A contingent doubts that the percent is really that high. It conducts its own survey. Out of 76 Americans surveyed, only two had seen or sensed the presence of an angel. As a result of the contingent\u2019s survey, would you agree with the <em data-effect=\"italics\">Newsweek<\/em> poll? In complete sentences, also give three reasons why the two polls might give different results.\r\n\r\n<\/div>\r\n<div id=\"fs-idm3310512\" data-type=\"solution\">\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"exercise\">\r\n<div id=\"id9401240\" data-type=\"problem\">\r\n<p id=\"element-140a\">12) The mean work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it\u2019s shorter. She asks ten engineering friends in start-ups for the lengths of their mean work weeks. Based on the results that follow, should she count on the mean work week to be shorter than 60 hours?<\/p>\r\n<p id=\"fs-idp187110944\">Data (length of mean work week): 70;\u00a0 \u00a045;\u00a0 \u00a055;\u00a0 \u00a060;\u00a0 \u00a065;\u00a0 \u00a055;\u00a0 \u00a055;\u00a0 \u00a060;\u00a0 \u00a050;\u00a0 \u00a055.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n&nbsp;\r\n<div id=\"element-761\" data-type=\"exercise\">\r\n<div id=\"id9401488\" data-type=\"problem\">\r\n<p id=\"element-926\">13) Use the \u201cLap time\u201d data for Lap 4 (see <a class=\"autogenerated-content\" href=\"\/contents\/3ef830bc-5247-460a-9007-e3fd762e5e93\">(Figure)<\/a>) to test the claim that Terri finishes Lap 4, on average, in less than 129 seconds. Use all twenty races given.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-idp13518384\" data-type=\"solution\">\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp191961952\" data-type=\"exercise\">\r\n<div id=\"id9401521\" data-type=\"problem\">\r\n\r\n&nbsp;\r\n<p id=\"element-564a\">14) Use the \u201cInitial Public Offering\u201d data (see <a class=\"autogenerated-content\" href=\"\/contents\/3ef830bc-5247-460a-9007-e3fd762e5e93\">(Figure)<\/a>) to test the claim that the mean offer price was \\$18 per share. Do not use all the data. Use your random number generator to randomly survey 15 prices.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"id9401550\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\r\n<div data-type=\"title\">Note<\/div>\r\n<\/div>\r\n<div data-type=\"exercise\">\r\n<div data-type=\"solution\">\r\n\r\n<strong>Answers to odd questions<\/strong>\r\n\r\n1)\r\n<ol id=\"fs-idp21859504\" type=\"a\">\r\n \t<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2265 50,000<\/li>\r\n \t<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &lt; 50,000<\/li>\r\n \t<li>Let \\(\\overline{X}\\) = the average lifespan of a brand of tires.<\/li>\r\n \t<li>normal distribution<\/li>\r\n \t<li><em data-effect=\"italics\">z<\/em> = -2.315<\/li>\r\n \t<li><em data-effect=\"italics\">p<\/em>-value = 0.0103<\/li>\r\n \t<li>Check student\u2019s solution.\r\n<ol id=\"fs-idm113459088\" type=\"i\">\r\n \t<li>alpha: 0.05<\/li>\r\n \t<li>Decision: Reject the null hypothesis.<\/li>\r\n \t<li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is less than 0.05.<\/li>\r\n \t<li>Conclusion: There is sufficient evidence to conclude that the mean lifespan of the tires is less than 50,000 miles.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>(43,537, 49,463)<\/li>\r\n<\/ol>\r\n3)\r\n<ol id=\"fs-idm17421600\" type=\"a\">\r\n \t<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = \\$1.00<\/li>\r\n \t<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2260 \\$1.00<\/li>\r\n \t<li>Let \\(\\overline{X}\\) = the average cost of a daily newspaper.<\/li>\r\n \t<li>normal distribution<\/li>\r\n \t<li><em data-effect=\"italics\">z<\/em> = \u20130.866<\/li>\r\n \t<li><em data-effect=\"italics\">p<\/em>-value = 0.3865<\/li>\r\n \t<li>Check student\u2019s solution.\r\n<ol id=\"fs-idm69728368\" type=\"i\">\r\n \t<li>Alpha: 0.01<\/li>\r\n \t<li>Decision: Do not reject the null hypothesis.<\/li>\r\n \t<li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is greater than 0.01.<\/li>\r\n \t<li>Conclusion: There is sufficient evidence to support the claim that the mean cost of daily papers is \\$1. The mean cost could be \\$1.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>(\\$0.84, \\$1.06)<\/li>\r\n<\/ol>\r\n5)\r\n<ol id=\"fs-idm8405344\" type=\"a\">\r\n \t<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 10<\/li>\r\n \t<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2260 10<\/li>\r\n \t<li>Let \\(\\overline{X}\\) the mean number of sick days an employee takes per year.<\/li>\r\n \t<li>Student\u2019s <em data-effect=\"italics\">t<\/em>-distribution<\/li>\r\n \t<li><em data-effect=\"italics\">t<\/em> = \u20131.12<\/li>\r\n \t<li><em data-effect=\"italics\">p<\/em>-value = 0.300<\/li>\r\n \t<li>Check student\u2019s solution.\r\n<ol id=\"fs-idm68554224\" type=\"i\">\r\n \t<li>Alpha: 0.05<\/li>\r\n \t<li>Decision: Do not reject the null hypothesis.<\/li>\r\n \t<li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is greater than 0.05.<\/li>\r\n \t<li>Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean number of sick days is not ten.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>(4.9443, 11.806)<\/li>\r\n<\/ol>\r\n7)\r\n<ol id=\"fs-idm150879872\" type=\"a\">\r\n \t<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2265 0.6<\/li>\r\n \t<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &lt; 0.6<\/li>\r\n \t<li>Let <em data-effect=\"italics\">P\u2032<\/em> = the proportion of students who feel more enriched as a result of taking Elementary Statistics.<\/li>\r\n \t<li>normal for a single proportion<\/li>\r\n \t<li>1.12<\/li>\r\n \t<li><em data-effect=\"italics\">p<\/em>-value = 0.1308<\/li>\r\n \t<li>Check student\u2019s solution.\r\n<ol id=\"fs-idp14435968\" type=\"i\">\r\n \t<li>Alpha: 0.05<\/li>\r\n \t<li>Decision: Do not reject the null hypothesis.<\/li>\r\n \t<li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is greater than 0.05.<\/li>\r\n \t<li>Conclusion: There is insufficient evidence to conclude that less than 60 percent of her students feel more enriched.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Confidence Interval: (0.409, 0.654)<span data-type=\"newline\">\r\n<\/span>The \u201cplus-4s\u201d confidence interval is (0.411, 0.648)<\/li>\r\n<\/ol>\r\n9)\r\n<ol id=\"fs-idm153124688\" type=\"a\">\r\n \t<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 4<\/li>\r\n \t<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2260 4<\/li>\r\n \t<li>Let \\(\\overline{X}\\) the average I.Q. of a set of brown trout.<\/li>\r\n \t<li>two-tailed Student's t-test<\/li>\r\n \t<li><em data-effect=\"italics\">t<\/em> = 1.95<\/li>\r\n \t<li><em data-effect=\"italics\">p<\/em>-value = 0.076<\/li>\r\n \t<li>Check student\u2019s solution.\r\n<ol id=\"fs-idp21135088\" type=\"i\">\r\n \t<li>Alpha: 0.05<\/li>\r\n \t<li>Decision: Reject the null hypothesis.<\/li>\r\n \t<li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is greater than 0.05<\/li>\r\n \t<li>Conclusion: There is insufficient evidence to conclude that the average IQ of brown trout is not four.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>(3.8865,5.9468)<\/li>\r\n<\/ol>\r\n11)\r\n<ol id=\"fs-idm3310256\" type=\"a\">\r\n \t<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2265 0.13<\/li>\r\n \t<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &lt; 0.13<\/li>\r\n \t<li>Let <em data-effect=\"italics\">P\u2032<\/em> = the proportion of Americans who have seen or sensed angels<\/li>\r\n \t<li>normal for a single proportion<\/li>\r\n \t<li>\u20132.688<\/li>\r\n \t<li><em data-effect=\"italics\">p<\/em>-value = 0.0036<\/li>\r\n \t<li>Check student\u2019s solution.\r\n<ol id=\"fs-idm15778400\" type=\"i\">\r\n \t<li>alpha: 0.05<\/li>\r\n \t<li>Decision: Reject the null hypothesis.<\/li>\r\n \t<li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is less than 0.05.<\/li>\r\n \t<li>Conclusion: There is sufficient evidence to conclude that the percentage of Americans who have seen or sensed an angel is less than 13%.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>(0, 0.0623).<span data-type=\"newline\">\r\n<\/span> The\u201cplus-4s\u201d confidence interval is (0.0022, 0.0978)<\/li>\r\n<\/ol>\r\n13)\r\n<ol id=\"fs-idp13518640\" type=\"a\">\r\n \t<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2265 129<\/li>\r\n \t<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &lt; 129<\/li>\r\n \t<li>Let \\(\\overline{X}\\) = the average time in seconds that Terri finishes Lap 4.<\/li>\r\n \t<li>Student's <em data-effect=\"italics\">t<\/em>-distribution<\/li>\r\n \t<li><em data-effect=\"italics\">t<\/em> = 1.209<\/li>\r\n \t<li>0.8792<\/li>\r\n \t<li>Check student\u2019s solution.\r\n<ol id=\"fs-idm130980080\" type=\"i\">\r\n \t<li>Alpha: 0.05<\/li>\r\n \t<li>Decision: Do not reject the null hypothesis.<\/li>\r\n \t<li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is greater than 0.05.<\/li>\r\n \t<li>Conclusion: There is insufficient evidence to conclude that Terri\u2019s mean lap time is less than 129 seconds.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>(128.63, 130.37)<\/li>\r\n<\/ol>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-idp118682672\" class=\"footnotes\" data-depth=\"1\">\r\n<h3 data-type=\"title\">References<\/h3>\r\nData from Amit Schitai. Director of Instructional Technology and Distance Learning. LBCC.\r\n\r\nData from <em data-effect=\"italics\">Bloomberg Businessweek<\/em>. Available online at http:\/\/www.businessweek.com\/news\/2011- 09-15\/nyc-smoking-rate-falls-to-record-low-of-14-bloomberg-says.html.\r\n\r\nData from energy.gov. Available online at http:\/\/energy.gov (accessed June 27. 2013).\r\n\r\nData from Gallup\u00ae. Available online at www.gallup.com (accessed June 27, 2013).\r\n\r\nData from <em data-effect=\"italics\">Growing by Degrees<\/em> by Allen and Seaman.\r\n\r\nData from La Leche League International. Available online at http:\/\/www.lalecheleague.org\/Law\/BAFeb01.html.\r\n\r\nData from the American Automobile Association. Available online at www.aaa.com (accessed June 27, 2013).\r\n\r\nData from the American Library Association. Available online at www.ala.org (accessed June 27, 2013).\r\n\r\nData from the Bureau of Labor Statistics. Available online at http:\/\/www.bls.gov\/oes\/current\/oes291111.htm.\r\n<p id=\"fs-idp1\">Data from the Centers for Disease Control and Prevention. Available online at www.cdc.gov (accessed June 27, 2013)<\/p>\r\nData from the U.S. Census Bureau, available online at http:\/\/quickfacts.census.gov\/qfd\/states\/00000.html (accessed June 27, 2013).\r\n\r\nData from the United States Census Bureau. Available online at http:\/\/www.census.gov\/hhes\/socdemo\/language\/.\r\n<p id=\"eip-79\">Data from Toastmasters International. Available online at http:\/\/toastmasters.org\/artisan\/detail.asp?CategoryID=1&amp;SubCategoryID=10&amp;ArticleID=429&amp;Page=1.<\/p>\r\nData from Weather Underground. Available online at www.wunderground.com (accessed June 27, 2013).\r\n\r\nFederal Bureau of Investigations. \u201cUniform Crime Reports and Index of Crime in Daviess in the State of Kentucky enforced by Daviess County from 1985 to 2005.\u201d Available online at http:\/\/www.disastercenter.com\/kentucky\/crime\/3868.htm (accessed June 27, 2013).\r\n\r\n\u201cFoothill-De Anza Community College District.\u201d De Anza College, Winter 2006. Available online at http:\/\/research.fhda.edu\/factbook\/DAdemofs\/Fact_sheet_da_2006w.pdf.\r\n\r\nJohansen, C., J. Boice, Jr., J. McLaughlin, J. Olsen. \u201cCellular Telephones and Cancer\u2014a Nationwide Cohort Study in Denmark.\u201d Institute of Cancer Epidemiology and the Danish Cancer Society, 93(3):203-7. Available online at http:\/\/www.ncbi.nlm.nih.gov\/pubmed\/11158188 (accessed June 27, 2013).\r\n\r\nRape, Abuse &amp; Incest National Network. \u201cHow often does sexual assault occur?\u201d RAINN, 2009. Available online at http:\/\/www.rainn.org\/get-information\/statistics\/frequency-of-sexual-assault (accessed June 27, 2013).\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\" data-type=\"glossary\">\r\n<h3 data-type=\"glossary-title\">Glossary<\/h3>\r\n<dl>\r\n \t<dt>Central Limit Theorem<\/dt>\r\n \t<dd>Given a random variable (RV) with known mean \\(\\mu \\) and known standard deviation \u03c3. We are sampling with size <em data-effect=\"italics\">n<\/em> and we are interested in two new RVs - the sample mean, \\(\\overline{X}\\), and the sample sum, \\(\\Sigma X\\). If the size <em data-effect=\"italics\">n<\/em> of the sample is sufficiently large, then \\(\\overline{X}~N\\left(\\mu \\text{,}\\frac{\\sigma }{\\sqrt{n}}\\right)\\) and \\(\\Sigma X~N\\left(n\\mu ,\\sqrt{n}\\sigma \\right)\\). If the size <em data-effect=\"italics\">n<\/em> of the sample is sufficiently large, then the distribution of the sample means and the distribution of the sample sums will approximate a normal distribution regardless of the shape of the population. The mean of the sample means will equal the population mean and the mean of the sample sums will equal <em data-effect=\"italics\">n<\/em> times the population mean. The standard deviation of the distribution of the sample means, \\(\\frac{\\sigma }{\\sqrt{n}}\\), is called the standard error of the mean.<\/dd>\r\n<\/dl>\r\n<\/div>","rendered":"<p>&nbsp;<\/p>\n<ul>\n<li>In a <span data-type=\"term\">hypothesis test<\/span> problem, you may see words such as &#8220;the level of significance is 1%.&#8221; The &#8220;1%&#8221; is the preconceived or preset <em data-effect=\"italics\">\u03b1<\/em>.<\/li>\n<li>The statistician setting up the hypothesis test selects the value of <em data-effect=\"italics\">\u03b1<\/em> to use <strong>before<\/strong> collecting the sample data.<\/li>\n<li><strong>If no level of significance is given, a common standard to use is <em data-effect=\"italics\">\u03b1<\/em> = 0.05.<\/strong><\/li>\n<li>When you calculate the <em data-effect=\"italics\">p<\/em>-value and draw the picture, the <em data-effect=\"italics\">p<\/em>-value is the area in the left tail, the right tail, or split evenly between the two tails. For this reason, we call the hypothesis test left, right, or two tailed.<\/li>\n<li>The <strong>alternative hypothesis<\/strong>, \\({H}_{a}\\), tells you if the test is left, right, or two-tailed. It is the <strong>key<\/strong> to conducting the appropriate test.<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em><strong>never<\/strong> has a symbol that contains an equal sign.<\/li>\n<li><strong>Thinking about the meaning of the<\/strong><span data-type=\"term\"><em data-effect=\"italics\">p<\/em>-value<\/span>: A data analyst (and anyone else) should have more confidence that he made the correct decision to reject the null hypothesis with a smaller <em data-effect=\"italics\">p<\/em>-value (for example, 0.001 as opposed to 0.04) even if using the 0.05 level for alpha. Similarly, for a large <em data-effect=\"italics\">p<\/em>-value such as 0.4, as opposed to a <em data-effect=\"italics\">p<\/em>-value of 0.056 (alpha = 0.05 is less than either number), a data analyst should have more confidence that she made the correct decision in not rejecting the null hypothesis. This makes the data analyst use judgment rather than mindlessly applying rules.<\/li>\n<\/ul>\n<p id=\"fs-idp42898080\">The following examples illustrate a left-, right-, and two-tailed test.<\/p>\n<div id=\"fs-idp41487520\" class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"fs-idp41487648\"><em data-effect=\"italics\">H<sub>o<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 5, <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &lt; 5<\/p>\n<p>Test of a single population mean. <em data-effect=\"italics\">H<sub>a<\/sub><\/em> tells you the test is left-tailed. The picture of the <em data-effect=\"italics\">p<\/em>-value is as follows:<\/p>\n<div id=\"fs-idm151356816\" class=\"bc-figure figure\"><span id=\"id14703034\" data-type=\"media\" data-alt=\"Normal distribution curve of a single population mean with a value of 5 on the x-axis and the p-value points to the area on the left tail of the curve.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/05\/fig-ch09_09_01-02-1.jpg\" alt=\"Normal distribution curve of a single population mean with a value of 5 on the x-axis and the p-value points to the area on the left tail of the curve.\" width=\"380\" data-media-type=\"image\/jpg\" \/><\/span><\/div>\n<\/div>\n<div id=\"fs-idp42408528\" class=\"statistics try\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<div data-type=\"title\">Try It<\/div>\n<div id=\"fs-idp67352928\" data-type=\"exercise\">\n<div data-type=\"problem\">\n<p id=\"eip-idm132446832\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 10, <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &lt; 10<\/p>\n<p id=\"eip-idp3861728\">Assume the <em data-effect=\"italics\">p<\/em>-value is 0.0935. What type of test is this? Draw the picture of the <em data-effect=\"italics\">p<\/em>-value.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\" data-type=\"example\">\n<p><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2264 0.2\u2003\u2003<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &gt; 0.2<\/p>\n<p id=\"fs-idp117549824\">This is a test of a single population proportion. <em data-effect=\"italics\">H<sub>a<\/sub><\/em> tells you the test is <strong>right-tailed<\/strong>. The picture of the <em data-effect=\"italics\">p<\/em>-value is as follows:<\/p>\n<div id=\"fs-idm2304848\" class=\"bc-figure figure\"><span id=\"id14703170\" data-type=\"media\" data-alt=\"Normal distribution curve of a single population proportion with the value of 0.2 on the x-axis. The p-value points to the area on the right tail of the curve.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/fig-ch09_09_02a-1.jpg\" alt=\"Normal distribution curve of a single population proportion with the value of 0.2 on the x-axis. The p-value points to the area on the right tail of the curve.\" width=\"380\" data-media-type=\"image\/jpg\" \/><\/span><\/div>\n<\/div>\n<div id=\"fs-idm126098448\" class=\"statistics try\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<div data-type=\"title\">Try It<\/div>\n<div id=\"eip-417\" data-type=\"exercise\">\n<div id=\"fs-idp131553552\" data-type=\"problem\">\n<p id=\"eip-idm89526848\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2264 1, <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &gt; 1<\/p>\n<p id=\"eip-idm24425280\">Assume the <em data-effect=\"italics\">p<\/em>-value is 0.1243. What type of test is this? Draw the picture of the <em data-effect=\"italics\">p<\/em>-value.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\" data-type=\"example\">\n<p id=\"element-38a\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 50\u2003\u2003<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 50<\/p>\n<p>This is a test of a single population mean. <em data-effect=\"italics\">H<sub>a<\/sub><\/em> tells you the test is <strong>two-tailed<\/strong>. The picture of the <em data-effect=\"italics\">p<\/em>-value is as follows.<\/p>\n<div id=\"fs-idm117972816\" class=\"bc-figure figure\"><span id=\"id4829633\" data-type=\"media\" data-alt=\"Normal distribution curve of a single population mean with a value of 50 on the x-axis. The p-value formulas, 1\/2(p-value), for a two-tailed test is shown for the areas on the left and right tails of the curve.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/fig-ch09_09_03N-1.jpg\" alt=\"Normal distribution curve of a single population mean with a value of 50 on the x-axis. The p-value formulas, 1\/2(p-value), for a two-tailed test is shown for the areas on the left and right tails of the curve.\" width=\"380\" data-media-type=\"image\/jpg\" \/><\/span><\/div>\n<\/div>\n<div id=\"fs-idm63250432\" class=\"statistics try\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<div data-type=\"title\">Try It<\/div>\n<div data-type=\"exercise\">\n<div data-type=\"problem\">\n<p id=\"eip-idp41687312\"><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.5, <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 0.5<\/p>\n<p id=\"eip-idm5668432\">Assume the <em data-effect=\"italics\">p<\/em>-value is 0.2564. What type of test is this? Draw the picture of the <em data-effect=\"italics\">p<\/em>-value.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idm12892576\" class=\"bc-section section\" data-depth=\"1\">\n<h3 data-type=\"title\">Full Hypothesis Test Examples<\/h3>\n<div class=\"textbox textbox--examples\" data-type=\"example\">\n<div data-type=\"exercise\">\n<div id=\"id5300933\" data-type=\"problem\">\n<p>Jeffrey, as an eight-year old, <strong>established a mean time of 16.43 seconds<\/strong> for swimming the 25-yard freestyle, with a <strong>standard deviation of 0.8 seconds<\/strong>. His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for <strong>15 25-yard freestyle swims<\/strong>. For the 15 swims, <strong>Jeffrey&#8217;s mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds.<\/strong> Conduct a hypothesis test using a preset <em data-effect=\"italics\">\u03b1<\/em> = 0.05. Assume that the swim times for the 25-yard freestyle are normal.<\/p>\n<\/div>\n<div id=\"id5036614\" data-type=\"solution\">\n<p>Set up the Hypothesis Test:<\/p>\n<p>Since the problem is about a mean, this is a <strong>test of a single population mean<\/strong>.<\/p>\n<p><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 16.43\u2003\u2003<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &lt; 16.43<\/p>\n<p id=\"element-241\">For Jeffrey to swim faster, his time will be less than 16.43 seconds. The &#8220;&lt;&#8221; tells you this is left-tailed.<\/p>\n<p>Determine the distribution needed:<\/p>\n<p><strong>Random variable: <\/strong>\\(\\overline{X}\\) = the mean time to swim the 25-yard freestyle.<\/p>\n<p><strong>Distribution for the test: <\/strong>\\(\\overline{X}\\) is normal (population <span data-type=\"term\">standard deviation<\/span> is known: <em data-effect=\"italics\">\u03c3<\/em> = 0.8)<\/p>\n<p>\\(\\overline{X}~N\\left(\\mu ,\\frac{{\\sigma }_{X}}{\\sqrt{n}}\\right)\\) Therefore, \\(\\overline{X}~N\\left(16.43,\\frac{0.8}{\\sqrt{15}}\\right)\\)<\/p>\n<p><em data-effect=\"italics\">\u03bc<\/em> = 16.43 comes from <em data-effect=\"italics\">H<sub>0<\/sub><\/em> and not the data. <em data-effect=\"italics\">\u03c3<\/em> = 0.8, and <em data-effect=\"italics\">n<\/em> = 15.<\/p>\n<p>Calculate the <em data-effect=\"italics\">p<\/em>-value using the normal distribution for a mean:<\/p>\n<p id=\"element-706\"><em data-effect=\"italics\">p<\/em>-value = <em data-effect=\"italics\">P<\/em>(\\(\\overline{x}\\) &lt; 16) = 0.0187 where the sample mean in the problem is given as 16.<\/p>\n<p><em data-effect=\"italics\">p<\/em>-value = 0.0187 (This is called the <strong>actual level of significance<\/strong>.) The <em data-effect=\"italics\">p<\/em>-value is the area to the left of the sample mean is given as 16.<\/p>\n<p><strong>Graph:<\/strong><\/p>\n<div id=\"hyptest11_ex1\" class=\"bc-figure figure\"><span id=\"id4282005\" data-type=\"media\" data-alt=\"Normal distribution curve for the average time to swim the 25-yard freestyle with values 16, as the sample mean, and 16.43 on the x-axis. A vertical upward line extends from 16 on the x-axis to the curve. An arrow points to the left tail of the curve.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/fig-ch09_11_01N-1.jpg\" alt=\"Normal distribution curve for the average time to swim the 25-yard freestyle with values 16, as the sample mean, and 16.43 on the x-axis. A vertical upward line extends from 16 on the x-axis to the curve. An arrow points to the left tail of the curve.\" width=\"380\" data-media-type=\"image\/jpg\" \/><\/span><\/div>\n<p><em data-effect=\"italics\">\u03bc<\/em> = 16.43 comes from <em data-effect=\"italics\">H<sub>0<\/sub><\/em>. Our assumption is <em data-effect=\"italics\">\u03bc<\/em> = 16.43.<\/p>\n<p><strong>Interpretation of the <em data-effect=\"italics\">p<\/em>-value: If <em data-effect=\"italics\">H<sub>0<\/sub><\/em> is true<\/strong>, there is a 0.0187 probability (1.87%)that Jeffrey&#8217;s mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87% chance is small, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event.<\/p>\n<p>Compare <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value:<\/p>\n<p id=\"element-890\"><em data-effect=\"italics\">\u03b1<\/em> = 0.05 <em data-effect=\"italics\">p<\/em>-value = 0.0187 <em data-effect=\"italics\">\u03b1<\/em> &gt; <em data-effect=\"italics\">p<\/em>-value<\/p>\n<p id=\"element-543\"><strong>Make a decision:<\/strong> Since <em data-effect=\"italics\">\u03b1<\/em> &gt; <em data-effect=\"italics\">p<\/em>-value, reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>.<\/p>\n<p>&nbsp;<\/p>\n<p>This means that you reject <em data-effect=\"italics\">\u03bc<\/em> = 16.43. In other words, you do not think Jeffrey swims the 25-yard freestyle in 16.43 seconds but faster with the new goggles.<\/p>\n<p><strong>Conclusion:<\/strong> At the 5% significance level, we conclude that Jeffrey swims faster using the new goggles. The sample data show there is sufficient evidence that Jeffrey&#8217;s mean time to swim the 25-yard freestyle is less than 16.43 seconds.<\/p>\n<p id=\"element-252\">The <em data-effect=\"italics\">p<\/em>-value can easily be calculated.<\/p>\n<div id=\"fs-idm31641280\" class=\"statistics calculator\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<p id=\"element-362\">Press <code>STAT<\/code> and arrow over to <code>TESTS<\/code>. Press <code>1:Z-Test<\/code>. Arrow over to <code>Stats<\/code> and press <code>ENTER<\/code>. Arrow down and enter 16.43 for <em data-effect=\"italics\">\u03bc<sub>0<\/sub><\/em> (null hypothesis), .8 for <em data-effect=\"italics\">\u03c3<\/em>, 16 for the sample mean, and 15 for <em data-effect=\"italics\">n<\/em>. Arrow down to <em data-effect=\"italics\">\u03bc<\/em> : (alternate hypothesis) and arrow over to &lt; <em data-effect=\"italics\">\u03bc<sub>0<\/sub><\/em>. Press <code>ENTER<\/code>. Arrow down to <code>Calculate<\/code> and press <code>ENTER<\/code>. The calculator not only calculates the <em data-effect=\"italics\">p<\/em>-value (<em data-effect=\"italics\">p<\/em> = 0.0187) but it also calculates the test statistic (<em data-effect=\"italics\">z<\/em>-score) for the sample mean. <em data-effect=\"italics\">\u03bc<\/em> &lt; 16.43 is the alternative hypothesis. Do this set of instructions again except arrow to <code>Draw<\/code>(instead of <code>Calculate<\/code>). Press <code>ENTER<\/code>. A shaded graph appears with <em data-effect=\"italics\">z<\/em> = -2.08 (test statistic) and <em data-effect=\"italics\">p<\/em> = 0.0187 (<em data-effect=\"italics\">p<\/em>-value). Make sure when you use <code>Draw<\/code> that no other equations are highlighted in <em data-effect=\"italics\">Y<\/em> = and the plots are turned off.<\/p>\n<\/div>\n<p>When the calculator does a <em data-effect=\"italics\">Z<\/em>-Test, the <code>Z-Test<\/code> function finds the <em data-effect=\"italics\">p<\/em>-value by doing a normal probability calculation using the <span data-type=\"term\">central limit theorem<\/span>:<\/p>\n<p id=\"fs-idp159988784\">\\(P\\left(\\overline{x}&lt;16\\right)=\\)<code>2nd DISTR normcdf<\/code>\\(\\left(-10^99,16,16.43,0.8\/\\sqrt{15}\\right)\\) .<\/p>\n<p>The Type I and Type II errors for this problem are as follows:<\/p>\n<p id=\"element-385\">The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.)<\/p>\n<p>The Type II error is that there is not evidence to conclude that Jeffrey swims the 25-yard free-style, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25-yard free-style, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.)<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idm112426112\" class=\"statistics try\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<div data-type=\"title\">Try It<\/div>\n<div id=\"fs-idp78508304\" data-type=\"exercise\">\n<div id=\"fs-idp117367632\" data-type=\"problem\">\n<p>The mean throwing distance of a football for Marco, a high school freshman quarterback, is 40 yards, with a standard deviation of two yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, Marco\u2019s mean distance was 45 yards. The coach thought the different grip helped Marco throw farther than 40 yards. Conduct a hypothesis test using a preset <em data-effect=\"italics\">\u03b1<\/em> = 0.05. Assume the throw distances for footballs are normal.<\/p>\n<p id=\"eip-idp30205504\">First, determine what type of test this is, set up the hypothesis test, find the <em data-effect=\"italics\">p<\/em>-value, sketch the graph, and state your conclusion.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idm24767312\" class=\"statistics calculator\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<p id=\"eip-idm96892800\">Press STAT and arrow over to TESTS. Press 1:Z-Test. Arrow over to Stats and press ENTER. Arrow down and enter 40 for <em data-effect=\"italics\">\u03bc<\/em>0 (null hypothesis), 2 for <em data-effect=\"italics\">\u03c3<\/em>, 45 for the sample mean, and 20 for <em data-effect=\"italics\">n<\/em>. Arrow down to <em data-effect=\"italics\">\u03bc<\/em>: (alternative hypothesis) and set it either as &lt;, \u2260, or &gt;. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the <em data-effect=\"italics\">p<\/em>-value but it also calculates the test statistic (<em data-effect=\"italics\">z<\/em>-score) for the sample mean. Select &lt;, \u2260, or &gt; for the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with test statistic and <em data-effect=\"italics\">p<\/em>-value. Make sure when you use Draw that no other equations are highlighted in <em data-effect=\"italics\">Y<\/em> = and the plots are turned off.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<div data-type=\"title\">Historical Note (<a class=\"autogenerated-content\" href=\"#fs-idp41487520\">(Figure)<\/a>)<\/div>\n<p id=\"eip-idp233123200\">The traditional way to compare the two probabilities, <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value, is to compare the critical value (<em data-effect=\"italics\">z<\/em>-score from <em data-effect=\"italics\">\u03b1<\/em>) to the test statistic (<em data-effect=\"italics\">z<\/em>-score from data). The calculated test statistic for the <em data-effect=\"italics\">p<\/em>-value is \u20132.08. (From the Central Limit Theorem, the test statistic formula is \\(z=\\frac{\\overline{x}-{\\mu }_{X}}{\\left(\\frac{{\\sigma }_{X}}{\\sqrt{n}}\\right)}\\). For this problem, \\(\\overline{x}\\) = 16, <em data-effect=\"italics\">\u03bc<sub>X<\/sub><\/em> = 16.43 from the null hypothes is, <em data-effect=\"italics\">\u03c3<sub>X<\/sub><\/em> = 0.8, and <em data-effect=\"italics\">n<\/em> = 15.) You can find the critical value for <em data-effect=\"italics\">\u03b1<\/em> = 0.05 in the normal table (see <strong>15.Tables<\/strong> in the Table of Contents). The <em data-effect=\"italics\">z<\/em>-score for an area to the left equal to 0.05 is midway between \u20131.65 and \u20131.64 (0.05 is midway between 0.0505 and 0.0495). The <em data-effect=\"italics\">z<\/em>-score is \u20131.645. Since \u20131.645 &gt; \u20132.08 (which demonstrates that \u03b1 &gt; <em data-effect=\"italics\">p<\/em>-value), reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>. Traditionally, the decision to reject or not reject was done in this way. Today, comparing the two probabilities <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value is very common. For this problem, the <em data-effect=\"italics\">p<\/em>-value, 0.0187 is considerably smaller than <em data-effect=\"italics\">\u03b1<\/em>, 0.05. You can be confident about your decision to reject. The graph shows <em data-effect=\"italics\">\u03b1<\/em>, the <em data-effect=\"italics\">p<\/em>-value, and the test statistic and the critical value.<\/p>\n<div id=\"eip-idm47467424\" class=\"bc-figure figure\"><span id=\"eip-idp126239824\" data-type=\"media\" data-alt=\"Distribution curve comparing the \u03b1 to the p-value. Values of -2.15 and -1.645 are on the x-axis. Vertical upward lines extend from both of these values to the curve. The p-value is equal to 0.0158 and points to the area to the left of -2.15. \u03b1 is equal to 0.05 and points to the area between the values of -2.15 and -1.645.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/fig-ch09_11_02-1.png\" alt=\"Distribution curve comparing the \u03b1 to the p-value. Values of -2.15 and -1.645 are on the x-axis. Vertical upward lines extend from both of these values to the curve. The p-value is equal to 0.0158 and points to the area to the left of -2.15. \u03b1 is equal to 0.05 and points to the area between the values of -2.15 and -1.645.\" width=\"380\" data-media-type=\"image\/png\" \/><\/span><\/div>\n<\/div>\n<div class=\"textbox textbox--examples\" data-type=\"example\">\n<div data-type=\"exercise\">\n<div id=\"id5291254\" data-type=\"problem\">\n<p>A college football coach records the mean weight that his players can bench press as <strong>275 pounds<\/strong>, with a <strong>standard deviation of 55 pounds<\/strong>. Three of his players thought that the mean weight was <strong>more than<\/strong> that amount. They asked <strong>30<\/strong> of their teammates for their estimated maximum lift on the bench press exercise. The data ranged from 205 pounds to 385 pounds. The actual different weights were (frequencies are in parentheses) <span id=\"set-1\" data-type=\"list\" data-list-type=\"labeled-item\" data-display=\"inline\"><span data-type=\"item\">205(3)\u00a0 \u00a0<\/span><span data-type=\"item\">215(3)\u00a0 \u00a0<\/span><span data-type=\"item\">225(1)\u00a0 \u00a0<\/span><span data-type=\"item\">241(2)\u00a0 \u00a0<\/span><span data-type=\"item\">252(2)\u00a0 \u00a0<\/span><span data-type=\"item\">265(2)\u00a0 \u00a0<\/span><span data-type=\"item\">275(2)\u00a0 \u00a0 <\/span><span data-type=\"item\">313(2)\u00a0 \u00a0<\/span><span data-type=\"item\">316(5)\u00a0 \u00a0<\/span><span data-type=\"item\">338(2)\u00a0 \u00a0<\/span><span data-type=\"item\">341(1)\u00a0 \u00a0<\/span><span data-type=\"item\">345(2)\u00a0 \u00a0<\/span><span data-type=\"item\">368(2)\u00a0 \u00a0<\/span><span data-type=\"item\">385(1)<\/span><\/span>.<\/p>\n<p>Conduct a hypothesis test using a 2.5% level of significance to determine if the bench press mean is <strong>more than 275 pounds<\/strong>.<\/p>\n<\/div>\n<div id=\"id5342777\" data-type=\"solution\">\n<p>Set up the Hypothesis Test:<\/p>\n<p>Since the problem is about a mean weight, this is a <strong>test of a single population mean<\/strong>.<\/p>\n<p><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 275<span data-type=\"newline\"><br \/>\n<\/span><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &gt; 275<span data-type=\"newline\"><br \/>\n<\/span>This is a right-tailed test.<\/p>\n<p>Calculating the distribution needed:<\/p>\n<p id=\"element-975\">Random variable: \\(\\overline{X}\\) = the mean weight, in pounds, lifted by the football players.<\/p>\n<p><strong>Distribution for the test:<\/strong> It is normal because <em data-effect=\"italics\">\u03c3<\/em> is known.<\/p>\n<p>\\(\\overline{X}~N\\left(275,\\frac{55}{\\sqrt{30}}\\right)\\)<\/p>\n<p>\\(\\overline{x}=286.2\\) pounds (from the data).<\/p>\n<p><em data-effect=\"italics\">\u03c3<\/em> = 55 pounds <strong>(Always use <em data-effect=\"italics\">\u03c3<\/em> if you know it.)<\/strong> We assume <em data-effect=\"italics\">\u03bc<\/em> = 275 pounds unless our data shows us otherwise.<\/p>\n<p>Calculate the <em data-effect=\"italics\">p<\/em>-value using the normal distribution for a mean and using the sample mean as input (see <a class=\"autogenerated-content\" href=\"\/contents\/d0ba1833-f0d2-4195-8765-3c436745f0fb\">(Figure)<\/a> for using the data as input):<\/p>\n<p>\\(p\\text{-value}=P\\left(\\overline{x}&gt;286.2\\right)=0.1323\\).<\/p>\n<p><strong>Interpretation of the <em data-effect=\"italics\">p<\/em>-value:<\/strong> If <em data-effect=\"italics\">H<sub>0<\/sub><\/em> is true, then there is a 0.1331 probability (13.23%) that the football players can lift a mean weight of 286.2 pounds or more. Because a 13.23% chance is large enough, a mean weight lift of 286.2 pounds or more is not a rare event.<\/p>\n<div id=\"hyptest11_ex3\" class=\"bc-figure figure\"><span id=\"id4317824\" data-type=\"media\" data-alt=\"Normal distribution curve of the average weight lifted by football players with values of 275 and 286.2 on the x-axis. A vertical upward line extends from 286.2 to the curve. The p-value points to the area to the right of 286.2.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/fig-ch09_11_03N-1.jpg\" alt=\"Normal distribution curve of the average weight lifted by football players with values of 275 and 286.2 on the x-axis. A vertical upward line extends from 286.2 to the curve. The p-value points to the area to the right of 286.2.\" width=\"380\" data-media-type=\"image\/jpg\" \/><\/span><\/div>\n<p>Compare <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value:<\/p>\n<p><em data-effect=\"italics\">\u03b1<\/em> = 0.025 <em data-effect=\"italics\">p<\/em>-value = 0.1323<\/p>\n<p><strong>Make a decision:<\/strong> Since <em data-effect=\"italics\">\u03b1<\/em> &lt;<em data-effect=\"italics\">p<\/em>-value, do not reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>.<\/p>\n<p><strong>Conclusion:<\/strong> At the 2.5% level of significance, from the sample data, there is not sufficient evidence to conclude that the true mean weight lifted is more than 275 pounds.<\/p>\n<p>The <em data-effect=\"italics\">p<\/em>-value can easily be calculated.<\/p>\n<div id=\"fs-idm79549936\" class=\"statistics calculator\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<p id=\"element-233\">Put the data and frequencies into lists. Press <code>STAT<\/code> and arrow over to <code>TESTS<\/code>. Press <code>1:Z-Test<\/code>. Arrow over to <code>Data<\/code> and press <code>ENTER<\/code>. Arrow down and enter 275 for <em data-effect=\"italics\">\u03bc<sub>0<\/sub><\/em>, 55 for <em data-effect=\"italics\">\u03c3<\/em>, the name of the list where you put the data, and the name of the list where you put the frequencies. Arrow down to <em data-effect=\"italics\">\u03bc:<\/em> and arrow over to &gt; <em data-effect=\"italics\">\u03bc<sub>0<\/sub><\/em>. Press <code>ENTER<\/code>. Arrow down to <code>Calculate<\/code> and press <code>ENTER<\/code>. The calculator not only calculates the <em data-effect=\"italics\">p<\/em>-value (<em data-effect=\"italics\">p<\/em> = 0.1331, a little different from the previous calculation &#8211; in it we used the sample mean rounded to one decimal place instead of the data) but it also calculates the test statistic (<em data-effect=\"italics\">z<\/em>-score) for the sample mean, the sample mean, and the sample standard deviation. <em data-effect=\"italics\">\u03bc<\/em> &gt; 275 is the alternative hypothesis. Do this set of instructions again except arrow to <code>Draw<\/code> (instead of <code>Calculate<\/code>). Press <code>ENTER<\/code>. A shaded graph appears with <em data-effect=\"italics\">z<\/em> = 1.112 (test statistic) and <em data-effect=\"italics\">p<\/em> = 0.1331 (<em data-effect=\"italics\">p<\/em>-value). Make sure when you use <code>Draw<\/code> that no other equations are highlighted in <em data-effect=\"italics\">Y<\/em> = and the plots are turned off.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"element-207\" class=\"textbox textbox--examples\" data-type=\"example\">\n<div data-type=\"exercise\">\n<div id=\"id4319892\" data-type=\"problem\">\n<p id=\"fs-idp90231232\">Statistics students believe that the mean score on the first statistics test is 65. A statistics instructor thinks the mean score is higher than 65. He samples ten statistics students and obtains the scores <span id=\"set-2\" data-type=\"list\" data-list-type=\"labeled-item\" data-display=\"inline\"><span data-type=\"item\">65\u00a0 \u00a0<\/span><span data-type=\"item\">65\u00a0 \u00a0<\/span><span data-type=\"item\">70\u00a0 \u00a0<\/span><span data-type=\"item\">67\u00a0 \u00a0<\/span><span data-type=\"item\">66\u00a0 \u00a0<\/span><span data-type=\"item\">63\u00a0 \u00a0<\/span><span data-type=\"item\">63\u00a0 \u00a0<\/span><span data-type=\"item\">68\u00a0 \u00a0<\/span><span data-type=\"item\">72\u00a0 \u00a0<\/span><span data-type=\"item\">71<\/span><\/span>. He performs a hypothesis test using a 5% level of significance. The data are assumed to be from a normal distribution.<\/p>\n<\/div>\n<div id=\"id5147242\" data-type=\"solution\">\n<p>Set up the hypothesis test:<\/p>\n<p>A 5% level of significance means that <em data-effect=\"italics\">\u03b1<\/em> = 0.05. This is a test of a <strong>single population mean<\/strong>.<\/p>\n<p><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 65\u2003\u2003<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &gt; 65<\/p>\n<p>Since the instructor thinks the average score is higher, use a &#8220;&gt;&#8221;. The &#8220;&gt;&#8221; means the test is right-tailed.<\/p>\n<p id=\"fs-idp165624976\">Determine the distribution needed:<\/p>\n<p><strong>Random variable:<\/strong>\\(\\overline{X}\\) = average score on the first statistics test.<\/p>\n<p><strong>Distribution for the test:<\/strong> If you read the problem carefully, you will notice that there is <strong>no population standard deviation given<\/strong>. You are only given <em data-effect=\"italics\">n<\/em> = 10 sample data values. Notice also that the data come from a normal distribution. This means that the distribution for the test is a student&#8217;s <em data-effect=\"italics\">t<\/em>.<\/p>\n<p id=\"element-240\">Use <em data-effect=\"italics\">t<\/em><sub>df<\/sub>. Therefore, the distribution for the test is <em data-effect=\"italics\">t<\/em><sub>9<\/sub> where <em data-effect=\"italics\">n<\/em> = 10 and <em data-effect=\"italics\">df<\/em> = 10 &#8211; 1 = 9.<\/p>\n<p>Calculate the <em data-effect=\"italics\">p<\/em>-value using the Student&#8217;s <em data-effect=\"italics\">t<\/em>-distribution:<\/p>\n<p><em data-effect=\"italics\">p<\/em>-value = <em data-effect=\"italics\">P<\/em>(\\(\\overline{x}\\) &gt; 67) = 0.0396 where the sample mean and sample standard deviation are calculated as 67 and 3.1972 from the data.<\/p>\n<p><strong>Interpretation of the <em data-effect=\"italics\">p<\/em>-value:<\/strong> If the null hypothesis is true, then there is a 0.0396 probability (3.96%) that the sample mean is 65 or more.<\/p>\n<div id=\"hyptest11_ex4\" class=\"bc-figure figure\"><span id=\"id5241154\" data-type=\"media\" data-alt=\"Normal distribution curve of average scores on the first statistic tests with 65 and 67 values on the x-axis. A vertical upward line extends from 67 to the curve. The p-value points to the area to the right of 67.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/fig-ch09_11_04N-1.jpg\" alt=\"Normal distribution curve of average scores on the first statistic tests with 65 and 67 values on the x-axis. A vertical upward line extends from 67 to the curve. The p-value points to the area to the right of 67.\" width=\"380\" data-media-type=\"image\/jpg\" \/><\/span><\/div>\n<p>Compare <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value:<\/p>\n<p>Since <em data-effect=\"italics\">\u03b1<\/em> = 0.05 and <em data-effect=\"italics\">p<\/em>-value = 0.0396. <em data-effect=\"italics\">\u03b1<\/em> &gt; <em data-effect=\"italics\">p<\/em>-value.<\/p>\n<p><strong>Make a decision:<\/strong> Since <em data-effect=\"italics\">\u03b1<\/em> &gt; <em data-effect=\"italics\">p<\/em>-value, reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>.<\/p>\n<p id=\"fs-idm21057040\">This means you reject <em data-effect=\"italics\">\u03bc<\/em> = 65. In other words, you believe the average test score is more than 65.<\/p>\n<p><strong>Conclusion:<\/strong> At a 5% level of significance, the sample data show sufficient evidence that the mean (average) test score is more than 65, just as the math instructor thinks.<\/p>\n<p>The <em data-effect=\"italics\">p<\/em>-value can easily be calculated.<\/p>\n<div id=\"fs-idm14472336\" class=\"statistics calculator\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<p>Put the data into a list. Press <code>STAT<\/code> and arrow over to <code>TESTS<\/code>. Press <code>2:T-Test<\/code>. Arrow over to <code>Data<\/code> and press <code>ENTER<\/code>. Arrow down and enter 65 for <em data-effect=\"italics\">\u03bc<\/em><sub>0<\/sub>, the name of the list where you put the data, and 1 for <code>Freq:<\/code>. Arrow down to <em data-effect=\"italics\">\u03bc<\/em>: and arrow over to &gt; <em data-effect=\"italics\">\u03bc<\/em><sub>0<\/sub>. Press <code>ENTER<\/code>. Arrow down to <code>Calculate<\/code> and press <code>ENTER<\/code>. The calculator not only calculates the <em data-effect=\"italics\">p<\/em>-value (<em data-effect=\"italics\">p<\/em> = 0.0396) but it also calculates the test statistic (<em data-effect=\"italics\">t<\/em>-score) for the sample mean, the sample mean, and the sample standard deviation. <em data-effect=\"italics\">\u03bc<\/em> &gt; 65 is the alternative hypothesis. Do this set of instructions again except arrow to <code>Draw<\/code> (instead of <code>Calculate<\/code>). Press <code>ENTER<\/code>. A shaded graph appears with <em data-effect=\"italics\">t<\/em> = 1.9781 (test statistic) and <em data-effect=\"italics\">p<\/em> = 0.0396 (<em data-effect=\"italics\">p<\/em>-value). Make sure when you use <code>Draw<\/code> that no other equations are highlighted in <em data-effect=\"italics\">Y<\/em> = and the plots are turned off.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idm186895712\" class=\"statistics try\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<div data-type=\"title\">Try It<\/div>\n<div data-type=\"exercise\">\n<div data-type=\"problem\">\n<p id=\"eip-idm59977568\">It is believed that a stock price for a particular company will grow at a rate of \\$5 per week with a standard deviation of \\$1. An investor believes the stock won\u2019t grow as quickly. The changes in stock price is recorded for ten weeks and are as follows: \\$4, \\$3, \\$2, \\$3, \\$1, \\$7, \\$2, \\$1, \\$1, \\$2. Perform a hypothesis test using a 5% level of significance. State the null and alternative hypotheses, find the <em data-effect=\"italics\">p<\/em>-value, state your conclusion, and identify the Type I and Type II errors.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\" data-type=\"example\">\n<div data-type=\"exercise\">\n<div id=\"id5304801\" data-type=\"problem\">\n<p>Joon believes that 50% of first-time brides in the United States are younger than their grooms. She performs a hypothesis test to determine if the percentage is <strong>the same or different from 50%<\/strong>. Joon samples <strong>100 first-time brides<\/strong> and <strong>53<\/strong> reply that they are younger than their grooms. For the hypothesis test, she uses a 1% level of significance.<\/p>\n<\/div>\n<div id=\"id4150001\" data-type=\"solution\">\n<p id=\"element-399\">Set up the hypothesis test:<\/p>\n<p>The 1% level of significance means that <em data-effect=\"italics\">\u03b1<\/em> = 0.01. This is a <strong>test of a single population proportion<\/strong>.<\/p>\n<p><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.50\u2003\u2003<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 0.50<\/p>\n<p>The words <strong>&#8220;is the same or different from&#8221;<\/strong> tell you this is a two-tailed test.<\/p>\n<p>Calculate the distribution needed:<\/p>\n<p id=\"element-783\"><strong>Random variable:<\/strong><em data-effect=\"italics\">P\u2032<\/em> = the percent of of first-time brides who are younger than their grooms.<\/p>\n<p id=\"element-581\"><strong>Distribution for the test:<\/strong> The problem contains no mention of a mean. The information is given in terms of percentages. Use the distribution for <em data-effect=\"italics\">P\u2032<\/em>, the estimated proportion.<\/p>\n<p id=\"fs-idm81154768\">\\({P}^{\\prime }~N\\left(p,\\sqrt{\\frac{p\\cdot q}{n}}\\right)\\) Therefore, \\({P}^{\\prime }~N\\left(0.5,\\sqrt{\\frac{0.5\\cdot 0.5}{100}}\\right)\\)<\/p>\n<p id=\"fs-idm107085552\">where <em data-effect=\"italics\">p<\/em> = 0.50, <em data-effect=\"italics\">q<\/em> = 1\u2212<em data-effect=\"italics\">p<\/em> = 0.50, and <em data-effect=\"italics\">n<\/em> = 100<\/p>\n<p id=\"fs-idp14246672\">Calculate the <em data-effect=\"italics\">p<\/em>-value using the normal distribution for proportions:<\/p>\n<p><em data-effect=\"italics\">p<\/em>-value = <em data-effect=\"italics\">P<\/em> (<em data-effect=\"italics\">p\u2032<\/em> &lt; 0.47 or <em data-effect=\"italics\">p\u2032<\/em> &gt; 0.53) = 0.5485<\/p>\n<p id=\"fs-idp133040416\">where <em data-effect=\"italics\">x<\/em> = 53, <em data-effect=\"italics\">p\u2032<\/em> = \\(\\frac{x}{n}\\text{\u00a0=\u00a0}\\frac{\\text{53}}{\\text{100}}\\) = 0.53.<\/p>\n<p><strong>Interpretation of the <em data-effect=\"italics\">p<\/em>-value:<\/strong> If the null hypothesis is true, there is 0.5485 probability (54.85%) that the sample (estimated) proportion \\(p\\text{&#8216;}\\) is 0.53 or more OR 0.47 or less (see the graph in <a class=\"autogenerated-content\" href=\"#hyptest11_ex5\">(Figure)<\/a>).<\/p>\n<div id=\"hyptest11_ex5\" class=\"bc-figure figure\"><span id=\"id5140475\" data-type=\"media\" data-alt=\"Normal distribution curve of the percent of first time brides who are younger than the groom with values of 0.47, 0.50, and 0.53 on the x-axis. Vertical upward lines extend from 0.47 and 0.53 to the curve. 1\/2(p-values) are calculated for the areas on outsides of 0.47 and 0.53.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/fig-ch09_11_05N-1.jpg\" alt=\"Normal distribution curve of the percent of first time brides who are younger than the groom with values of 0.47, 0.50, and 0.53 on the x-axis. Vertical upward lines extend from 0.47 and 0.53 to the curve. 1\/2(p-values) are calculated for the areas on outsides of 0.47 and 0.53.\" width=\"380\" data-media-type=\"image\/jpg\" \/><\/span><\/div>\n<p id=\"element-327\"><em data-effect=\"italics\">\u03bc<\/em> = <em data-effect=\"italics\">p<\/em> = 0.50 comes from <em data-effect=\"italics\">H<sub>0<\/sub><\/em>, the null hypothesis.<\/p>\n<p><em data-effect=\"italics\">p\u2032<\/em> = 0.53. Since the curve is symmetrical and the test is two-tailed, the <em data-effect=\"italics\">p\u2032<\/em> for the left tail is equal to 0.50 \u2013 0.03 = 0.47 where <em data-effect=\"italics\">\u03bc<\/em> = <em data-effect=\"italics\">p<\/em> = 0.50. (0.03 is the difference between 0.53 and 0.50.)<\/p>\n<p>Compare <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value:<\/p>\n<p>Since <em data-effect=\"italics\">\u03b1<\/em> = 0.01 and <em data-effect=\"italics\">p<\/em>-value = 0.5485. <em data-effect=\"italics\">\u03b1<\/em> &lt; <em data-effect=\"italics\">p<\/em>-value.<\/p>\n<p><strong>Make a decision:<\/strong> Since <em data-effect=\"italics\">\u03b1<\/em> &lt; <em data-effect=\"italics\">p<\/em>-value, you cannot reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>.<\/p>\n<p><strong>Conclusion:<\/strong> At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of first-time brides who are younger than their grooms is different from 50%.<\/p>\n<p>The <em data-effect=\"italics\">p<\/em>-value can easily be calculated.<\/p>\n<div id=\"fs-idm35225600\" class=\"statistics calculator\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<p>Press <code>STAT<\/code> and arrow over to <code>TESTS<\/code>. Press <code>5:1-PropZTest<\/code>. Enter .5 for <em data-effect=\"italics\">p<\/em><sub>0<\/sub>, 53 for <em data-effect=\"italics\">x<\/em> and 100 for <em data-effect=\"italics\">n<\/em>. Arrow down to <code>Prop<\/code> and arrow to <code>not equals<\/code> <em data-effect=\"italics\">p<\/em><sub>0<\/sub>. Press <code>ENTER<\/code>. Arrow down to <code>Calculate<\/code> and press <code>ENTER<\/code>. The calculator calculates the <em data-effect=\"italics\">p<\/em>-value (<em data-effect=\"italics\">p<\/em> = 0.5485) and the test statistic (<em data-effect=\"italics\">z<\/em>-score). <code>Prop not equals<\/code> .5 is the alternate hypothesis. Do this set of instructions again except arrow to <code>Draw<\/code> (instead of <code>Calculate<\/code>). Press <code>ENTER<\/code>. A shaded graph appears with <em data-effect=\"italics\">z<\/em> = 0.6 (test statistic) and <em data-effect=\"italics\">p<\/em> = 0.5485 (<em data-effect=\"italics\">p<\/em>-value). Make sure when you use <code>Draw<\/code> that no other equations are highlighted in <em data-effect=\"italics\">Y<\/em> = and the plots are turned off.<\/p>\n<\/div>\n<p>The Type I and Type II errors are as follows:<\/p>\n<p>The Type I error is to conclude that the proportion of first-time brides who are younger than their grooms is different from 50% when, in fact, the proportion is actually 50%. (Reject the null hypothesis when the null hypothesis is true).<\/p>\n<p id=\"fs-idp134008544\">The Type II error is there is not enough evidence to conclude that the proportion of first time brides who are younger than their grooms differs from 50% when, in fact, the proportion does differ from 50%. (Do not reject the null hypothesis when the null hypothesis is false.)<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idp8711616\" class=\"statistics try\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<div data-type=\"title\">Try It<\/div>\n<div data-type=\"exercise\">\n<div data-type=\"problem\">\n<p id=\"eip-idp103490704\">A teacher believes that 85% of students in the class will want to go on a field trip to the local zoo. She performs a hypothesis test to determine if the percentage is the same or different from 85%. The teacher samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1% level of significance.<\/p>\n<p id=\"eip-idp10599024\">First, determine what type of test this is, set up the hypothesis test, find the <em data-effect=\"italics\">p<\/em>-value, sketch the graph, and state your conclusion.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\" data-type=\"example\">\n<div id=\"element-312\" data-type=\"exercise\">\n<div id=\"id5318755\" data-type=\"problem\">\n<p>Suppose a consumer group suspects that the proportion of households that have three cell phones is 30%. A cell phone company has reason to believe that the proportion is not 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three cell phones.<\/p>\n<\/div>\n<div id=\"id4213559\" data-type=\"solution\">\n<p>Set up the Hypothesis Test:<\/p>\n<p><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.30 <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 0.30<\/p>\n<p>Determine the distribution needed:<\/p>\n<p>The <strong>random variable<\/strong> is <em data-effect=\"italics\">P\u2032<\/em> = proportion of households that have three cell phones.<\/p>\n<p>The <strong>distribution<\/strong> for the hypothesis test is \\(P\\text{&#8216;}~N\\left(0.30,\\sqrt{\\frac{\\left(0.30\\right)\\cdot \\left(0.70\\right)}{150}}\\right)\\)<\/p>\n<p>&nbsp;<\/p>\n<div data-type=\"exercise\">\n<div id=\"id4146441\" data-type=\"problem\">\n<p>a. The value that helps determine the <em data-effect=\"italics\">p<\/em>-value is <em data-effect=\"italics\">p\u2032<\/em>. Calculate <em data-effect=\"italics\">p\u2032<\/em>.<\/p>\n<\/div>\n<div id=\"id4274849\" data-type=\"solution\" data-print-placement=\"end\">\n<p>a. <em data-effect=\"italics\">p\u2032<\/em> = \\(\\frac{x}{n}\\) where <em data-effect=\"italics\">x<\/em> is the number of successes and <em data-effect=\"italics\">n<\/em> is the total number in the sample.<\/p>\n<p><em data-effect=\"italics\">x<\/em> = 43, <em data-effect=\"italics\">n<\/em> = 150<\/p>\n<p><em data-effect=\"italics\">p\u2032<\/em> = \\(\\frac{\\text{43}}{\\text{150}}\\)<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\">\n<div id=\"id5150803\" data-type=\"problem\">\n<p>b. What is a <strong>success<\/strong> for this problem?<\/p>\n<\/div>\n<div id=\"id5036464\" data-type=\"solution\" data-print-placement=\"end\">\n<p>b. A success is having three cell phones in a household.<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<div id=\"element-519\" data-type=\"exercise\">\n<div id=\"id4292096\" data-type=\"problem\">\n<p id=\"element-781\">c. What is the level of significance?<\/p>\n<\/div>\n<div id=\"id5242701\" data-type=\"solution\" data-print-placement=\"end\">\n<p>c. The level of significance is the preset <em data-effect=\"italics\">\u03b1<\/em>. Since <em data-effect=\"italics\">\u03b1<\/em> is not given, assume that <em data-effect=\"italics\">\u03b1<\/em> = 0.05.<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\">\n<div id=\"id4673389\" data-type=\"problem\">\n<p>d. Draw the graph for this problem. Draw the horizontal axis. Label and shade appropriately.<span data-type=\"newline\"><br \/>\n<\/span>Calculate the <em data-effect=\"italics\">p<\/em>-value.<\/p>\n<\/div>\n<div id=\"id5212164\" data-type=\"solution\" data-print-placement=\"end\">\n<p>d. <em data-effect=\"italics\">p<\/em>-value = 0.7216<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\">\n<div id=\"id5104601\" data-type=\"problem\">\n<p>e. Make a decision. _____________(Reject\/Do not reject) <em data-effect=\"italics\">H<sub>0<\/sub><\/em> because____________.<\/p>\n<\/div>\n<div id=\"id5137008\" data-type=\"solution\" data-print-placement=\"end\">\n<p>e. Assuming that <em data-effect=\"italics\">\u03b1<\/em> = 0.05, <em data-effect=\"italics\">\u03b1<\/em> &lt; <em data-effect=\"italics\">p<\/em>-value. The decision is do not reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em> because there is not sufficient evidence to conclude that the proportion of households that have three cell phones is not 30%.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idm14663600\" class=\"statistics try\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<div data-type=\"title\">Try It<\/div>\n<div data-type=\"exercise\">\n<div data-type=\"problem\">\n<p id=\"eip-idm17000432\">Marketers believe that 92% of adults in the United States own a cell phone. A cell phone manufacturer believes that number is actually lower. 200 American adults are surveyed, of which, 174 report having cell phones. Use a 5% level of significance. State the null and alternative hypothesis, find the <em data-effect=\"italics\">p<\/em>-value, state your conclusion, and identify the Type I and Type II errors.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-idp32935184\">The next example is a poem written by a statistics student named Nicole Hart. The solution to the problem follows the poem. Notice that the hypothesis test is for a single population proportion. This means that the null and alternate hypotheses use the parameter <em data-effect=\"italics\">p<\/em>. The distribution for the test is normal. The estimated proportion <em data-effect=\"italics\">p<\/em>\u2032 is the proportion of fleas killed to the total fleas found on Fido. This is sample information. The problem gives a preconceived <em data-effect=\"italics\">\u03b1<\/em> = 0.01, for comparison, and a 95% confidence interval computation. The poem is clever and humorous, so please enjoy it!<\/p>\n<div class=\"textbox textbox--examples\" data-type=\"example\">\n<div id=\"fs-idp65427488\" data-type=\"exercise\">\n<div id=\"id5257658\" data-type=\"problem\">\n<p id=\"eip-idm151173424\">My dog has so many fleas,<span data-type=\"newline\"><br \/>\n<\/span> They do not come off with ease.<span data-type=\"newline\"><br \/>\n<\/span> As for shampoo, I have tried many types<span data-type=\"newline\"><br \/>\n<\/span> Even one called Bubble Hype,<span data-type=\"newline\"><br \/>\n<\/span> Which only killed 25% of the fleas,<span data-type=\"newline\"><br \/>\n<\/span> Unfortunately I was not pleased.<\/p>\n<p>I&#8217;ve used all kinds of soap,<span data-type=\"newline\"><br \/>\n<\/span> Until I had given up hope<span data-type=\"newline\"><br \/>\n<\/span> Until one day I saw<span data-type=\"newline\"><br \/>\n<\/span> An ad that put me in awe.<\/p>\n<p>A shampoo used for dogs<span data-type=\"newline\"><br \/>\n<\/span> Called GOOD ENOUGH to Clean a Hog<span data-type=\"newline\"><br \/>\n<\/span> Guaranteed to kill more fleas.<\/p>\n<p>I gave Fido a bath<span data-type=\"newline\"><br \/>\n<\/span> And after doing the math<span data-type=\"newline\"><br \/>\n<\/span> His number of fleas<span data-type=\"newline\"><br \/>\n<\/span> Started dropping by 3&#8217;s!<\/p>\n<p>Before his shampoo<span data-type=\"newline\"><br \/>\n<\/span> I counted 42.<span data-type=\"newline\"><br \/>\n<\/span> At the end of his bath,<span data-type=\"newline\"><br \/>\n<\/span> I redid the math<span data-type=\"newline\"><br \/>\n<\/span> And the new shampoo had killed 17 fleas.<span data-type=\"newline\"><br \/>\n<\/span> So now I was pleased.<\/p>\n<p>Now it is time for you to have some fun<span data-type=\"newline\"><br \/>\n<\/span> With the level of significance being .01,<span data-type=\"newline\"><br \/>\n<\/span> You must help me figure out<span data-type=\"newline\"><br \/>\n<\/span> Use the new shampoo or go without?<\/p>\n<\/div>\n<div id=\"id5100583\" data-type=\"solution\">\n<p>Set up the hypothesis test:<\/p>\n<p><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2264 0.25 \u2003\u2003<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &gt; 0.25<\/p>\n<p>Determine the distribution needed:<\/p>\n<p>In words, CLEARLY state what your random variable \\(\\overline{X}\\) or <em data-effect=\"italics\">P\u2032<\/em> represents.<\/p>\n<p id=\"fs-idp88897968\"><em data-effect=\"italics\">P\u2032<\/em> = The proportion of fleas that are killed by the new shampoo<\/p>\n<p>State the distribution to use for the test.<\/p>\n<p><strong>Normal:<\/strong>\\(N\\left(0.25,\\sqrt{\\frac{\\left(0.25\\right)\\left(1-0.25\\right)}{42}}\\right)\\)<\/p>\n<p><strong>Test Statistic:<\/strong><em data-effect=\"italics\">z<\/em> = 2.3163<\/p>\n<p>Calculate the <em data-effect=\"italics\">p<\/em>-value using the normal distribution for proportions:<\/p>\n<p><em data-effect=\"italics\">p<\/em>-value = 0.0103<\/p>\n<p>In one to two complete sentences, explain what the <em data-effect=\"italics\">p<\/em>-value means for this problem.<\/p>\n<p>If the null hypothesis is true (the proportion is 0.25), then there is a 0.0103 probability that the sample (estimated) proportion is 0.4048 \\(\\left(\\frac{17}{42}\\right)\\) or more.<\/p>\n<p>Use the previous information to sketch a picture of this situation. CLEARLY, label and scale the horizontal axis and shade the region(s) corresponding to the <em data-effect=\"italics\">p<\/em>-value.<\/p>\n<div id=\"hyptest11_ex6\" class=\"bc-figure figure\"><span id=\"id4295819\" data-type=\"media\" data-alt=\"Normal distribution graph of the proportion of fleas killed by the new shampoo with values of 0.25 and 0.4048 on the x-axis. A vertical upward line extends from 0.4048 to the curve and the area to the left of this is shaded in. The test statistic of the sample proportion is listed.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/fig-ch09_11_06-1.jpg\" alt=\"Normal distribution graph of the proportion of fleas killed by the new shampoo with values of 0.25 and 0.4048 on the x-axis. A vertical upward line extends from 0.4048 to the curve and the area to the left of this is shaded in. The test statistic of the sample proportion is listed.\" width=\"380\" data-media-type=\"image\/jpg\" \/><\/span><\/div>\n<p id=\"element-372\">Compare <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value:<\/p>\n<p>Indicate the correct decision (\u201creject\u201d or \u201cdo not reject\u201d the null hypothesis), the reason for it, and write an appropriate conclusion, using complete sentences.<\/p>\n<table summary=\"\">\n<thead valign=\"middle\">\n<tr>\n<th data-align=\"center\">alpha<\/th>\n<th data-align=\"center\">decision<\/th>\n<th data-align=\"center\">reason for decision<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td data-align=\"center\">0.01<\/td>\n<td data-align=\"center\">Do not reject \\({H}_{0}\\)<\/td>\n<td data-align=\"center\"><em data-effect=\"italics\">\u03b1<\/em> &lt; <em data-effect=\"italics\">p<\/em>-value<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>Conclusion:<\/strong> At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of fleas that are killed by the new shampoo is more than 25%.<\/p>\n<p>Construct a 95% confidence interval for the true mean or proportion. Include a sketch of the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval.<\/p>\n<div id=\"hyptest11_ex7\" class=\"bc-figure figure\"><span id=\"id5244822\" data-type=\"media\" data-alt=\"Normal distribution graph of the proportion of fleas killed by the new shampoo with values of 0.26, 17\/42, and 0.55 on the x-axis. A vertical upward line extends from 0.26 and 0.55. The area between these two points is equal to 0.95.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/fig-ch09_11_07-1.jpg\" alt=\"Normal distribution graph of the proportion of fleas killed by the new shampoo with values of 0.26, 17\/42, and 0.55 on the x-axis. A vertical upward line extends from 0.26 and 0.55. The area between these two points is equal to 0.95.\" width=\"380\" data-media-type=\"image\/jpg\" \/><\/span><\/div>\n<p><strong>Confidence Interval:<\/strong> (0.26,0.55) We are 95% confident that the true population proportion <em data-effect=\"italics\">p<\/em> of fleas that are killed by the new shampoo is between 26% and 55%.<\/p>\n<div id=\"id5328435\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<div data-type=\"title\">Note<\/div>\n<p id=\"eip-idm81920704\">This test result is not very definitive since the <em data-effect=\"italics\">p<\/em>-value is very close to alpha. In reality, one would probably do more tests by giving the dog another bath after the fleas have had a chance to return.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\" data-type=\"example\">\n<div data-type=\"exercise\">\n<div data-type=\"problem\">\n<p id=\"eip-idm29000496\">The National Institute of Standards and Technology provides exact data on conductivity properties of materials. Following are conductivity measurements for 11 randomly selected pieces of a particular type of glass.<\/p>\n<p id=\"eip-idm46067920\">1.11;\u00a0 \u00a01.07;\u00a0 \u00a01.11;\u00a0 \u00a01.07;\u00a0 \u00a01.12;\u00a0 \u00a01.08;\u00a0 \u00a00.98;\u00a0 \u00a00.98;\u00a0 \u00a01.02;\u00a0 \u00a00.95;\u00a0 \u00a00.95 <span data-type=\"newline\"><br \/>\n<\/span>Is there convincing evidence that the average conductivity of this type of glass is greater than one? Use a significance level of 0.05. Assume the population is normal.<\/p>\n<\/div>\n<div data-type=\"solution\">\n<p id=\"eip-idp69766080\">Let\u2019s follow a four-step process to answer this statistical question.<\/p>\n<ol id=\"eip-idp69766512\" type=\"1\">\n<li><strong>State the Question<\/strong>: We need to determine if, at a 0.05 significance level, the average conductivity of the selected glass is greater than one. Our hypotheses will be\n<ol id=\"eip-idp129554720\" type=\"a\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2264 1<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &gt; 1<\/li>\n<\/ol>\n<\/li>\n<li><strong>Plan<\/strong>: We are testing a sample mean without a known population standard deviation. Therefore, we need to use a Student&#8217;s-t distribution. Assume the underlying population is normal.<\/li>\n<li><strong>Do the calculations<\/strong>: We will input the sample data into the TI-83 as follows.\n<div id=\"eip-idp122053552\" class=\"bc-figure figure\"><span id=\"eip-idp167220912\" data-type=\"media\" data-alt=\"\"><img decoding=\"async\" src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/CNX_Stats_C09_M09_001a-1.jpg\" alt=\"\" width=\"250\" data-media-type=\"image\/jpeg\" \/><\/span><\/div>\n<div id=\"eip-idp29927184\" class=\"bc-figure figure\"><span id=\"eip-idp29927312\" data-type=\"media\" data-alt=\"\"><img decoding=\"async\" src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/CNX_Stats_C09_M09_001b-1.jpg\" alt=\"\" width=\"250\" data-media-type=\"image\/jpeg\" \/><\/span><\/div>\n<div id=\"eip-idp126675984\" class=\"bc-figure figure\"><span id=\"eip-idp156983600\" data-type=\"media\" data-alt=\"\"><img decoding=\"async\" src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/CNX_Stats_C09_M09_001c-1.jpg\" alt=\"\" width=\"250\" data-media-type=\"image\/jpeg\" \/><\/span><\/div>\n<div id=\"eip-idp154420944\" class=\"bc-figure figure\"><span id=\"eip-idp154421072\" data-type=\"media\" data-alt=\"\"><img decoding=\"async\" src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/CNX_Stats_C09_M09_001d-1.jpg\" alt=\"\" width=\"250\" data-media-type=\"image\/jpeg\" \/><\/span><\/div>\n<\/li>\n<li><strong>State the Conclusions<\/strong>: Since the <em data-effect=\"italics\">p<\/em>-value* (<em data-effect=\"italics\">p<\/em> = 0.036) is less than our alpha value, we will reject the null hypothesis. It is reasonable to state that the data supports the claim that the average conductivity level is greater than one.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"eip-767\" class=\"textbox textbox--examples\" data-type=\"example\">\n<div id=\"eip-178\" data-type=\"exercise\">\n<div data-type=\"problem\">\n<p id=\"eip-idp173526416\">In a study of 420,019 cell phone users, 172 of the subjects developed brain cancer. Test the claim that cell phone users developed brain cancer at a greater rate than that for non-cell phone users (the rate of brain cancer for non-cell phone users is 0.0340%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.<\/p>\n<\/div>\n<div data-type=\"solution\">\n<p id=\"eip-idp3953440\">We will follow the four-step process.<\/p>\n<ol id=\"eip-idp191218624\" type=\"1\">\n<li>We need to conduct a hypothesis test on the claimed cancer rate. Our hypotheses will be\n<ol id=\"eip-idp171226608\" type=\"a\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2264 0.00034<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &gt; 0.00034<\/li>\n<\/ol>\n<p id=\"eip-idm39752176\" class=\"finger\">If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes cancer-causing environments, we want to minimize the chances of incorrectly identifying causes of cancer.<\/p>\n<\/li>\n<li>We will be testing a sample proportion with <em data-effect=\"italics\">x<\/em> = 172 and <em data-effect=\"italics\">n<\/em> = 420,019. The sample is sufficiently large because we have <em data-effect=\"italics\">np<\/em> = 420,019(0.00034) = 142.8, <em data-effect=\"italics\">nq<\/em> = 420,019(0.99966) = 419,876.2, two independent outcomes, and a fixed probability of success <em data-effect=\"italics\">p<\/em> = 0.00034. Thus we will be able to generalize our results to the population.<\/li>\n<li>The associated TI results are <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-idm80776672\" class=\"bc-figure figure\"><span id=\"eip-idp39356816\" data-type=\"media\" data-alt=\"\"><img decoding=\"async\" src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/CNX_Stats_C09_M09_002a-1.jpg\" alt=\"\" width=\"250\" data-media-type=\"image\/jpeg\" \/><\/span><\/div>\n<div id=\"eip-idp38098208\" class=\"bc-figure figure\"><span id=\"eip-idp38098336\" data-type=\"media\" data-alt=\"\"><img decoding=\"async\" src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/CNX_Stats_C09_M09_002b-1.jpg\" alt=\"\" width=\"250\" data-media-type=\"image\/jpeg\" \/><\/span><\/div>\n<\/li>\n<li>Since the <em data-effect=\"italics\">p<\/em>-value = 0.0073 is greater than our alpha value = 0.005, we cannot reject the null. Therefore, we conclude that there is not enough evidence to support the claim of higher brain cancer rates for the cell phone users.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\" data-type=\"example\">\n<div data-type=\"exercise\">\n<div data-type=\"problem\">\n<p id=\"eip-idp47042256\">According to the US Census there are approximately 268,608,618 residents aged 12 and older. Statistics from the Rape, Abuse, and Incest National Network indicate that, on average, 207,754 rapes occur each year (male and female) for persons aged 12 and older. This translates into a percentage of sexual assaults of 0.078%. In Daviess County, KY, there were reported 11 rapes for a population of 37,937. Conduct an appropriate hypothesis test to determine if there is a statistically significant difference between the local sexual assault percentage and the national sexual assault percentage. Use a significance level of 0.01.<\/p>\n<\/div>\n<div data-type=\"solution\">\n<p>We will follow the four-step plan.<\/p>\n<ol id=\"eip-idp113522624\" type=\"1\">\n<li>We need to test whether the proportion of sexual assaults in Daviess County, KY is significantly different from the national average.<\/li>\n<li>Since we are presented with proportions, we will use a one-proportion <em data-effect=\"italics\">z<\/em>-test. The hypotheses for the test will be\n<ol id=\"eip-idp17854992\" type=\"a\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.00078<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 0.00078<\/li>\n<\/ol>\n<\/li>\n<li>The following screen shots display the summary statistics from the hypothesis test.\n<div id=\"fs-idm3544720\" class=\"bc-figure figure\"><span id=\"eip-idp26791072\" data-type=\"media\" data-alt=\"\"><img decoding=\"async\" src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/CNX_Stats_C09_M09_003a-1.jpg\" alt=\"\" width=\"250\" data-media-type=\"image\/jpeg\" \/><\/span><\/div>\n<div id=\"eip-idp38471600\" class=\"bc-figure figure\"><span id=\"eip-idp38471728\" data-type=\"media\" data-alt=\"\"><img decoding=\"async\" src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/CNX_Stats_C09_M09_003b-1.jpg\" alt=\"\" width=\"250\" data-media-type=\"image\/jpeg\" \/><\/span><\/div>\n<\/li>\n<li>Since the <em data-effect=\"italics\">p<\/em>-value, <em data-effect=\"italics\">p<\/em> = 0.00063, is less than the alpha level of 0.01, the sample data indicates that we should reject the null hypothesis. In conclusion, the sample data support the claim that the proportion of sexual assaults in Daviess County, Kentucky is different from the national average proportion.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idp69608272\" class=\"summary\" data-depth=\"1\">\n<h3 data-type=\"title\">Chapter Review<\/h3>\n<p id=\"fs-idp168110512\">The <span data-type=\"term\">hypothesis test<\/span> itself has an established process. This can be summarized as follows:<\/p>\n<ol>\n<li>Determine <em data-effect=\"italics\">H<sub>0<\/sub><\/em> and <em data-effect=\"italics\">H<sub>a<\/sub><\/em>. Remember, they are contradictory.<\/li>\n<li>Determine the random variable.<\/li>\n<li>Determine the distribution for the test.<\/li>\n<li>Draw a graph, calculate the test statistic, and use the test statistic to calculate the <em data-effect=\"italics\">p<\/em>-value. (A <em data-effect=\"italics\">z<\/em>-score and a <em data-effect=\"italics\">t<\/em>-score are examples of test statistics.)<\/li>\n<li>Compare the preconceived <em data-effect=\"italics\">\u03b1<\/em> with the <em data-effect=\"italics\">p<\/em>-value, make a decision (reject or do not reject <em data-effect=\"italics\">H<sub>0<\/sub><\/em>), and write a clear conclusion using English sentences.<\/li>\n<\/ol>\n<p>Notice that in performing the hypothesis test, you use <em data-effect=\"italics\">\u03b1<\/em> and not <em data-effect=\"italics\">\u03b2<\/em>. <em data-effect=\"italics\">\u03b2<\/em> is needed to help determine the sample size of the data that is used in calculating the <em data-effect=\"italics\">p<\/em>-value. Remember that the quantity 1 \u2013 <em data-effect=\"italics\">\u03b2<\/em> is called the <strong>Power of the Test<\/strong>. A high power is desirable. If the power is too low, statisticians typically increase the sample size while keeping <em data-effect=\"italics\">\u03b1<\/em> the same.If the power is low, the null hypothesis might not be rejected when it should be.<\/p>\n<\/div>\n<div class=\"practice\" data-depth=\"1\">\n<div data-type=\"exercise\">\n<div id=\"eip-135\" data-type=\"problem\">\n<p>Assume <em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 9 and <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &lt; 9. Is this a left-tailed, right-tailed, or two-tailed test?<\/p>\n<\/div>\n<div data-type=\"solution\">\n<p>This is a left-tailed test.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\">\n<div id=\"eip-286\" data-type=\"problem\">\n<p>Assume <em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2264 6 and <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &gt; 6. Is this a left-tailed, right-tailed, or two-tailed test?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idp45725152\" data-type=\"exercise\">\n<div id=\"eip-116\" data-type=\"problem\">\n<p>Assume <em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> = 0.25 and <em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2260 0.25. Is this a left-tailed, right-tailed, or two-tailed test?<\/p>\n<\/div>\n<div data-type=\"solution\">\n<p>This is a two-tailed test.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\">\n<div data-type=\"problem\">\n<p id=\"eip-195\">Draw the general graph of a left-tailed test.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\">\n<div data-type=\"problem\">\n<p>Draw the graph of a two-tailed test.<\/p>\n<\/div>\n<div data-type=\"solution\">\n<div id=\"fs-idm149132800\" class=\"bc-figure figure\"><span id=\"eip-idp62459136\" data-type=\"media\" data-alt=\"\"><img decoding=\"async\" src=\"https:\/\/pressbooks.ccconline.org\/acccomposition1\/wp-content\/uploads\/sites\/83\/2022\/08\/CNX_Stats_C09_M09_item002annoN-1.jpg\" alt=\"\" width=\"380\" data-media-type=\"image\/jpeg\" \/><\/span><\/div>\n<\/div>\n<\/div>\n<div data-type=\"exercise\">\n<div id=\"eip-648\" data-type=\"problem\">\n<p>A bottle of water is labeled as containing 16 fluid ounces of water. You believe it is less than that. What type of test would you use?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\">\n<div data-type=\"problem\">\n<p>Your friend claims that his mean golf score is 63. You want to show that it is higher than that. What type of test would you use?<\/p>\n<\/div>\n<div data-type=\"solution\">\n<p id=\"eip-68\">a right-tailed test<\/p>\n<\/div>\n<\/div>\n<div id=\"eip-729\" data-type=\"exercise\">\n<div id=\"eip-491\" data-type=\"problem\">\n<p>A bathroom scale claims to be able to identify correctly any weight within a pound. You think that it cannot be that accurate. What type of test would you use?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\">\n<div data-type=\"problem\">\n<p id=\"eip-458\">You flip a coin and record whether it shows heads or tails. You know the probability of getting heads is 50%, but you think it is less for this particular coin. What type of test would you use?<\/p>\n<\/div>\n<div id=\"fs-idp66689312\" data-type=\"solution\">\n<p id=\"eip-139\">a left-tailed test<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\">\n<div id=\"eip-519\" data-type=\"problem\">\n<p>If the alternative hypothesis has a not equals ( \u2260 ) symbol, you know to use which type of test?<\/p>\n<\/div>\n<\/div>\n<div id=\"eip-504\" data-type=\"exercise\">\n<div id=\"eip-634\" data-type=\"problem\">\n<p>Assume the null hypothesis states that the mean is at least 18. Is this a left-tailed, right-tailed, or two-tailed test?<\/p>\n<\/div>\n<div data-type=\"solution\">\n<p>This is a left-tailed test.<\/p>\n<\/div>\n<\/div>\n<div id=\"eip-320\" data-type=\"exercise\">\n<div data-type=\"problem\">\n<p>Assume the null hypothesis states that the mean is at most 12. Is this a left-tailed, right-tailed, or two-tailed test?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\">\n<div data-type=\"problem\">\n<p>Assume the null hypothesis states that the mean is equal to 88. The alternative hypothesis states that the mean is not equal to 88. Is this a left-tailed, right-tailed, or two-tailed test?<\/p>\n<\/div>\n<div data-type=\"solution\">\n<p>This is a two-tailed test.<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div id=\"fs-idp139727497473120\" class=\"free-response\" data-depth=\"1\">\n<h3 data-type=\"title\">Homework<\/h3>\n<p id=\"eip-idm30094416\"><em data-effect=\"italics\">For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in <a class=\"autogenerated-content\" href=\"\/contents\/c0449c55-aa47-4f1c-bd5f-0521652f0e82\">(Figure)<\/a>. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.<\/em><\/p>\n<div id=\"id8395739\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<div data-type=\"title\">Note<\/div>\n<p id=\"eip-idm120632832\">If you are using a Student&#8217;s-<em data-effect=\"italics\">t<\/em> distribution for one of the following homework problems, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, however.)<\/p>\n<\/div>\n<div data-type=\"exercise\">\n<div id=\"id8395759\" data-type=\"problem\">\n<p id=\"fs-idp44498592\">1) A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using alpha = 0.05, is the data highly inconsistent with the claim?<\/p>\n<\/div>\n<div id=\"fs-idm164410224\" data-type=\"solution\">\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\">\n<div id=\"id8395793\" data-type=\"problem\">\n<p>2) From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant of around 2.1 years. A survey of 40 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level?<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\">\n<div id=\"id9424298\" data-type=\"problem\">\n<p>3) The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of 20\u00a2. A study was done to test the claim that the mean cost of a daily newspaper is \\$1.00. Twelve costs yield a mean cost of 95\u00a2 with a standard deviation of 18\u00a2. Do the data support the claim at the 1% level?<\/p>\n<\/div>\n<div id=\"fs-idm17421856\" data-type=\"solution\">\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\">\n<div id=\"id9424338\" data-type=\"problem\">\n<p id=\"element-977\">4) An article in the <em data-effect=\"italics\">San Jose Mercury News<\/em> stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 49 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1% level?<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<div id=\"element-932a\" data-type=\"exercise\">\n<div id=\"id9424572\" data-type=\"problem\">\n<p>5) The mean number of sick days an employee takes per year is believed to be about ten. Members of a personnel department do not believe this figure. They randomly survey eight employees. The number of sick days they took for the past year are as follows: 12;\u00a0 4;\u00a0 15;\u00a0 3;\u00a0 11;\u00a0 8;\u00a0 6;\u00a0 8. Let <em data-effect=\"italics\">x<\/em> = the number of sick days they took for the past year. Should the personnel team believe that the mean number is ten?<\/p>\n<\/div>\n<div id=\"fs-idm143535568\" data-type=\"solution\">\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\">\n<div id=\"id9424624\" data-type=\"problem\">\n<p>6) In 1955, <em data-effect=\"italics\">Life Magazine<\/em> reported that the 25 year-old mother of three worked, on average, an 80 hour week. Recently, many groups have been studying whether or not the women&#8217;s movement has, in fact, resulted in an increase in the average work week for women (combining employment and at-home work). Suppose a study was done to determine if the mean work week has increased. 81 women were surveyed with the following results. The sample mean was 83; the sample standard deviation was ten. Does it appear that the mean work week has increased for women at the 5% level?<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\">\n<div id=\"id9400577\" data-type=\"problem\">\n<p>7) Your statistics instructor claims that 60 percent of the students who take her Elementary Statistics class go through life feeling more enriched. For some reason that she can&#8217;t quite figure out, most people don&#8217;t believe her. You decide to check this out on your own. You randomly survey 64 of her past Elementary Statistics students and find that 34 feel more enriched as a result of her class. Now, what do you think?<\/p>\n<\/div>\n<div id=\"fs-idm68403536\" data-type=\"solution\">\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\">\n<div id=\"id9400611\" data-type=\"problem\">\n<p>8) A Nissan Motor Corporation advertisement read, \u201cThe average man\u2019s I.Q. is 107. The average brown trout\u2019s I.Q. is 4. So why can\u2019t man catch brown trout?\u201d Suppose you believe that the brown trout\u2019s mean I.Q. is greater than four. You catch 12 brown trout. A fish psychologist determines the I.Q.s as follows: 5;\u00a0 \u00a04;\u00a0 \u00a07;\u00a0 \u00a03;\u00a0 \u00a06;\u00a0 \u00a04;\u00a0 \u00a05;\u00a0 \u00a03;\u00a0 \u00a06;\u00a0 \u00a03;\u00a0 \u00a08;\u00a0 \u00a05. Conduct a hypothesis test of your belief.<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\">\n<div id=\"id9400954\" data-type=\"problem\">\n<p>9) Refer to <a href=\"#element-410\">Exercise 9.119<\/a>. Conduct a hypothesis test to see if your decision and conclusion would change if your belief were that the brown trout\u2019s mean I.Q. is <strong>not<\/strong> four.<\/p>\n<\/div>\n<div id=\"fs-idm153124944\" data-type=\"solution\">\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\">\n<div id=\"id9400994\" data-type=\"problem\">\n<p>10) According to an article in <em data-effect=\"italics\">Newsweek<\/em>, the natural ratio of girls to boys is 100:105. In China, the birth ratio is 100: 114 (46.7% girls). Suppose you don\u2019t believe the reported figures of the percent of girls born in China. You conduct a study. In this study, you count the number of girls and boys born in 150 randomly chosen recent births. There are 60 girls and 90 boys born of the 150. Based on your study, do you believe that the percent of girls born in China is 46.7?<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<div id=\"element-658\" data-type=\"exercise\">\n<div id=\"id9401185\" data-type=\"problem\">\n<p>11) A poll done for <em data-effect=\"italics\">Newsweek<\/em> found that 13% of Americans have seen or sensed the presence of an angel. A contingent doubts that the percent is really that high. It conducts its own survey. Out of 76 Americans surveyed, only two had seen or sensed the presence of an angel. As a result of the contingent\u2019s survey, would you agree with the <em data-effect=\"italics\">Newsweek<\/em> poll? In complete sentences, also give three reasons why the two polls might give different results.<\/p>\n<\/div>\n<div id=\"fs-idm3310512\" data-type=\"solution\">\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\">\n<div id=\"id9401240\" data-type=\"problem\">\n<p id=\"element-140a\">12) The mean work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it\u2019s shorter. She asks ten engineering friends in start-ups for the lengths of their mean work weeks. Based on the results that follow, should she count on the mean work week to be shorter than 60 hours?<\/p>\n<p id=\"fs-idp187110944\">Data (length of mean work week): 70;\u00a0 \u00a045;\u00a0 \u00a055;\u00a0 \u00a060;\u00a0 \u00a065;\u00a0 \u00a055;\u00a0 \u00a055;\u00a0 \u00a060;\u00a0 \u00a050;\u00a0 \u00a055.<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div id=\"element-761\" data-type=\"exercise\">\n<div id=\"id9401488\" data-type=\"problem\">\n<p id=\"element-926\">13) Use the \u201cLap time\u201d data for Lap 4 (see <a class=\"autogenerated-content\" href=\"\/contents\/3ef830bc-5247-460a-9007-e3fd762e5e93\">(Figure)<\/a>) to test the claim that Terri finishes Lap 4, on average, in less than 129 seconds. Use all twenty races given.<\/p>\n<\/div>\n<div id=\"fs-idp13518384\" data-type=\"solution\">\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idp191961952\" data-type=\"exercise\">\n<div id=\"id9401521\" data-type=\"problem\">\n<p>&nbsp;<\/p>\n<p id=\"element-564a\">14) Use the \u201cInitial Public Offering\u201d data (see <a class=\"autogenerated-content\" href=\"\/contents\/3ef830bc-5247-460a-9007-e3fd762e5e93\">(Figure)<\/a>) to test the claim that the mean offer price was \\$18 per share. Do not use all the data. Use your random number generator to randomly survey 15 prices.<\/p>\n<\/div>\n<\/div>\n<div id=\"id9401550\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<div data-type=\"title\">Note<\/div>\n<\/div>\n<div data-type=\"exercise\">\n<div data-type=\"solution\">\n<p><strong>Answers to odd questions<\/strong><\/p>\n<p>1)<\/p>\n<ol id=\"fs-idp21859504\" type=\"a\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2265 50,000<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &lt; 50,000<\/li>\n<li>Let \\(\\overline{X}\\) = the average lifespan of a brand of tires.<\/li>\n<li>normal distribution<\/li>\n<li><em data-effect=\"italics\">z<\/em> = -2.315<\/li>\n<li><em data-effect=\"italics\">p<\/em>-value = 0.0103<\/li>\n<li>Check student\u2019s solution.\n<ol id=\"fs-idm113459088\" type=\"i\">\n<li>alpha: 0.05<\/li>\n<li>Decision: Reject the null hypothesis.<\/li>\n<li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is less than 0.05.<\/li>\n<li>Conclusion: There is sufficient evidence to conclude that the mean lifespan of the tires is less than 50,000 miles.<\/li>\n<\/ol>\n<\/li>\n<li>(43,537, 49,463)<\/li>\n<\/ol>\n<p>3)<\/p>\n<ol id=\"fs-idm17421600\" type=\"a\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = \\$1.00<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2260 \\$1.00<\/li>\n<li>Let \\(\\overline{X}\\) = the average cost of a daily newspaper.<\/li>\n<li>normal distribution<\/li>\n<li><em data-effect=\"italics\">z<\/em> = \u20130.866<\/li>\n<li><em data-effect=\"italics\">p<\/em>-value = 0.3865<\/li>\n<li>Check student\u2019s solution.\n<ol id=\"fs-idm69728368\" type=\"i\">\n<li>Alpha: 0.01<\/li>\n<li>Decision: Do not reject the null hypothesis.<\/li>\n<li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is greater than 0.01.<\/li>\n<li>Conclusion: There is sufficient evidence to support the claim that the mean cost of daily papers is \\$1. The mean cost could be \\$1.<\/li>\n<\/ol>\n<\/li>\n<li>(\\$0.84, \\$1.06)<\/li>\n<\/ol>\n<p>5)<\/p>\n<ol id=\"fs-idm8405344\" type=\"a\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 10<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2260 10<\/li>\n<li>Let \\(\\overline{X}\\) the mean number of sick days an employee takes per year.<\/li>\n<li>Student\u2019s <em data-effect=\"italics\">t<\/em>-distribution<\/li>\n<li><em data-effect=\"italics\">t<\/em> = \u20131.12<\/li>\n<li><em data-effect=\"italics\">p<\/em>-value = 0.300<\/li>\n<li>Check student\u2019s solution.\n<ol id=\"fs-idm68554224\" type=\"i\">\n<li>Alpha: 0.05<\/li>\n<li>Decision: Do not reject the null hypothesis.<\/li>\n<li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is greater than 0.05.<\/li>\n<li>Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean number of sick days is not ten.<\/li>\n<\/ol>\n<\/li>\n<li>(4.9443, 11.806)<\/li>\n<\/ol>\n<p>7)<\/p>\n<ol id=\"fs-idm150879872\" type=\"a\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2265 0.6<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &lt; 0.6<\/li>\n<li>Let <em data-effect=\"italics\">P\u2032<\/em> = the proportion of students who feel more enriched as a result of taking Elementary Statistics.<\/li>\n<li>normal for a single proportion<\/li>\n<li>1.12<\/li>\n<li><em data-effect=\"italics\">p<\/em>-value = 0.1308<\/li>\n<li>Check student\u2019s solution.\n<ol id=\"fs-idp14435968\" type=\"i\">\n<li>Alpha: 0.05<\/li>\n<li>Decision: Do not reject the null hypothesis.<\/li>\n<li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is greater than 0.05.<\/li>\n<li>Conclusion: There is insufficient evidence to conclude that less than 60 percent of her students feel more enriched.<\/li>\n<\/ol>\n<\/li>\n<li>Confidence Interval: (0.409, 0.654)<span data-type=\"newline\"><br \/>\n<\/span>The \u201cplus-4s\u201d confidence interval is (0.411, 0.648)<\/li>\n<\/ol>\n<p>9)<\/p>\n<ol id=\"fs-idm153124688\" type=\"a\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> = 4<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2260 4<\/li>\n<li>Let \\(\\overline{X}\\) the average I.Q. of a set of brown trout.<\/li>\n<li>two-tailed Student&#8217;s t-test<\/li>\n<li><em data-effect=\"italics\">t<\/em> = 1.95<\/li>\n<li><em data-effect=\"italics\">p<\/em>-value = 0.076<\/li>\n<li>Check student\u2019s solution.\n<ol id=\"fs-idp21135088\" type=\"i\">\n<li>Alpha: 0.05<\/li>\n<li>Decision: Reject the null hypothesis.<\/li>\n<li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is greater than 0.05<\/li>\n<li>Conclusion: There is insufficient evidence to conclude that the average IQ of brown trout is not four.<\/li>\n<\/ol>\n<\/li>\n<li>(3.8865,5.9468)<\/li>\n<\/ol>\n<p>11)<\/p>\n<ol id=\"fs-idm3310256\" type=\"a\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> \u2265 0.13<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">p<\/em> &lt; 0.13<\/li>\n<li>Let <em data-effect=\"italics\">P\u2032<\/em> = the proportion of Americans who have seen or sensed angels<\/li>\n<li>normal for a single proportion<\/li>\n<li>\u20132.688<\/li>\n<li><em data-effect=\"italics\">p<\/em>-value = 0.0036<\/li>\n<li>Check student\u2019s solution.\n<ol id=\"fs-idm15778400\" type=\"i\">\n<li>alpha: 0.05<\/li>\n<li>Decision: Reject the null hypothesis.<\/li>\n<li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is less than 0.05.<\/li>\n<li>Conclusion: There is sufficient evidence to conclude that the percentage of Americans who have seen or sensed an angel is less than 13%.<\/li>\n<\/ol>\n<\/li>\n<li>(0, 0.0623).<span data-type=\"newline\"><br \/>\n<\/span> The\u201cplus-4s\u201d confidence interval is (0.0022, 0.0978)<\/li>\n<\/ol>\n<p>13)<\/p>\n<ol id=\"fs-idp13518640\" type=\"a\">\n<li><em data-effect=\"italics\">H<sub>0<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> \u2265 129<\/li>\n<li><em data-effect=\"italics\">H<sub>a<\/sub><\/em>: <em data-effect=\"italics\">\u03bc<\/em> &lt; 129<\/li>\n<li>Let \\(\\overline{X}\\) = the average time in seconds that Terri finishes Lap 4.<\/li>\n<li>Student&#8217;s <em data-effect=\"italics\">t<\/em>-distribution<\/li>\n<li><em data-effect=\"italics\">t<\/em> = 1.209<\/li>\n<li>0.8792<\/li>\n<li>Check student\u2019s solution.\n<ol id=\"fs-idm130980080\" type=\"i\">\n<li>Alpha: 0.05<\/li>\n<li>Decision: Do not reject the null hypothesis.<\/li>\n<li>Reason for decision: The <em data-effect=\"italics\">p<\/em>-value is greater than 0.05.<\/li>\n<li>Conclusion: There is insufficient evidence to conclude that Terri\u2019s mean lap time is less than 129 seconds.<\/li>\n<\/ol>\n<\/li>\n<li>(128.63, 130.37)<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-idp118682672\" class=\"footnotes\" data-depth=\"1\">\n<h3 data-type=\"title\">References<\/h3>\n<p>Data from Amit Schitai. Director of Instructional Technology and Distance Learning. LBCC.<\/p>\n<p>Data from <em data-effect=\"italics\">Bloomberg Businessweek<\/em>. Available online at http:\/\/www.businessweek.com\/news\/2011- 09-15\/nyc-smoking-rate-falls-to-record-low-of-14-bloomberg-says.html.<\/p>\n<p>Data from energy.gov. Available online at http:\/\/energy.gov (accessed June 27. 2013).<\/p>\n<p>Data from Gallup\u00ae. Available online at www.gallup.com (accessed June 27, 2013).<\/p>\n<p>Data from <em data-effect=\"italics\">Growing by Degrees<\/em> by Allen and Seaman.<\/p>\n<p>Data from La Leche League International. Available online at http:\/\/www.lalecheleague.org\/Law\/BAFeb01.html.<\/p>\n<p>Data from the American Automobile Association. Available online at www.aaa.com (accessed June 27, 2013).<\/p>\n<p>Data from the American Library Association. Available online at www.ala.org (accessed June 27, 2013).<\/p>\n<p>Data from the Bureau of Labor Statistics. Available online at http:\/\/www.bls.gov\/oes\/current\/oes291111.htm.<\/p>\n<p id=\"fs-idp1\">Data from the Centers for Disease Control and Prevention. Available online at www.cdc.gov (accessed June 27, 2013)<\/p>\n<p>Data from the U.S. Census Bureau, available online at http:\/\/quickfacts.census.gov\/qfd\/states\/00000.html (accessed June 27, 2013).<\/p>\n<p>Data from the United States Census Bureau. Available online at http:\/\/www.census.gov\/hhes\/socdemo\/language\/.<\/p>\n<p id=\"eip-79\">Data from Toastmasters International. Available online at http:\/\/toastmasters.org\/artisan\/detail.asp?CategoryID=1&amp;SubCategoryID=10&amp;ArticleID=429&amp;Page=1.<\/p>\n<p>Data from Weather Underground. Available online at www.wunderground.com (accessed June 27, 2013).<\/p>\n<p>Federal Bureau of Investigations. \u201cUniform Crime Reports and Index of Crime in Daviess in the State of Kentucky enforced by Daviess County from 1985 to 2005.\u201d Available online at http:\/\/www.disastercenter.com\/kentucky\/crime\/3868.htm (accessed June 27, 2013).<\/p>\n<p>\u201cFoothill-De Anza Community College District.\u201d De Anza College, Winter 2006. Available online at http:\/\/research.fhda.edu\/factbook\/DAdemofs\/Fact_sheet_da_2006w.pdf.<\/p>\n<p>Johansen, C., J. Boice, Jr., J. McLaughlin, J. Olsen. \u201cCellular Telephones and Cancer\u2014a Nationwide Cohort Study in Denmark.\u201d Institute of Cancer Epidemiology and the Danish Cancer Society, 93(3):203-7. Available online at http:\/\/www.ncbi.nlm.nih.gov\/pubmed\/11158188 (accessed June 27, 2013).<\/p>\n<p>Rape, Abuse &amp; Incest National Network. \u201cHow often does sexual assault occur?\u201d RAINN, 2009. Available online at http:\/\/www.rainn.org\/get-information\/statistics\/frequency-of-sexual-assault (accessed June 27, 2013).<\/p>\n<\/div>\n<div class=\"textbox shaded\" data-type=\"glossary\">\n<h3 data-type=\"glossary-title\">Glossary<\/h3>\n<dl>\n<dt>Central Limit Theorem<\/dt>\n<dd>Given a random variable (RV) with known mean \\(\\mu \\) and known standard deviation \u03c3. We are sampling with size <em data-effect=\"italics\">n<\/em> and we are interested in two new RVs &#8211; the sample mean, \\(\\overline{X}\\), and the sample sum, \\(\\Sigma X\\). If the size <em data-effect=\"italics\">n<\/em> of the sample is sufficiently large, then \\(\\overline{X}~N\\left(\\mu \\text{,}\\frac{\\sigma }{\\sqrt{n}}\\right)\\) and \\(\\Sigma X~N\\left(n\\mu ,\\sqrt{n}\\sigma \\right)\\). If the size <em data-effect=\"italics\">n<\/em> of the sample is sufficiently large, then the distribution of the sample means and the distribution of the sample sums will approximate a normal distribution regardless of the shape of the population. The mean of the sample means will equal the population mean and the mean of the sample sums will equal <em data-effect=\"italics\">n<\/em> times the population mean. The standard deviation of the distribution of the sample means, \\(\\frac{\\sigma }{\\sqrt{n}}\\), is called the standard error of the mean.<\/dd>\n<\/dl>\n<\/div>\n","protected":false},"author":32,"menu_order":63,"template":"","meta":{"pb_show_title":"","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-375","chapter","type-chapter","status-publish","hentry"],"part":343,"_links":{"self":[{"href":"https:\/\/pressbooks.ccconline.org\/accintrostats\/wp-json\/pressbooks\/v2\/chapters\/375","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.ccconline.org\/accintrostats\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.ccconline.org\/accintrostats\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.ccconline.org\/accintrostats\/wp-json\/wp\/v2\/users\/32"}],"version-history":[{"count":2,"href":"https:\/\/pressbooks.ccconline.org\/accintrostats\/wp-json\/pressbooks\/v2\/chapters\/375\/revisions"}],"predecessor-version":[{"id":689,"href":"https:\/\/pressbooks.ccconline.org\/accintrostats\/wp-json\/pressbooks\/v2\/chapters\/375\/revisions\/689"}],"part":[{"href":"https:\/\/pressbooks.ccconline.org\/accintrostats\/wp-json\/pressbooks\/v2\/parts\/343"}],"metadata":[{"href":"https:\/\/pressbooks.ccconline.org\/accintrostats\/wp-json\/pressbooks\/v2\/chapters\/375\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.ccconline.org\/accintrostats\/wp-json\/wp\/v2\/media?parent=375"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.ccconline.org\/accintrostats\/wp-json\/pressbooks\/v2\/chapter-type?post=375"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.ccconline.org\/accintrostats\/wp-json\/wp\/v2\/contributor?post=375"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.ccconline.org\/accintrostats\/wp-json\/wp\/v2\/license?post=375"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}