{"id":333,"date":"2022-05-18T16:38:42","date_gmt":"2022-05-18T16:38:42","guid":{"rendered":"https:\/\/pressbooks.ccconline.org\/accintrostats\/chapter\/a-population-proportion\/"},"modified":"2022-11-09T16:21:23","modified_gmt":"2022-11-09T16:21:23","slug":"a-population-proportion","status":"publish","type":"chapter","link":"https:\/\/pressbooks.ccconline.org\/accintrostats\/chapter\/a-population-proportion\/","title":{"raw":"Chapter 9.2: A Population Proportion","rendered":"Chapter 9.2: A Population Proportion"},"content":{"raw":" \r\n\r\nDuring an election year, we see articles in the newspaper that state confidence intervals<\/span> in terms of proportions or percentages. For example, a poll for a particular candidate running for president might show that the candidate has 40% of the vote within three percentage points (if the sample is large enough). Often, election polls are calculated with 95% confidence, so, the pollsters would be 95% confident that the true proportion of voters who favored the candidate would be between 0.37 and 0.43: (0.40 \u2013 0.03,0.40 + 0.03).\r\n\r\nInvestors in the stock market are interested in the true proportion of stocks that go up and down each week. Businesses that sell personal computers are interested in the proportion of households in the United States that own personal computers. Confidence intervals can be calculated for the true proportion of stocks that go up or down each week and for the true proportion of households in the United States that own personal computers.\r\n\r\nThe procedure to find the confidence interval, the sample size, the error bound<\/span>, and the confidence level<\/span> for a proportion is similar to that for the population mean, but the formulas are different.\r\n

How do you know you are dealing with a proportion problem?<\/strong> First, the underlying distribution is a<\/strong> binomial distribution<\/span>. (There is no mention of a mean or average.) If X<\/em> is a binomial random variable, then X<\/em> ~ B<\/em>(n<\/em>, p<\/em>) where n<\/em> is the number of trials and p<\/em> is the probability of a success. To form a proportion, take X<\/em>, the random variable for the number of successes and divide it by n<\/em>, the number of trials (or the sample size). The random variable P\u2032<\/em> (read \"P prime\") is that proportion,<\/p>\r\n\\({P}^{\\prime }=\\frac{X}{n}\\)\r\n\r\n(Sometimes the random variable is denoted as \\(\\stackrel{^}{P}\\), read \"P hat\".)\r\n\r\nWhen n<\/em> is large and p<\/em> is not close to zero or one, we can use the normal distribution<\/span> to approximate the binomial.\r\n\r\n\\(X~N\\left(np,\\sqrt{npq}\\right)\\)\r\n\r\nIf we divide the random variable, the mean, and the standard deviation by n<\/em>, we get a normal distribution of proportions with P\u2032<\/em>, called the estimated proportion, as the random variable. (Recall that a proportion as the number of successes divided by n<\/em>.)\r\n

\\(\\frac{X}{n}={P}^{\\prime }\\text{~\u00a0}N\\left(\\frac{np}{n},\\frac{\\sqrt{npq}}{n}\\right)\\)<\/p>\r\nUsing algebra to simplify : \\(\\frac{\\sqrt{npq}}{n}=\\sqrt{\\frac{pq}{n}}\\)\r\n\r\nP\u2032<\/em> follows a normal distribution for proportions<\/strong>: \\(\\frac{X}{n}={P}^{\\prime }\\text{~\u00a0}N\\left(\\frac{np}{n},\\frac{\\sqrt{npq}}{n}\\right)\\)\r\n\r\nThe confidence interval has the form (p\u2032<\/em> \u2013 EBP<\/em>, p\u2032<\/em> + EBP<\/em>). EBP<\/em> is error bound for the proportion.\r\n\r\np\u2032<\/em> = \\(\\frac{x}{n}\\)\r\n\r\np\u2032<\/em> = the estimated proportion<\/strong> of successes (p\u2032<\/em> is a point estimate<\/strong> for p<\/em>, the true proportion.)\r\n\r\nx<\/em> = the number<\/strong> of successes\r\n\r\nn<\/em> = the size of the sample\r\n\r\nThe error bound for a proportion is<\/strong>\r\n

\\(EBP=\\left({z}_{\\frac{\\alpha }{2}}\\right)\\left(\\sqrt{\\frac{{p}^{\\prime }{q}^{\\prime }}{n}}\\right)\\) where q\u2032<\/em> = 1 \u2013 p\u2032<\/em><\/p>\r\nThis formula is similar to the error bound formula for a mean, except that the \"appropriate standard deviation\" is different. For a mean, when the population standard deviation is known, the appropriate standard deviation that we use is \\(\\frac{\\sigma }{\\sqrt{n}}\\). For a proportion, the appropriate standard deviation is \\(\\sqrt{\\frac{pq}{n}}\\).\r\n\r\nHowever, in the error bound formula, we use \\(\\sqrt{\\frac{{p}^{\\prime }{q}^{\\prime }}{n}}\\) as the standard deviation, instead of \\(\\sqrt{\\frac{pq}{n}}\\).\r\n\r\nIn the error bound formula, the sample proportions p\u2032<\/em> and q\u2032<\/em> are estimates of the unknown population proportions p<\/em> and q<\/em><\/strong>. The estimated proportions p\u2032<\/em> and q\u2032<\/em> are used because p<\/em> and q<\/em> are not known. The sample proportions p\u2032<\/em> and q\u2032<\/em> are calculated from the data: p\u2032<\/em> is the estimated proportion of successes, and q\u2032<\/em> is the estimated proportion of failures.\r\n\r\nThe confidence interval can be used only if the number of successes np\u2032<\/em> and the number of failures nq\u2032<\/em> are both greater than five.\r\n

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Note<\/div>\r\n

For the normal distribution of proportions, the z<\/em>-score formula is as follows.<\/p>\r\nIf \\({P}^{\\prime }\\text{~}N\\left(p,\\sqrt{\\frac{pq}{n}}\\right)\\) then the z<\/em>-score formula is \\(z=\\frac{{p}^{\\prime }-p}{\\sqrt{\\frac{pq}{n}}}\\)\r\n\r\n<\/div>\r\n

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\r\n\r\nSuppose that a market research firm is hired to estimate the percent of adults living in a large city who have cell phones. Five hundred randomly selected adult residents in this city are surveyed to determine whether they have cell phones. Of the 500 people surveyed, 421 responded yes - they own cell phones. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of adult residents of this city who have cell phones.\r\n\r\n<\/div>\r\n
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Solution A<\/div>\r\n