{"id":273,"date":"2021-07-14T15:59:04","date_gmt":"2021-07-14T15:59:04","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/a-population-proportion\/"},"modified":"2023-12-05T09:29:33","modified_gmt":"2023-12-05T09:29:33","slug":"a-population-proportion","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/a-population-proportion\/","title":{"raw":"Estimating a Population Proportion","rendered":"Estimating a Population Proportion"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<section>\r\n<ul id=\"list12315\">\r\n \t<li>Using the formula for creating a confidence interval or technology, construct a confidence interval for a population proportion<\/li>\r\n<\/ul>\r\n<\/section><\/div>\r\nDuring an election year, we see articles in the newspaper that state\u00a0<strong>confidence intervals<\/strong> 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\u00a0<strong>error bound<\/strong>, and the <strong>confidence level<\/strong> for a proportion is similar to that for the population mean, but the formulas are different.\r\n\r\n<iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7115058&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=3ReWri_jh3M&amp;video_target=tpm-plugin-h575qprm-3ReWri_jh3M\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe>\r\n\r\n<strong>How do you know you are dealing with a proportion problem? <\/strong>First, the underlying <strong>distribution is a binomial distribution<\/strong>. (There is no mention of a mean or average.) If <em>X<\/em> is a binomial random variable, then <em>X<\/em> ~ <em>B<\/em>(<em>n<\/em>, <em>p<\/em>) where <em>n\u00a0<\/em>is the number of trials and <em>p<\/em> is the probability of a success. To form a proportion, take <em>X<\/em>, the random variable for the number of successes and divide it by <em>n<\/em>, the number of trials (or the sample size). The random variable <em>P\u2032<\/em>(read \"P prime\") is that proportion,\r\n\r\n[latex]\\displaystyle{P'}=\\frac{{X}}{{n}}[\/latex]\r\n\r\n(Sometimes the random variable is denoted as\u00a0[latex]\\displaystyle\\hat{P}[\/latex], read \"P hat.\")\r\n\r\nWhen\u00a0<em>n<\/em> is large and <em>p<\/em> is not close to zero or one, we can use the <strong>normal distribution<\/strong> to approximate the binomial.\r\n\r\n[latex]\\displaystyle{X}[\/latex]~[latex]{N}{({n}{p},\\sqrt{{{n}{p}{q}}})}[\/latex]\r\n\r\nIf we divide the random variable, the mean, and the standard deviation by\u00a0<em>n<\/em>, we get a normal distribution of proportions with <em>P\u2032<\/em>, called the estimated proportion, as the random variable. (Recall that a proportion as the number of successes divided by <em>n<\/em>.)\r\n\r\n[latex]\\displaystyle\\frac{{X}}{{n}}={P'}{\\sim}{N}{(\\frac{{{n}{p}}}{{n}},\\frac{{\\sqrt{{{n}{p}{q}}}}}{{n}})}[\/latex]\r\n\r\nUsing algebra to simplify: [latex]\\displaystyle\\frac{{\\sqrt{{{n}{p}{q}}}}}{{n}}=\\sqrt{{\\frac{{{p}{q}}}{{n}}}}[\/latex]\r\n\r\n<strong><em>P\u2032<\/em> follows a normal distribution for proportions:\u00a0<\/strong>[latex]\\displaystyle\\frac{{X}}{{n}}={P'}{\\sim}{N}{(\\frac{{{n}{p}}}{{n}},\\frac{{\\sqrt{{{n}{p}{q}}}}}{{n}})}[\/latex]\r\n\r\nThe confidence interval has the form (<em>p\u2032<\/em> \u2013 <em>EBP<\/em>, <em>p\u2032<\/em> + <em>EBP<\/em>). <em>EBP<\/em> is error bound for the proportion.\r\n\r\n[latex]\\displaystyle{p'}=\\frac{{x}}{{n}}[\/latex]\r\n\r\n<em>p\u2032<\/em> = the <strong>estimated proportion<\/strong> of successes (<em>p\u2032<\/em> is a <strong>point estimate<\/strong> for <em>p<\/em>, the true proportion.)\r\n\r\n<em>x<\/em> = the <strong>number<\/strong> of successes\r\n\r\n<em>n<\/em> = the size of the sample\r\n\r\n<strong>The error bound for a proportion is<\/strong>\r\n\r\n<span class=\"s1\"><em>EBP<\/em> = [latex]\\displaystyle({z}_{\\frac{{\\alpha}}{{2}}})(\\sqrt{\\frac{{p'q'}}{{n}}})[\/latex] where q' = 1 - p'.<\/span>\r\n<p class=\"p1\">This 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\u00a0[latex]\\displaystyle\\frac{{\\sigma}}{{\\sqrt{n}}}[\/latex].\u00a0For a proportion, the appropriate standard deviation is\u00a0[latex]\\displaystyle\\sqrt{\\frac{{pq}}{{n}}}[\/latex].<\/p>\r\n<p class=\"p1\">However, in the error bound formula, we use\u00a0[latex]\\displaystyle\\sqrt{\\frac{{p'q'}}{{n}}}[\/latex]\u00a0as the standard deviation, instead of\u00a0[latex]\\displaystyle\\sqrt{\\frac{{pq}}{{n}}}[\/latex].<\/p>\r\nIn the error bound formula, <strong>the sample proportions <em data-redactor-tag=\"em\">p\u2032<\/em> and <em data-redactor-tag=\"em\">q\u2032<\/em> are estimates of the unknown population proportions <em data-redactor-tag=\"em\">p<\/em> and <em data-redactor-tag=\"em\">q<\/em>.<\/strong> The estimated proportions\u00a0<em data-redactor-tag=\"em\">p\u2032<\/em> and <em data-redactor-tag=\"em\">q\u2032<\/em> are used because <em data-redactor-tag=\"em\">p<\/em> and <em data-redactor-tag=\"em\">q<\/em> are not known. The sample proportions <em data-redactor-tag=\"em\">p\u2032 <\/em>and <em data-redactor-tag=\"em\">q\u2032<\/em> are calculated from the data: <em data-redactor-tag=\"em\">p\u2032<\/em> is the estimated proportion of successes, and <em data-redactor-tag=\"em\">q\u2032<\/em> is the estimated proportion of failures.\r\n\r\nThe confidence interval can be used only if the number of successes <em data-redactor-tag=\"em\">np\u2032<\/em> and the number of failures <em data-redactor-tag=\"em\">nq\u2032<\/em> are both greater than five.\r\n<h4>Note<\/h4>\r\nFor the normal distribution of proportions, the <em data-redactor-tag=\"em\">z<\/em>-score formula is as follows. If [latex]\\displaystyle{P'}{\\sim}{N}[\/latex](p,\u00a0[latex]\\displaystyle\\sqrt{\\frac{{pq}}{{n}}}[\/latex])\u00a0then the <em data-redactor-tag=\"em\">z<\/em>-score formula is z =\u00a0[latex]\\dfrac{p' - p}{\\sqrt{\\dfrac{pq}{n}}}[\/latex]\r\n<div class=\"textbox exercises\">\r\n<h3>Example 1<\/h3>\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&nbsp;\r\n\r\nThe first solution is step-by-step.\r\n\r\n[reveal-answer q=\"851998\"]Show Solution A[\/reveal-answer]\r\n[hidden-answer a=\"851998\"]\r\n\r\nLet <em data-redactor-tag=\"em\">X<\/em> = the number of people in the sample who have cell phones. <em data-redactor-tag=\"em\">X<\/em> is binomial. X ~ B(500, [latex]\\displaystyle\\frac{{421}}{{500}}[\/latex])\r\n\r\nTo calculate the confidence interval, you must find <em data-redactor-tag=\"em\">p\u2032<\/em>, <em data-redactor-tag=\"em\">q\u2032<\/em>, and\u00a0<em data-redactor-tag=\"em\">EBP<\/em>.\r\n\r\n<em data-redactor-tag=\"em\">n<\/em> = 500\r\n\r\n<em data-redactor-tag=\"em\">x<\/em> = the number of successes = 421\r\n\r\np' = \u00a0[latex]\\displaystyle\\frac{{x}}{{n}} =\\frac{{421}}{{500}}[\/latex] = 0.842\r\n\r\n<em data-redactor-tag=\"em\">p\u2032<\/em> = 0.842 is the sample proportion; this is the point estimate of the population proportion.\r\n\r\n<em data-redactor-tag=\"em\">q\u2032<\/em> = 1 \u2013 <em data-redactor-tag=\"em\">p\u2032<\/em> = 1 \u2013 0.842 = 0.158\r\n\r\nSince <em data-redactor-tag=\"em\">CL<\/em> = 0.95, then <em data-redactor-tag=\"em\">\u03b1<\/em> = 1 \u2013 <em data-redactor-tag=\"em\">CL<\/em> = 1 \u2013 0.95 = 0.05 [latex](\\dfrac{a}{2})[\/latex] = 0.025.\r\n\r\nThen\u00a0[latex]\\displaystyle{z}_{\\frac{{\\alpha}}{{2}}}={z}_{0.025}[\/latex] = 1.96\r\n\r\nUse the TI-83, 83+, or 84+ calculator command invNorm(0.975,0,1) to find <em data-redactor-tag=\"em\">z<sub data-redactor-tag=\"sub\">0.025<\/sub><\/em>. Remember that the area to the right of <em data-redactor-tag=\"em\">z<sub data-redactor-tag=\"sub\">0.025<\/sub><\/em> is 0.025 and the area to the left of <em data-redactor-tag=\"em\">z<sub data-redactor-tag=\"sub\">0.025<\/sub><\/em> is 0.975. This can also be found using appropriate commands on other calculators, using a computer, or using a Standard Normal probability table.\r\n\r\n<em>EBP<\/em> = [latex]\\displaystyle({z}_{\\frac{{\\alpha}}{{2}}})(\\sqrt{\\frac{{p'q'}}{{n}}})[\/latex] = (1.96 [latex]\\displaystyle\\sqrt{\\frac{{(0.842)(0.158)}}{{500}}}[\/latex] = 0.032\r\n\r\n<span id=\"MathJax-Span-45357\" class=\"mrow\"><span id=\"MathJax-Span-45358\" class=\"mrow\"><span id=\"MathJax-Span-45359\" class=\"mtable\"><span id=\"MathJax-Span-45406\" class=\"mtd\"><span id=\"MathJax-Span-45407\" class=\"mrow\"><span id=\"MathJax-Span-45408\" class=\"mi\">p<\/span><span id=\"MathJax-Span-45409\" class=\"mo\">'\u00a0<\/span><span id=\"MathJax-Span-45410\" class=\"mo\">\u2212\u00a0<\/span><em><span id=\"MathJax-Span-45411\" class=\"mi\">E<\/span><span id=\"MathJax-Span-45412\" class=\"mi\">B<\/span><span id=\"MathJax-Span-45413\" class=\"mi\">P\u00a0<\/span><\/em><span id=\"MathJax-Span-45414\" class=\"mo\">=\u00a0<\/span><span id=\"MathJax-Span-45415\" class=\"mn\">0.842\u00a0<\/span><span id=\"MathJax-Span-45416\" class=\"mo\">\u2212\u00a0<\/span><span id=\"MathJax-Span-45417\" class=\"mn\">0.032\u00a0<\/span><span id=\"MathJax-Span-45418\" class=\"mo\">=\u00a0<\/span><span id=\"MathJax-Span-45419\" class=\"mn\">0.81<\/span><\/span><\/span><\/span><\/span><\/span>\r\n\r\n<span class=\"mrow\"><span class=\"mtable\"><span id=\"MathJax-Span-45420\" class=\"mtd\"><span id=\"MathJax-Span-45421\" class=\"mrow\"><span id=\"MathJax-Span-45422\" class=\"mi\">p<\/span><span id=\"MathJax-Span-45423\" class=\"mo\">\u2032\u00a0<\/span><span id=\"MathJax-Span-45424\" class=\"mo\">+\u00a0<\/span><em><span id=\"MathJax-Span-45425\" class=\"mi\">E<\/span><span id=\"MathJax-Span-45426\" class=\"mi\">B<\/span><span id=\"MathJax-Span-45427\" class=\"mi\">P\u00a0<\/span><\/em><span id=\"MathJax-Span-45428\" class=\"mo\">=\u00a0<\/span><span id=\"MathJax-Span-45429\" class=\"mn\">0.842\u00a0<\/span><span id=\"MathJax-Span-45430\" class=\"mo\">+\u00a0<\/span><span id=\"MathJax-Span-45431\" class=\"mn\">0.032\u00a0<\/span><span id=\"MathJax-Span-45432\" class=\"mo\">=\u00a0<\/span><span id=\"MathJax-Span-45433\" class=\"mn\">0.874<\/span><\/span><\/span><\/span><\/span>\r\n\r\nThe confidence interval for the true binomial population proportion is (<em data-redactor-tag=\"em\">p\u2032<\/em> \u2013 <em data-redactor-tag=\"em\">EBP<\/em>, <em data-redactor-tag=\"em\">p\u2032<\/em> + <em data-redactor-tag=\"em\">EBP<\/em>) = (0.810, 0.874).\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\nThe second solution uses a function of the TI-83, 83+ or 84 calculators.\r\n[reveal-answer q=\"31628\"]Show Solution B[\/reveal-answer]\r\n[hidden-answer a=\"31628\"]\r\n\r\nPress <code data-redactor-tag=\"code\">STAT<\/code> and arrow over to\u00a0<code data-redactor-tag=\"code\">TESTS<\/code>.\r\n\r\nArrow down to <code data-redactor-tag=\"code\">A:1-PropZint<\/code>. Press <code data-redactor-tag=\"code\">ENTER<\/code>.\r\n\r\nArrow down to\u00a0<em>x<\/em> and enter 421.\r\n\r\nArrow down to <em>n\u00a0<\/em>and enter 500.\r\n\r\nArrow down to <code data-redactor-tag=\"code\">C-Level<\/code> and enter .95.\r\n\r\nArrow down to <code data-redactor-tag=\"code\">Calculate<\/code> and press <code data-redactor-tag=\"code\">ENTER<\/code>.\r\n\r\nThe confidence interval is (0.81003, 0.87397).\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\nInterpretation and explanation:\r\n[reveal-answer q=\"505886\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"505886\"]\r\n\r\n<strong>Interpretation<\/strong>\r\nWe estimate with 95% confidence that between 81% and 87.4% of all adult residents of this city have cell phones.\r\n\r\n<strong>Explanation of 95% Confidence Level<\/strong>\r\nNinety-five percent of the confidence intervals constructed in this way would contain the true value for the population proportion of all adult residents of this city who have cell phones.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it 1<\/h3>\r\nSuppose 250 randomly selected people are surveyed to determine if they own a tablet. Of the 250 surveyed, 98 reported owning a tablet. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of people who own tablets.\r\n[reveal-answer q=\"529236\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"529236\"]\r\n\r\n(0.3315, 0.4525)\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3><\/h3>\r\n<div class=\"textbox exercises\">\r\n<h3>Example 2<\/h3>\r\nFor a class project, a political science student at a large university wants to estimate the percent of students who are registered voters. He surveys 500 students and finds that 300 are registered voters. Compute a 90% confidence interval for the true percent of students who are registered voters, and interpret the confidence interval.\r\n\r\n&nbsp;\r\n\r\nThe first solution is step-by-step.\r\n[reveal-answer q=\"304347\"]Show Solution A[\/reveal-answer]\r\n[hidden-answer a=\"304347\"]\r\n\r\n<em data-redactor-tag=\"em\">x<\/em> = 300 and <em data-redactor-tag=\"em\">n<\/em> = 500\r\n\r\n<em>p'<\/em> = [latex]\\displaystyle\\frac{{x}}{{n}} = \\frac{{300}}{{500}}[\/latex] = 0.600\r\n\r\n<em>q'<\/em> = 1 - <em>p'<\/em> - 1 - 0.600 = 0.400\r\n\r\nSince <em data-redactor-tag=\"em\">CL<\/em> = 0.90, then <em data-redactor-tag=\"em\">\u03b1<\/em> = 1 \u2013 <em data-redactor-tag=\"em\">CL<\/em> = 1 \u2013 0.90 = 0.10[latex](\\dfrac{\\alpha}{2})[\/latex] = 0.05\r\n\r\n[latex]\\displaystyle{z}_{\\frac{{\\alpha}}{{2}}}[\/latex] =\u00a0[latex]\\displaystyle{z}_{0.05}[\/latex] = 1.645\r\n<p class=\"p1\">Use the TI-83, 83+, or 84+ calculator command invNorm(0.95,0,1) to find <em data-redactor-tag=\"em\">z<sub data-redactor-tag=\"sub\">0.05<\/sub><\/em>. Remember that the area to the right of <em data-redactor-tag=\"em\">z<sub data-redactor-tag=\"sub\">0.05<\/sub><\/em> is 0.05 and the area to the left of <em data-redactor-tag=\"em\">z<sub data-redactor-tag=\"sub\">0.05<\/sub><\/em> is 0.95. This can also be found using appropriate commands on other calculators, using a computer, or using a standard normal probability table.<\/p>\r\n<em>EBP<\/em> = [latex]\\displaystyle({z}_{\\frac{{\\alpha}}{{2}}})(\\sqrt{\\frac{{p'q'}}{{n}}})[\/latex] = (1.645)[latex]\\displaystyle\\sqrt{\\frac{{(0.60)(0.40)}}{{500}}}[\/latex] = 0.036\r\n\r\n<em>p'<\/em> - <em>EBP<\/em> = 0.60 - 0.036 = 0.564\r\n\r\n<em><span id=\"MathJax-Span-3826\" class=\"mrow\"><span id=\"MathJax-Span-3827\" class=\"semantics\"><span id=\"MathJax-Span-3828\" class=\"mrow\"><span id=\"MathJax-Span-3829\" class=\"mrow\"><span id=\"MathJax-Span-3833\" class=\"mo\">p' + EBP<\/span><\/span><\/span><\/span><\/span><\/em><span id=\"MathJax-Span-3826\" class=\"mrow\"><span id=\"MathJax-Span-3827\" class=\"semantics\"><span id=\"MathJax-Span-3828\" class=\"mrow\"><span id=\"MathJax-Span-3829\" class=\"mrow\"><span id=\"MathJax-Span-3833\" class=\"mo\">\u00a0<\/span><\/span><\/span><\/span><\/span><em><span id=\"MathJax-Span-3826\" class=\"mrow\"><span id=\"MathJax-Span-3827\" class=\"semantics\"><span id=\"MathJax-Span-3828\" class=\"mrow\"><span id=\"MathJax-Span-3829\" class=\"mrow\"><span id=\"MathJax-Span-3837\" class=\"mo\">=\u00a0<\/span><\/span><\/span><\/span><\/span><\/em><span id=\"MathJax-Span-3826\" class=\"mrow\"><span id=\"MathJax-Span-3827\" class=\"semantics\"><span id=\"MathJax-Span-3828\" class=\"mrow\"><span id=\"MathJax-Span-3829\" class=\"mrow\"><span id=\"MathJax-Span-3838\" class=\"mn\">0.60\u00a0<\/span><\/span><\/span><\/span><\/span><span id=\"MathJax-Span-3826\" class=\"mrow\"><span id=\"MathJax-Span-3827\" class=\"semantics\"><span id=\"MathJax-Span-3828\" class=\"mrow\"><span id=\"MathJax-Span-3829\" class=\"mrow\"><span id=\"MathJax-Span-3839\" class=\"mo\">+\u00a0<\/span><\/span><\/span><\/span><\/span><span id=\"MathJax-Span-3826\" class=\"mrow\"><span id=\"MathJax-Span-3827\" class=\"semantics\"><span id=\"MathJax-Span-3828\" class=\"mrow\"><span id=\"MathJax-Span-3829\" class=\"mrow\"><span id=\"MathJax-Span-3840\" class=\"mn\">0.036\u00a0<\/span><\/span><\/span><\/span><\/span><span id=\"MathJax-Span-3826\" class=\"mrow\"><span id=\"MathJax-Span-3827\" class=\"semantics\"><span id=\"MathJax-Span-3828\" class=\"mrow\"><span id=\"MathJax-Span-3829\" class=\"mrow\"><span id=\"MathJax-Span-3841\" class=\"mo\">=\u00a0<\/span><span id=\"MathJax-Span-3842\" class=\"mn\">0.636<\/span><\/span><\/span><\/span><\/span>\r\n\r\nThe confidence interval for the true binomial population proportion is (p\u2032 - EBP, p\u2032 + EBP) = (0.564,0.636).\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\nThe second solution uses a function of the TI-83, 83+, or 84 calculators.\r\n[reveal-answer q=\"879572\"]Show Solution B[\/reveal-answer]\r\n[hidden-answer a=\"879572\"]\r\n\r\nPress <code data-redactor-tag=\"code\">STAT<\/code> and arrow over to\u00a0<code data-redactor-tag=\"code\">TESTS<\/code>.\r\n\r\nArrow down to <code data-redactor-tag=\"code\">A:1-PropZint<\/code>. Press <code data-redactor-tag=\"code\">ENTER<\/code>.\r\n\r\nArrow down to <em>x\u00a0<\/em>and enter 300.\r\n\r\nArrow down to\u00a0<em>n<\/em> and enter 500.\r\n\r\nArrow down to <code data-redactor-tag=\"code\">C-Level<\/code> and enter 0.90.\r\n\r\nArrow down to <code data-redactor-tag=\"code\">Calculate<\/code> and press <code data-redactor-tag=\"code\">ENTER<\/code>.\r\n\r\nThe confidence interval is (0.564, 0.636).\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\nInterpretation and explanation:\r\n[reveal-answer q=\"784833\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"784833\"]\r\n<h4>Interpretation<\/h4>\r\n<ul>\r\n \t<li>We estimate with 90% confidence that the true percent of all students that are registered voters is between 56.4% and 63.6%.<\/li>\r\n \t<li>Alternate Wording: We estimate with 90% confidence that between 56.4% and 63.6% of ALL students are registered voters.<\/li>\r\n<\/ul>\r\n<h4>Explanation of 90% Confidence Level<\/h4>\r\nNinety percent of all confidence intervals constructed in this way contain the true value for the population percent of students that are registered voters.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it 2<\/h3>\r\nA student polls his school to see if students in the school district are for or against the new legislation regarding school uniforms. She surveys 600 students and finds that 480 are against the new legislation.\r\n\r\n1. Compute a 90% confidence interval for the true percent of students who are against the new legislation, and interpret the confidence interval.\r\n[reveal-answer q=\"161068\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"161068\"]\r\n\r\n(0.7731, 0.8269); We estimate with 90% confidence that the true percent of all students in the district who are against the new legislation is between 77.31% and 82.69%.\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n2. In a sample of 300 students, 68% said they own an iPod and a smart phone. Compute a 97% confidence interval for the true percent of students who own an iPod and a smartphone.\r\n\r\nThe first solution is step-by-step.\r\n[reveal-answer q=\"505992\"]Show Solution A[\/reveal-answer]\r\n[hidden-answer a=\"505992\"]\r\n<ul>\r\n \t<li>Sixty-eight percent (68%) of students own an iPod and a smart phone.\u00a0<span id=\"MathJax-Span-45554\" class=\"mi\">p<\/span><span id=\"MathJax-Span-45555\" class=\"mo\">\u2032<\/span><span id=\"MathJax-Span-45556\" class=\"mo\">=<\/span><span id=\"MathJax-Span-45557\" class=\"mn\">0.68, q<\/span><span id=\"MathJax-Span-45561\" class=\"mo\">\u2032<\/span><span id=\"MathJax-Span-45562\" class=\"mo\">=<\/span><span id=\"MathJax-Span-45563\" class=\"mn\">1<\/span><span id=\"MathJax-Span-45566\" class=\"mo\">\u2032<\/span><span id=\"MathJax-Span-45567\" class=\"mo\">=<\/span><span id=\"MathJax-Span-45568\" class=\"mn\">1<\/span><span id=\"MathJax-Span-45569\" class=\"mo\">\u2013<\/span><span id=\"MathJax-Span-45570\" class=\"mn\">0.68<\/span><span id=\"MathJax-Span-45571\" class=\"mo\">=<\/span><span id=\"MathJax-Span-45572\" class=\"mn\">0.32<\/span><\/li>\r\n \t<li>Since <em data-redactor-tag=\"em\">CL<\/em> = 0.97, we know <span id=\"MathJax-Element-1749-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-45573\" class=\"math\"><span id=\"MathJax-Span-45574\" class=\"mrow\"><span id=\"MathJax-Span-45575\" class=\"mo\">\u03b1<\/span><span id=\"MathJax-Span-45576\" class=\"mo\">=<\/span><span id=\"MathJax-Span-45577\" class=\"mn\">1<\/span><span id=\"MathJax-Span-45578\" class=\"mo\">\u2013<\/span><span id=\"MathJax-Span-45579\" class=\"mn\">0.97<\/span><span id=\"MathJax-Span-45580\" class=\"mo\">=<\/span><span id=\"MathJax-Span-45581\" class=\"mn\">0.03<\/span><\/span><\/span><\/span><\/li>\r\n \t<li>The area to the left of <em data-redactor-tag=\"em\">z<\/em><sub data-redactor-tag=\"sub\">0.015<\/sub> is 0.015, and the area to the right of <em data-redactor-tag=\"em\">z<\/em><sub data-redactor-tag=\"sub\">0.015<\/sub> is 1 \u2013 0.015 = 0.985.<\/li>\r\n \t<li>Using the TI 83, 83+, or 84+ calculator function InvNorm(.985,0,1), <em data-redactor-tag=\"em\">z<\/em><sub data-redactor-tag=\"sub\">0.015<\/sub> = 2.17<\/li>\r\n \t<li>EBP = [latex]\\displaystyle({z}_{\\frac{{\\alpha}}{{2}}})(\\sqrt{\\frac{{p'q'}}{{n}}})[\/latex] = (1.645)[latex]\\displaystyle\\sqrt{\\frac{{(0.68)(0.32)}}{{300}}}[\/latex] = 0.0269<\/li>\r\n<\/ul>\r\n<ul>\r\n \t<li>We are 97% confident that the true proportion of all students who own an iPod and a smart phone is between 0.6531 and 0.7069.<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n\r\nThe second solution uses a function of the TI-83, 83+, or 84 calculators.\r\n[reveal-answer q=\"575764\"]Show Solution B[\/reveal-answer]\r\n[hidden-answer a=\"575764\"]\r\n<ul>\r\n \t<li>Press STAT and arrow over to TESTS.<\/li>\r\n \t<li>Arrow down to A:1-PropZint.<\/li>\r\n \t<li>Press ENTER.<\/li>\r\n \t<li>Arrow down to x and enter 300*0.68.<\/li>\r\n \t<li>Arrow down to n and enter 300.<\/li>\r\n \t<li>Arrow down to C-Level and enter 0.97.<\/li>\r\n \t<li>Arrow down to Calculate and press ENTER.<\/li>\r\n \t<li>The confidence interval is (0.6531, 0.7069).<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<section>\n<ul id=\"list12315\">\n<li>Using the formula for creating a confidence interval or technology, construct a confidence interval for a population proportion<\/li>\n<\/ul>\n<\/section>\n<\/div>\n<p>During an election year, we see articles in the newspaper that state\u00a0<strong>confidence intervals<\/strong> 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).<\/p>\n<p>Investors 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.<\/p>\n<p>The procedure to find the confidence interval, the sample size, the\u00a0<strong>error bound<\/strong>, and the <strong>confidence level<\/strong> for a proportion is similar to that for the population mean, but the formulas are different.<\/p>\n<p><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=7115058&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=3ReWri_jh3M&amp;video_target=tpm-plugin-h575qprm-3ReWri_jh3M\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/p>\n<p><strong>How do you know you are dealing with a proportion problem? <\/strong>First, the underlying <strong>distribution is a binomial distribution<\/strong>. (There is no mention of a mean or average.) If <em>X<\/em> is a binomial random variable, then <em>X<\/em> ~ <em>B<\/em>(<em>n<\/em>, <em>p<\/em>) where <em>n\u00a0<\/em>is the number of trials and <em>p<\/em> is the probability of a success. To form a proportion, take <em>X<\/em>, the random variable for the number of successes and divide it by <em>n<\/em>, the number of trials (or the sample size). The random variable <em>P\u2032<\/em>(read &#8220;P prime&#8221;) is that proportion,<\/p>\n<p>[latex]\\displaystyle{P'}=\\frac{{X}}{{n}}[\/latex]<\/p>\n<p>(Sometimes the random variable is denoted as\u00a0[latex]\\displaystyle\\hat{P}[\/latex], read &#8220;P hat.&#8221;)<\/p>\n<p>When\u00a0<em>n<\/em> is large and <em>p<\/em> is not close to zero or one, we can use the <strong>normal distribution<\/strong> to approximate the binomial.<\/p>\n<p>[latex]\\displaystyle{X}[\/latex]~[latex]{N}{({n}{p},\\sqrt{{{n}{p}{q}}})}[\/latex]<\/p>\n<p>If we divide the random variable, the mean, and the standard deviation by\u00a0<em>n<\/em>, we get a normal distribution of proportions with <em>P\u2032<\/em>, called the estimated proportion, as the random variable. (Recall that a proportion as the number of successes divided by <em>n<\/em>.)<\/p>\n<p>[latex]\\displaystyle\\frac{{X}}{{n}}={P'}{\\sim}{N}{(\\frac{{{n}{p}}}{{n}},\\frac{{\\sqrt{{{n}{p}{q}}}}}{{n}})}[\/latex]<\/p>\n<p>Using algebra to simplify: [latex]\\displaystyle\\frac{{\\sqrt{{{n}{p}{q}}}}}{{n}}=\\sqrt{{\\frac{{{p}{q}}}{{n}}}}[\/latex]<\/p>\n<p><strong><em>P\u2032<\/em> follows a normal distribution for proportions:\u00a0<\/strong>[latex]\\displaystyle\\frac{{X}}{{n}}={P'}{\\sim}{N}{(\\frac{{{n}{p}}}{{n}},\\frac{{\\sqrt{{{n}{p}{q}}}}}{{n}})}[\/latex]<\/p>\n<p>The confidence interval has the form (<em>p\u2032<\/em> \u2013 <em>EBP<\/em>, <em>p\u2032<\/em> + <em>EBP<\/em>). <em>EBP<\/em> is error bound for the proportion.<\/p>\n<p>[latex]\\displaystyle{p'}=\\frac{{x}}{{n}}[\/latex]<\/p>\n<p><em>p\u2032<\/em> = the <strong>estimated proportion<\/strong> of successes (<em>p\u2032<\/em> is a <strong>point estimate<\/strong> for <em>p<\/em>, the true proportion.)<\/p>\n<p><em>x<\/em> = the <strong>number<\/strong> of successes<\/p>\n<p><em>n<\/em> = the size of the sample<\/p>\n<p><strong>The error bound for a proportion is<\/strong><\/p>\n<p><span class=\"s1\"><em>EBP<\/em> = [latex]\\displaystyle({z}_{\\frac{{\\alpha}}{{2}}})(\\sqrt{\\frac{{p'q'}}{{n}}})[\/latex] where q&#8217; = 1 &#8211; p&#8217;.<\/span><\/p>\n<p class=\"p1\">This formula is similar to the error bound formula for a mean, except that the &#8220;appropriate standard deviation&#8221; is different. For a mean, when the population standard deviation is known, the appropriate standard deviation that we use is\u00a0[latex]\\displaystyle\\frac{{\\sigma}}{{\\sqrt{n}}}[\/latex].\u00a0For a proportion, the appropriate standard deviation is\u00a0[latex]\\displaystyle\\sqrt{\\frac{{pq}}{{n}}}[\/latex].<\/p>\n<p class=\"p1\">However, in the error bound formula, we use\u00a0[latex]\\displaystyle\\sqrt{\\frac{{p'q'}}{{n}}}[\/latex]\u00a0as the standard deviation, instead of\u00a0[latex]\\displaystyle\\sqrt{\\frac{{pq}}{{n}}}[\/latex].<\/p>\n<p>In the error bound formula, <strong>the sample proportions <em data-redactor-tag=\"em\">p\u2032<\/em> and <em data-redactor-tag=\"em\">q\u2032<\/em> are estimates of the unknown population proportions <em data-redactor-tag=\"em\">p<\/em> and <em data-redactor-tag=\"em\">q<\/em>.<\/strong> The estimated proportions\u00a0<em data-redactor-tag=\"em\">p\u2032<\/em> and <em data-redactor-tag=\"em\">q\u2032<\/em> are used because <em data-redactor-tag=\"em\">p<\/em> and <em data-redactor-tag=\"em\">q<\/em> are not known. The sample proportions <em data-redactor-tag=\"em\">p\u2032 <\/em>and <em data-redactor-tag=\"em\">q\u2032<\/em> are calculated from the data: <em data-redactor-tag=\"em\">p\u2032<\/em> is the estimated proportion of successes, and <em data-redactor-tag=\"em\">q\u2032<\/em> is the estimated proportion of failures.<\/p>\n<p>The confidence interval can be used only if the number of successes <em data-redactor-tag=\"em\">np\u2032<\/em> and the number of failures <em data-redactor-tag=\"em\">nq\u2032<\/em> are both greater than five.<\/p>\n<h4>Note<\/h4>\n<p>For the normal distribution of proportions, the <em data-redactor-tag=\"em\">z<\/em>-score formula is as follows. If [latex]\\displaystyle{P'}{\\sim}{N}[\/latex](p,\u00a0[latex]\\displaystyle\\sqrt{\\frac{{pq}}{{n}}}[\/latex])\u00a0then the <em data-redactor-tag=\"em\">z<\/em>-score formula is z =\u00a0[latex]\\dfrac{p' - p}{\\sqrt{\\dfrac{pq}{n}}}[\/latex]<\/p>\n<div class=\"textbox exercises\">\n<h3>Example 1<\/h3>\n<p>Suppose 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 &#8211; 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.<\/p>\n<p>&nbsp;<\/p>\n<p>The first solution is step-by-step.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q851998\">Show Solution A<\/span><\/p>\n<div id=\"q851998\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let <em data-redactor-tag=\"em\">X<\/em> = the number of people in the sample who have cell phones. <em data-redactor-tag=\"em\">X<\/em> is binomial. X ~ B(500, [latex]\\displaystyle\\frac{{421}}{{500}}[\/latex])<\/p>\n<p>To calculate the confidence interval, you must find <em data-redactor-tag=\"em\">p\u2032<\/em>, <em data-redactor-tag=\"em\">q\u2032<\/em>, and\u00a0<em data-redactor-tag=\"em\">EBP<\/em>.<\/p>\n<p><em data-redactor-tag=\"em\">n<\/em> = 500<\/p>\n<p><em data-redactor-tag=\"em\">x<\/em> = the number of successes = 421<\/p>\n<p>p&#8217; = \u00a0[latex]\\displaystyle\\frac{{x}}{{n}} =\\frac{{421}}{{500}}[\/latex] = 0.842<\/p>\n<p><em data-redactor-tag=\"em\">p\u2032<\/em> = 0.842 is the sample proportion; this is the point estimate of the population proportion.<\/p>\n<p><em data-redactor-tag=\"em\">q\u2032<\/em> = 1 \u2013 <em data-redactor-tag=\"em\">p\u2032<\/em> = 1 \u2013 0.842 = 0.158<\/p>\n<p>Since <em data-redactor-tag=\"em\">CL<\/em> = 0.95, then <em data-redactor-tag=\"em\">\u03b1<\/em> = 1 \u2013 <em data-redactor-tag=\"em\">CL<\/em> = 1 \u2013 0.95 = 0.05 [latex](\\dfrac{a}{2})[\/latex] = 0.025.<\/p>\n<p>Then\u00a0[latex]\\displaystyle{z}_{\\frac{{\\alpha}}{{2}}}={z}_{0.025}[\/latex] = 1.96<\/p>\n<p>Use the TI-83, 83+, or 84+ calculator command invNorm(0.975,0,1) to find <em data-redactor-tag=\"em\">z<sub data-redactor-tag=\"sub\">0.025<\/sub><\/em>. Remember that the area to the right of <em data-redactor-tag=\"em\">z<sub data-redactor-tag=\"sub\">0.025<\/sub><\/em> is 0.025 and the area to the left of <em data-redactor-tag=\"em\">z<sub data-redactor-tag=\"sub\">0.025<\/sub><\/em> is 0.975. This can also be found using appropriate commands on other calculators, using a computer, or using a Standard Normal probability table.<\/p>\n<p><em>EBP<\/em> = [latex]\\displaystyle({z}_{\\frac{{\\alpha}}{{2}}})(\\sqrt{\\frac{{p'q'}}{{n}}})[\/latex] = (1.96 [latex]\\displaystyle\\sqrt{\\frac{{(0.842)(0.158)}}{{500}}}[\/latex] = 0.032<\/p>\n<p><span id=\"MathJax-Span-45357\" class=\"mrow\"><span id=\"MathJax-Span-45358\" class=\"mrow\"><span id=\"MathJax-Span-45359\" class=\"mtable\"><span id=\"MathJax-Span-45406\" class=\"mtd\"><span id=\"MathJax-Span-45407\" class=\"mrow\"><span id=\"MathJax-Span-45408\" class=\"mi\">p<\/span><span id=\"MathJax-Span-45409\" class=\"mo\">&#8216;\u00a0<\/span><span id=\"MathJax-Span-45410\" class=\"mo\">\u2212\u00a0<\/span><em><span id=\"MathJax-Span-45411\" class=\"mi\">E<\/span><span id=\"MathJax-Span-45412\" class=\"mi\">B<\/span><span id=\"MathJax-Span-45413\" class=\"mi\">P\u00a0<\/span><\/em><span id=\"MathJax-Span-45414\" class=\"mo\">=\u00a0<\/span><span id=\"MathJax-Span-45415\" class=\"mn\">0.842\u00a0<\/span><span id=\"MathJax-Span-45416\" class=\"mo\">\u2212\u00a0<\/span><span id=\"MathJax-Span-45417\" class=\"mn\">0.032\u00a0<\/span><span id=\"MathJax-Span-45418\" class=\"mo\">=\u00a0<\/span><span id=\"MathJax-Span-45419\" class=\"mn\">0.81<\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<p><span class=\"mrow\"><span class=\"mtable\"><span id=\"MathJax-Span-45420\" class=\"mtd\"><span id=\"MathJax-Span-45421\" class=\"mrow\"><span id=\"MathJax-Span-45422\" class=\"mi\">p<\/span><span id=\"MathJax-Span-45423\" class=\"mo\">\u2032\u00a0<\/span><span id=\"MathJax-Span-45424\" class=\"mo\">+\u00a0<\/span><em><span id=\"MathJax-Span-45425\" class=\"mi\">E<\/span><span id=\"MathJax-Span-45426\" class=\"mi\">B<\/span><span id=\"MathJax-Span-45427\" class=\"mi\">P\u00a0<\/span><\/em><span id=\"MathJax-Span-45428\" class=\"mo\">=\u00a0<\/span><span id=\"MathJax-Span-45429\" class=\"mn\">0.842\u00a0<\/span><span id=\"MathJax-Span-45430\" class=\"mo\">+\u00a0<\/span><span id=\"MathJax-Span-45431\" class=\"mn\">0.032\u00a0<\/span><span id=\"MathJax-Span-45432\" class=\"mo\">=\u00a0<\/span><span id=\"MathJax-Span-45433\" class=\"mn\">0.874<\/span><\/span><\/span><\/span><\/span><\/p>\n<p>The confidence interval for the true binomial population proportion is (<em data-redactor-tag=\"em\">p\u2032<\/em> \u2013 <em data-redactor-tag=\"em\">EBP<\/em>, <em data-redactor-tag=\"em\">p\u2032<\/em> + <em data-redactor-tag=\"em\">EBP<\/em>) = (0.810, 0.874).<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>The second solution uses a function of the TI-83, 83+ or 84 calculators.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q31628\">Show Solution B<\/span><\/p>\n<div id=\"q31628\" class=\"hidden-answer\" style=\"display: none\">\n<p>Press <code data-redactor-tag=\"code\">STAT<\/code> and arrow over to\u00a0<code data-redactor-tag=\"code\">TESTS<\/code>.<\/p>\n<p>Arrow down to <code data-redactor-tag=\"code\">A:1-PropZint<\/code>. Press <code data-redactor-tag=\"code\">ENTER<\/code>.<\/p>\n<p>Arrow down to\u00a0<em>x<\/em> and enter 421.<\/p>\n<p>Arrow down to <em>n\u00a0<\/em>and enter 500.<\/p>\n<p>Arrow down to <code data-redactor-tag=\"code\">C-Level<\/code> and enter .95.<\/p>\n<p>Arrow down to <code data-redactor-tag=\"code\">Calculate<\/code> and press <code data-redactor-tag=\"code\">ENTER<\/code>.<\/p>\n<p>The confidence interval is (0.81003, 0.87397).<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>Interpretation and explanation:<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q505886\">Show Answer<\/span><\/p>\n<div id=\"q505886\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong>Interpretation<\/strong><br \/>\nWe estimate with 95% confidence that between 81% and 87.4% of all adult residents of this city have cell phones.<\/p>\n<p><strong>Explanation of 95% Confidence Level<\/strong><br \/>\nNinety-five percent of the confidence intervals constructed in this way would contain the true value for the population proportion of all adult residents of this city who have cell phones.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it 1<\/h3>\n<p>Suppose 250 randomly selected people are surveyed to determine if they own a tablet. Of the 250 surveyed, 98 reported owning a tablet. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of people who own tablets.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q529236\">Show Answer<\/span><\/p>\n<div id=\"q529236\" class=\"hidden-answer\" style=\"display: none\">\n<p>(0.3315, 0.4525)<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3><\/h3>\n<div class=\"textbox exercises\">\n<h3>Example 2<\/h3>\n<p>For a class project, a political science student at a large university wants to estimate the percent of students who are registered voters. He surveys 500 students and finds that 300 are registered voters. Compute a 90% confidence interval for the true percent of students who are registered voters, and interpret the confidence interval.<\/p>\n<p>&nbsp;<\/p>\n<p>The first solution is step-by-step.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q304347\">Show Solution A<\/span><\/p>\n<div id=\"q304347\" class=\"hidden-answer\" style=\"display: none\">\n<p><em data-redactor-tag=\"em\">x<\/em> = 300 and <em data-redactor-tag=\"em\">n<\/em> = 500<\/p>\n<p><em>p&#8217;<\/em> = [latex]\\displaystyle\\frac{{x}}{{n}} = \\frac{{300}}{{500}}[\/latex] = 0.600<\/p>\n<p><em>q&#8217;<\/em> = 1 &#8211; <em>p&#8217;<\/em> &#8211; 1 &#8211; 0.600 = 0.400<\/p>\n<p>Since <em data-redactor-tag=\"em\">CL<\/em> = 0.90, then <em data-redactor-tag=\"em\">\u03b1<\/em> = 1 \u2013 <em data-redactor-tag=\"em\">CL<\/em> = 1 \u2013 0.90 = 0.10[latex](\\dfrac{\\alpha}{2})[\/latex] = 0.05<\/p>\n<p>[latex]\\displaystyle{z}_{\\frac{{\\alpha}}{{2}}}[\/latex] =\u00a0[latex]\\displaystyle{z}_{0.05}[\/latex] = 1.645<\/p>\n<p class=\"p1\">Use the TI-83, 83+, or 84+ calculator command invNorm(0.95,0,1) to find <em data-redactor-tag=\"em\">z<sub data-redactor-tag=\"sub\">0.05<\/sub><\/em>. Remember that the area to the right of <em data-redactor-tag=\"em\">z<sub data-redactor-tag=\"sub\">0.05<\/sub><\/em> is 0.05 and the area to the left of <em data-redactor-tag=\"em\">z<sub data-redactor-tag=\"sub\">0.05<\/sub><\/em> is 0.95. This can also be found using appropriate commands on other calculators, using a computer, or using a standard normal probability table.<\/p>\n<p><em>EBP<\/em> = [latex]\\displaystyle({z}_{\\frac{{\\alpha}}{{2}}})(\\sqrt{\\frac{{p'q'}}{{n}}})[\/latex] = (1.645)[latex]\\displaystyle\\sqrt{\\frac{{(0.60)(0.40)}}{{500}}}[\/latex] = 0.036<\/p>\n<p><em>p&#8217;<\/em> &#8211; <em>EBP<\/em> = 0.60 &#8211; 0.036 = 0.564<\/p>\n<p><em><span id=\"MathJax-Span-3826\" class=\"mrow\"><span id=\"MathJax-Span-3827\" class=\"semantics\"><span id=\"MathJax-Span-3828\" class=\"mrow\"><span id=\"MathJax-Span-3829\" class=\"mrow\"><span id=\"MathJax-Span-3833\" class=\"mo\">p&#8217; + EBP<\/span><\/span><\/span><\/span><\/span><\/em><span id=\"MathJax-Span-3826\" class=\"mrow\"><span id=\"MathJax-Span-3827\" class=\"semantics\"><span id=\"MathJax-Span-3828\" class=\"mrow\"><span id=\"MathJax-Span-3829\" class=\"mrow\"><span id=\"MathJax-Span-3833\" class=\"mo\">\u00a0<\/span><\/span><\/span><\/span><\/span><em><span id=\"MathJax-Span-3826\" class=\"mrow\"><span id=\"MathJax-Span-3827\" class=\"semantics\"><span id=\"MathJax-Span-3828\" class=\"mrow\"><span id=\"MathJax-Span-3829\" class=\"mrow\"><span id=\"MathJax-Span-3837\" class=\"mo\">=\u00a0<\/span><\/span><\/span><\/span><\/span><\/em><span id=\"MathJax-Span-3826\" class=\"mrow\"><span id=\"MathJax-Span-3827\" class=\"semantics\"><span id=\"MathJax-Span-3828\" class=\"mrow\"><span id=\"MathJax-Span-3829\" class=\"mrow\"><span id=\"MathJax-Span-3838\" class=\"mn\">0.60\u00a0<\/span><\/span><\/span><\/span><\/span><span id=\"MathJax-Span-3826\" class=\"mrow\"><span id=\"MathJax-Span-3827\" class=\"semantics\"><span id=\"MathJax-Span-3828\" class=\"mrow\"><span id=\"MathJax-Span-3829\" class=\"mrow\"><span id=\"MathJax-Span-3839\" class=\"mo\">+\u00a0<\/span><\/span><\/span><\/span><\/span><span id=\"MathJax-Span-3826\" class=\"mrow\"><span id=\"MathJax-Span-3827\" class=\"semantics\"><span id=\"MathJax-Span-3828\" class=\"mrow\"><span id=\"MathJax-Span-3829\" class=\"mrow\"><span id=\"MathJax-Span-3840\" class=\"mn\">0.036\u00a0<\/span><\/span><\/span><\/span><\/span><span id=\"MathJax-Span-3826\" class=\"mrow\"><span id=\"MathJax-Span-3827\" class=\"semantics\"><span id=\"MathJax-Span-3828\" class=\"mrow\"><span id=\"MathJax-Span-3829\" class=\"mrow\"><span id=\"MathJax-Span-3841\" class=\"mo\">=\u00a0<\/span><span id=\"MathJax-Span-3842\" class=\"mn\">0.636<\/span><\/span><\/span><\/span><\/span><\/p>\n<p>The confidence interval for the true binomial population proportion is (p\u2032 &#8211; EBP, p\u2032 + EBP) = (0.564,0.636).<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>The second solution uses a function of the TI-83, 83+, or 84 calculators.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q879572\">Show Solution B<\/span><\/p>\n<div id=\"q879572\" class=\"hidden-answer\" style=\"display: none\">\n<p>Press <code data-redactor-tag=\"code\">STAT<\/code> and arrow over to\u00a0<code data-redactor-tag=\"code\">TESTS<\/code>.<\/p>\n<p>Arrow down to <code data-redactor-tag=\"code\">A:1-PropZint<\/code>. Press <code data-redactor-tag=\"code\">ENTER<\/code>.<\/p>\n<p>Arrow down to <em>x\u00a0<\/em>and enter 300.<\/p>\n<p>Arrow down to\u00a0<em>n<\/em> and enter 500.<\/p>\n<p>Arrow down to <code data-redactor-tag=\"code\">C-Level<\/code> and enter 0.90.<\/p>\n<p>Arrow down to <code data-redactor-tag=\"code\">Calculate<\/code> and press <code data-redactor-tag=\"code\">ENTER<\/code>.<\/p>\n<p>The confidence interval is (0.564, 0.636).<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>Interpretation and explanation:<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q784833\">Show Answer<\/span><\/p>\n<div id=\"q784833\" class=\"hidden-answer\" style=\"display: none\">\n<h4>Interpretation<\/h4>\n<ul>\n<li>We estimate with 90% confidence that the true percent of all students that are registered voters is between 56.4% and 63.6%.<\/li>\n<li>Alternate Wording: We estimate with 90% confidence that between 56.4% and 63.6% of ALL students are registered voters.<\/li>\n<\/ul>\n<h4>Explanation of 90% Confidence Level<\/h4>\n<p>Ninety percent of all confidence intervals constructed in this way contain the true value for the population percent of students that are registered voters.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it 2<\/h3>\n<p>A student polls his school to see if students in the school district are for or against the new legislation regarding school uniforms. She surveys 600 students and finds that 480 are against the new legislation.<\/p>\n<p>1. Compute a 90% confidence interval for the true percent of students who are against the new legislation, and interpret the confidence interval.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q161068\">Show Answer<\/span><\/p>\n<div id=\"q161068\" class=\"hidden-answer\" style=\"display: none\">\n<p>(0.7731, 0.8269); We estimate with 90% confidence that the true percent of all students in the district who are against the new legislation is between 77.31% and 82.69%.<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>2. In a sample of 300 students, 68% said they own an iPod and a smart phone. Compute a 97% confidence interval for the true percent of students who own an iPod and a smartphone.<\/p>\n<p>The first solution is step-by-step.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q505992\">Show Solution A<\/span><\/p>\n<div id=\"q505992\" class=\"hidden-answer\" style=\"display: none\">\n<ul>\n<li>Sixty-eight percent (68%) of students own an iPod and a smart phone.\u00a0<span id=\"MathJax-Span-45554\" class=\"mi\">p<\/span><span id=\"MathJax-Span-45555\" class=\"mo\">\u2032<\/span><span id=\"MathJax-Span-45556\" class=\"mo\">=<\/span><span id=\"MathJax-Span-45557\" class=\"mn\">0.68, q<\/span><span id=\"MathJax-Span-45561\" class=\"mo\">\u2032<\/span><span id=\"MathJax-Span-45562\" class=\"mo\">=<\/span><span id=\"MathJax-Span-45563\" class=\"mn\">1<\/span><span id=\"MathJax-Span-45566\" class=\"mo\">\u2032<\/span><span id=\"MathJax-Span-45567\" class=\"mo\">=<\/span><span id=\"MathJax-Span-45568\" class=\"mn\">1<\/span><span id=\"MathJax-Span-45569\" class=\"mo\">\u2013<\/span><span id=\"MathJax-Span-45570\" class=\"mn\">0.68<\/span><span id=\"MathJax-Span-45571\" class=\"mo\">=<\/span><span id=\"MathJax-Span-45572\" class=\"mn\">0.32<\/span><\/li>\n<li>Since <em data-redactor-tag=\"em\">CL<\/em> = 0.97, we know <span id=\"MathJax-Element-1749-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-45573\" class=\"math\"><span id=\"MathJax-Span-45574\" class=\"mrow\"><span id=\"MathJax-Span-45575\" class=\"mo\">\u03b1<\/span><span id=\"MathJax-Span-45576\" class=\"mo\">=<\/span><span id=\"MathJax-Span-45577\" class=\"mn\">1<\/span><span id=\"MathJax-Span-45578\" class=\"mo\">\u2013<\/span><span id=\"MathJax-Span-45579\" class=\"mn\">0.97<\/span><span id=\"MathJax-Span-45580\" class=\"mo\">=<\/span><span id=\"MathJax-Span-45581\" class=\"mn\">0.03<\/span><\/span><\/span><\/span><\/li>\n<li>The area to the left of <em data-redactor-tag=\"em\">z<\/em><sub data-redactor-tag=\"sub\">0.015<\/sub> is 0.015, and the area to the right of <em data-redactor-tag=\"em\">z<\/em><sub data-redactor-tag=\"sub\">0.015<\/sub> is 1 \u2013 0.015 = 0.985.<\/li>\n<li>Using the TI 83, 83+, or 84+ calculator function InvNorm(.985,0,1), <em data-redactor-tag=\"em\">z<\/em><sub data-redactor-tag=\"sub\">0.015<\/sub> = 2.17<\/li>\n<li>EBP = [latex]\\displaystyle({z}_{\\frac{{\\alpha}}{{2}}})(\\sqrt{\\frac{{p'q'}}{{n}}})[\/latex] = (1.645)[latex]\\displaystyle\\sqrt{\\frac{{(0.68)(0.32)}}{{300}}}[\/latex] = 0.0269<\/li>\n<\/ul>\n<ul>\n<li>We are 97% confident that the true proportion of all students who own an iPod and a smart phone is between 0.6531 and 0.7069.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<p>The second solution uses a function of the TI-83, 83+, or 84 calculators.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q575764\">Show Solution B<\/span><\/p>\n<div id=\"q575764\" class=\"hidden-answer\" style=\"display: none\">\n<ul>\n<li>Press STAT and arrow over to TESTS.<\/li>\n<li>Arrow down to A:1-PropZint.<\/li>\n<li>Press ENTER.<\/li>\n<li>Arrow down to x and enter 300*0.68.<\/li>\n<li>Arrow down to n and enter 300.<\/li>\n<li>Arrow down to C-Level and enter 0.97.<\/li>\n<li>Arrow down to Calculate and press ENTER.<\/li>\n<li>The confidence interval is (0.6531, 0.7069).<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-273\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>A Population Proportion. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/introductory-statistics\/pages\/8-3-a-population-proportion\">https:\/\/openstax.org\/books\/introductory-statistics\/pages\/8-3-a-population-proportion<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction<\/li><li>Introductory Statistics. <strong>Authored by<\/strong>: Barbara Illowsky, Susan Dean. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction\">https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">All rights reserved content<\/div><ul class=\"citation-list\"><li>Confidence Intervals for Population Proportions. <strong>Authored by<\/strong>: StatisticsLectures.com. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/3ReWri_jh3M\">https:\/\/youtu.be\/3ReWri_jh3M<\/a>. <strong>License<\/strong>: <em>All Rights Reserved<\/em>. <strong>License Terms<\/strong>: Standard YouTube License<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":169134,"menu_order":17,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"A Population Proportion\",\"author\":\"\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/introductory-statistics\/pages\/8-3-a-population-proportion\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction\"},{\"type\":\"copyrighted_video\",\"description\":\"Confidence Intervals for Population Proportions\",\"author\":\"StatisticsLectures.com\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/3ReWri_jh3M\",\"project\":\"\",\"license\":\"arr\",\"license_terms\":\"Standard YouTube License\"},{\"type\":\"cc\",\"description\":\"Introductory Statistics\",\"author\":\"Barbara Illowsky, Susan Dean\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-273","chapter","type-chapter","status-publish","hentry"],"part":269,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapters\/273","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/wp\/v2\/users\/169134"}],"version-history":[{"count":51,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapters\/273\/revisions"}],"predecessor-version":[{"id":3778,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapters\/273\/revisions\/3778"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/parts\/269"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapters\/273\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/wp\/v2\/media?parent=273"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapter-type?post=273"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/wp\/v2\/contributor?post=273"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/wp\/v2\/license?post=273"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}