{"id":1176,"date":"2023-06-22T01:59:42","date_gmt":"2023-06-22T01:59:42","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/introstatstest\/chapter\/confidence-interval-for-proportions-dig-deeper\/"},"modified":"2025-05-16T03:12:34","modified_gmt":"2025-05-16T03:12:34","slug":"confidence-interval-for-proportions-dig-deeper","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/introstatstest\/chapter\/confidence-interval-for-proportions-dig-deeper\/","title":{"raw":"Confidence Interval for Proportions: Fresh Take","rendered":"Confidence Interval for Proportions: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li class=\"li1\">Check the conditions for creating a confidence interval for population proportion.<\/li>\r\n\t<li class=\"li1\">Describe the connection between the confidence level and the confidence interval.<\/li>\r\n\t<li class=\"li1\">Calculate a confidence interval for a population proportion.<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Normal Approximation to the Binomial<\/h2>\r\n<p>The condition for creating a confidence interval for population proportion uses binomial probabilities within its central limit theorem.<\/p>\r\n<p>To compute the normal approximation to the binomial distribution, take a simple random sample from a population. You must meet these conditions for a <strong data-redactor-tag=\"strong\">binomial distribution<\/strong>:<\/p>\r\n<ul>\r\n\t<li>There are a certain number [latex]n[\/latex] of independent trials<\/li>\r\n\t<li>The outcomes of any trial are success or failure<\/li>\r\n\t<li>Each trial has the same probability of a success [latex]p[\/latex]<\/li>\r\n<\/ul>\r\n<p>The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities [latex]np[\/latex] and [latex]n(1-p)[\/latex] must both be greater than [latex]10[\/latex].<\/p>\r\n<p>The central limit theorem (CLT) is not for calculating probabilities involving an individual value. As the sample size gets larger, the mean of the sample means approaches the population mean. This is due to the law of large numbers.<\/p>\r\n<section class=\"textbox example\">Suppose [latex]250[\/latex] randomly selected people are surveyed to determine if they own a tablet. Of the [latex]250[\/latex] surveyed, [latex]98[\/latex] reported owning a tablet. Is the shape of the sampling distribution of the sample proportion approximately normal?[reveal-answer q=\"147724\"]Show answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"147724\"]From the question, we know that [latex]n=250[\/latex] and [latex]p=\\dfrac{98}{250}[\/latex].[latex]np=250 \\times (\\dfrac{98}{250})=98 &gt; 10[\/latex]\r\n\r\n<p>[latex]n(1-p)=250 \\times (1-\\dfrac{98}{250})=250 \\times (\\dfrac{152}{250})=152 &gt; 10[\/latex]<\/p>\r\n<p>Therefore, the shape of the sampling distribution of the sample proportion is approximately normal.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox recall\"><strong><strong>Sampling Distribution of the Sample Proportion<\/strong><\/strong><br \/>\r\n<br \/>\r\nWhen taking random samples of size [latex]n[\/latex] from a population distribution with proportion [latex]p[\/latex]:\r\n\r\n<ul>\r\n\t<li>The estimated <strong>mean<\/strong> of the distribution of sample proportions is [latex]\\hat{p}[\/latex].<\/li>\r\n\t<li>The estimated <strong>standard deviation<\/strong> of the distribution of sample proportions is called the <strong>standard error<\/strong>:[latex] SE = \\sqrt{\\dfrac{\\hat{p}(1-\\hat{p})}{n}} [\/latex].<\/li>\r\n\t<li>If [latex]n\\hat{p} \\geq 10[\/latex] and [latex]n(1-\\hat{p}) \\geq 10[\/latex], then the <strong>Central Limit Theorem<\/strong> (CLT) states that the distribution of the sample proportions follows an approximate normal distribution.<\/li>\r\n<\/ul>\r\n<p><strong>The confidence interval for a population proportion<\/strong> is:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\hat{p} \\pm z^{*} \\cdot\\sqrt{\\frac{\\hat{p}(1-\\hat{p})}{n}} [\/latex]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">Suppose [latex]250[\/latex] randomly selected people are surveyed to determine if they own a tablet. Of the [latex]250[\/latex] surveyed, [latex]98[\/latex] reported owning a tablet. Using a [latex]95\\%[\/latex] confidence level, compute and interpret a confidence interval estimate for the true proportion of people who own tablets.[reveal-answer q=\"886455\"]Show answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"886455\"]Recall that [latex]z^{*}[\/latex] for a [latex]95\\%[\/latex] confidence interval is [latex]1.96[\/latex].\r\n\r\n<p>The [latex]95\\%[\/latex] CI: [latex]\\dfrac{98}{250} \\pm 1.96 \\cdot \\sqrt{\\dfrac{\\dfrac{98}{250}(1-\\dfrac{98}{250})}{250}} = (0.3315, 0.4525)[\/latex].<\/p>\r\n<p>Interpretation: Based on the sample, we are [latex]95\\%[\/latex] confident that the true proportion of people who own tablets is between [latex]0.3315[\/latex] and [latex]0.4525[\/latex].<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li class=\"li1\">Check the conditions for creating a confidence interval for population proportion.<\/li>\n<li class=\"li1\">Describe the connection between the confidence level and the confidence interval.<\/li>\n<li class=\"li1\">Calculate a confidence interval for a population proportion.<\/li>\n<\/ul>\n<\/section>\n<h2>Normal Approximation to the Binomial<\/h2>\n<p>The condition for creating a confidence interval for population proportion uses binomial probabilities within its central limit theorem.<\/p>\n<p>To compute the normal approximation to the binomial distribution, take a simple random sample from a population. You must meet these conditions for a <strong data-redactor-tag=\"strong\">binomial distribution<\/strong>:<\/p>\n<ul>\n<li>There are a certain number [latex]n[\/latex] of independent trials<\/li>\n<li>The outcomes of any trial are success or failure<\/li>\n<li>Each trial has the same probability of a success [latex]p[\/latex]<\/li>\n<\/ul>\n<p>The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities [latex]np[\/latex] and [latex]n(1-p)[\/latex] must both be greater than [latex]10[\/latex].<\/p>\n<p>The central limit theorem (CLT) is not for calculating probabilities involving an individual value. As the sample size gets larger, the mean of the sample means approaches the population mean. This is due to the law of large numbers.<\/p>\n<section class=\"textbox example\">Suppose [latex]250[\/latex] randomly selected people are surveyed to determine if they own a tablet. Of the [latex]250[\/latex] surveyed, [latex]98[\/latex] reported owning a tablet. Is the shape of the sampling distribution of the sample proportion approximately normal?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q147724\">Show answer<\/button><\/p>\n<div id=\"q147724\" class=\"hidden-answer\" style=\"display: none\">From the question, we know that [latex]n=250[\/latex] and [latex]p=\\dfrac{98}{250}[\/latex].[latex]np=250 \\times (\\dfrac{98}{250})=98 > 10[\/latex]<\/p>\n<p>[latex]n(1-p)=250 \\times (1-\\dfrac{98}{250})=250 \\times (\\dfrac{152}{250})=152 > 10[\/latex]<\/p>\n<p>Therefore, the shape of the sampling distribution of the sample proportion is approximately normal.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox recall\"><strong><strong>Sampling Distribution of the Sample Proportion<\/strong><\/strong><\/p>\n<p>When taking random samples of size [latex]n[\/latex] from a population distribution with proportion [latex]p[\/latex]:<\/p>\n<ul>\n<li>The estimated <strong>mean<\/strong> of the distribution of sample proportions is [latex]\\hat{p}[\/latex].<\/li>\n<li>The estimated <strong>standard deviation<\/strong> of the distribution of sample proportions is called the <strong>standard error<\/strong>:[latex]SE = \\sqrt{\\dfrac{\\hat{p}(1-\\hat{p})}{n}}[\/latex].<\/li>\n<li>If [latex]n\\hat{p} \\geq 10[\/latex] and [latex]n(1-\\hat{p}) \\geq 10[\/latex], then the <strong>Central Limit Theorem<\/strong> (CLT) states that the distribution of the sample proportions follows an approximate normal distribution.<\/li>\n<\/ul>\n<p><strong>The confidence interval for a population proportion<\/strong> is:<\/p>\n<p style=\"text-align: center;\">[latex]\\hat{p} \\pm z^{*} \\cdot\\sqrt{\\frac{\\hat{p}(1-\\hat{p})}{n}}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\">Suppose [latex]250[\/latex] randomly selected people are surveyed to determine if they own a tablet. Of the [latex]250[\/latex] surveyed, [latex]98[\/latex] reported owning a tablet. Using a [latex]95\\%[\/latex] confidence level, compute and interpret a confidence interval estimate for the true proportion of people who own tablets.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q886455\">Show answer<\/button><\/p>\n<div id=\"q886455\" class=\"hidden-answer\" style=\"display: none\">Recall that [latex]z^{*}[\/latex] for a [latex]95\\%[\/latex] confidence interval is [latex]1.96[\/latex].<\/p>\n<p>The [latex]95\\%[\/latex] CI: [latex]\\dfrac{98}{250} \\pm 1.96 \\cdot \\sqrt{\\dfrac{\\dfrac{98}{250}(1-\\dfrac{98}{250})}{250}} = (0.3315, 0.4525)[\/latex].<\/p>\n<p>Interpretation: Based on the sample, we are [latex]95\\%[\/latex] confident that the true proportion of people who own tablets is between [latex]0.3315[\/latex] and [latex]0.4525[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":8,"menu_order":15,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":1163,"module-header":"fresh_take","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/1176"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/users\/8"}],"version-history":[{"count":5,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/1176\/revisions"}],"predecessor-version":[{"id":6728,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/1176\/revisions\/6728"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/parts\/1163"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/1176\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/media?parent=1176"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapter-type?post=1176"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/contributor?post=1176"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/license?post=1176"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}