{"id":1193,"date":"2023-06-22T01:59:55","date_gmt":"2023-06-22T01:59:55","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/introstatstest\/chapter\/sample-size-for-proportions-dig-deeper\/"},"modified":"2025-05-16T03:17:03","modified_gmt":"2025-05-16T03:17:03","slug":"sample-size-for-proportions-dig-deeper","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/introstatstest\/chapter\/sample-size-for-proportions-dig-deeper\/","title":{"raw":"Sample Size for Proportions: Fresh Take","rendered":"Sample Size for Proportions: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li class=\"li1\">Find the required sample size for a desired margin of error and population's confidence interval.<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Calculate the sample size [latex]n[\/latex]<\/h2>\r\n<p>Recall the margin of error formula: [latex] ME = z^{*} \\cdot \\sqrt{\\frac{\\hat{p}(1-\\hat{p})}{n}}[\/latex], to determine the minimum sample size needed to produce a given margin of error simply by solving for [latex]n[\/latex].<\/p>\r\n<p style=\"text-align: center;\">[latex] ME = z^{*} \\cdot \\sqrt{\\frac{\\hat{p}(1-\\hat{p})}{n}}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex] \\frac{ME}{z^{*}} = \\sqrt{\\frac{\\hat{p}(1-\\hat{p})}{n}}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex] (\\frac{ME}{z^{*}})^2 = \\frac{\\hat{p}(1-\\hat{p})}{n}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex] n \\cdot (\\frac{ME}{z^{*}})^2 = \\hat{p}(1-\\hat{p})[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex] n = \\hat{p}(1-\\hat{p})(\\frac{z^{*}}{ME})^{2}[\/latex]<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>sample size formula<\/h3>\r\n<p>The rearranged formula to find the sample size needed for proportion:<\/p>\r\n<p style=\"text-align: center;\">[latex] n = \\hat{p}(1-\\hat{p})(\\frac{z^{*}}{ME})^{2}[\/latex]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">Suppose a cell phone company wants to determine the current percentage of customers aged 50 and above who use text messaging. How many customers aged 50+ should the company survey in order to be 90% confident that the estimated (sample) proportion is within three percentage points of the true population proportion of customers aged 50+ who use text messaging on their cell phones? <strong>Answer: 752 cell phone customers<\/strong><br \/>\r\n[reveal-answer q=\"930757\"]Detailed solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"930757\"]<br \/>\r\nFrom the problem, we know that [latex]ME=3\\%=0.03[\/latex].\r\n\r\n<p>Recall that the [latex]z[\/latex] critical value for [latex]90\\%[\/latex] is [latex]1.645[\/latex]. You may use the <a href=\"https:\/\/lumen-learning.shinyapps.io\/normaldist\/\">Normal Distribution statistical tool\u00a0<\/a>to find the value of the [latex]z[\/latex] critical value.<\/p>\r\n<p>Since we were not given the value of [latex]p[\/latex], we can use the conservative value [latex]\\hat{p}=0.5[\/latex].<\/p>\r\n<p>Therefore:<\/p>\r\n<p>[latex] n = \\hat{p}(1-\\hat{p})(\\frac{z^{*}}{ME})^{2}[\/latex]<\/p>\r\n<p>[latex] n = 0.5(1-0.5)(\\frac{1.645}{0.03})^{2}[\/latex]<\/p>\r\n<p>[latex] n = 751.7[\/latex]<\/p>\r\n<p>Round the answer to the next higher value. The sample size should be 752 cell phone customers aged 50+ in order to be 90% confident that the estimated (sample) proportion is within three percentage points of the true population proportion of all customers aged 50+ who use text messaging on their cell phones.[\/hidden-answer]<\/p>\r\n<\/section>\r\n<p>Feel free to check the solution using our statistical tool below.<\/p>\r\n<p><iframe src=\"https:\/\/lumen-learning.shinyapps.io\/inference_prop\/\" width=\"100%\" height=\"850\"><\/iframe><\/p>\r\n<p>[<a href=\"https:\/\/lumen-learning.shinyapps.io\/inference_prop\/\" target=\"_blank\" rel=\"noopener\">Trouble viewing? Click to open in a new tab.<\/a>]<\/p>\r\n<p>Let's try finding the necessary sample size on another example.<\/p>\r\n<section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]1665[\/ohm2_question]<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li class=\"li1\">Find the required sample size for a desired margin of error and population&#8217;s confidence interval.<\/li>\n<\/ul>\n<\/section>\n<h2>Calculate the sample size [latex]n[\/latex]<\/h2>\n<p>Recall the margin of error formula: [latex]ME = z^{*} \\cdot \\sqrt{\\frac{\\hat{p}(1-\\hat{p})}{n}}[\/latex], to determine the minimum sample size needed to produce a given margin of error simply by solving for [latex]n[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]ME = z^{*} \\cdot \\sqrt{\\frac{\\hat{p}(1-\\hat{p})}{n}}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{ME}{z^{*}} = \\sqrt{\\frac{\\hat{p}(1-\\hat{p})}{n}}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex](\\frac{ME}{z^{*}})^2 = \\frac{\\hat{p}(1-\\hat{p})}{n}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]n \\cdot (\\frac{ME}{z^{*}})^2 = \\hat{p}(1-\\hat{p})[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]n = \\hat{p}(1-\\hat{p})(\\frac{z^{*}}{ME})^{2}[\/latex]<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>sample size formula<\/h3>\n<p>The rearranged formula to find the sample size needed for proportion:<\/p>\n<p style=\"text-align: center;\">[latex]n = \\hat{p}(1-\\hat{p})(\\frac{z^{*}}{ME})^{2}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\">Suppose a cell phone company wants to determine the current percentage of customers aged 50 and above who use text messaging. How many customers aged 50+ should the company survey in order to be 90% confident that the estimated (sample) proportion is within three percentage points of the true population proportion of customers aged 50+ who use text messaging on their cell phones? <strong>Answer: 752 cell phone customers<\/strong><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q930757\">Detailed solution<\/button><\/p>\n<div id=\"q930757\" class=\"hidden-answer\" style=\"display: none\">\nFrom the problem, we know that [latex]ME=3\\%=0.03[\/latex].<\/p>\n<p>Recall that the [latex]z[\/latex] critical value for [latex]90\\%[\/latex] is [latex]1.645[\/latex]. You may use the <a href=\"https:\/\/lumen-learning.shinyapps.io\/normaldist\/\">Normal Distribution statistical tool\u00a0<\/a>to find the value of the [latex]z[\/latex] critical value.<\/p>\n<p>Since we were not given the value of [latex]p[\/latex], we can use the conservative value [latex]\\hat{p}=0.5[\/latex].<\/p>\n<p>Therefore:<\/p>\n<p>[latex]n = \\hat{p}(1-\\hat{p})(\\frac{z^{*}}{ME})^{2}[\/latex]<br \/>\n[latex]n = 0.5(1-0.5)(\\frac{1.645}{0.03})^{2}[\/latex]<br \/>\n[latex]n = 751.7[\/latex]<\/p>\n<p>Round the answer to the next higher value. The sample size should be 752 cell phone customers aged 50+ in order to be 90% confident that the estimated (sample) proportion is within three percentage points of the true population proportion of all customers aged 50+ who use text messaging on their cell phones.<\/p><\/div>\n<\/div>\n<\/section>\n<p>Feel free to check the solution using our statistical tool below.<\/p>\n<p><iframe loading=\"lazy\" src=\"https:\/\/lumen-learning.shinyapps.io\/inference_prop\/\" width=\"100%\" height=\"850\"><\/iframe><\/p>\n<p>[<a href=\"https:\/\/lumen-learning.shinyapps.io\/inference_prop\/\" target=\"_blank\" rel=\"noopener\">Trouble viewing? Click to open in a new tab.<\/a>]<\/p>\n<p>Let&#8217;s try finding the necessary sample size on another example.<\/p>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm1665\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=1665&theme=lumen&iframe_resize_id=ohm1665&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":8,"menu_order":27,"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\/1193"}],"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":10,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/1193\/revisions"}],"predecessor-version":[{"id":6737,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/1193\/revisions\/6737"}],"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\/1193\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/media?parent=1193"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapter-type?post=1193"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/contributor?post=1193"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/license?post=1193"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}