{"id":1419,"date":"2023-06-22T02:23:00","date_gmt":"2023-06-22T02:23:00","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/introstatstest\/chapter\/chi-square-test-of-homogeneity-fresh-take\/"},"modified":"2025-05-16T23:43:58","modified_gmt":"2025-05-16T23:43:58","slug":"chi-square-test-of-homogeneity-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/introstatstest\/chapter\/chi-square-test-of-homogeneity-fresh-take\/","title":{"raw":"Chi-Square Test of Homogeneity \u2013 Fresh Take","rendered":"Chi-Square Test of Homogeneity \u2013 Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Complete a chi-square test of homogeneity&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:12801,&quot;3&quot;:{&quot;1&quot;:0},&quot;12&quot;:0,&quot;15&quot;:&quot;arial&quot;,&quot;16&quot;:9}\">Complete a chi-square test of homogeneity<\/span><\/li>\r\n\t<li><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Write the conclusion of a chi-square test of homogeneity in context of the problem&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:12801,&quot;3&quot;:{&quot;1&quot;:0},&quot;12&quot;:0,&quot;15&quot;:&quot;arial&quot;,&quot;16&quot;:9}\">Write the conclusion of a chi-square test of homogeneity in context of the problem<\/span><\/li>\r\n<\/ul>\r\n<\/section>\r\n<section class=\"textbox recall\">A <strong>chi-square test of homogeneity<\/strong> determines if two or more populations (or subgroups of a population) have the same distribution of a single categorical variable. The formula for the chi-square test statistic:\r\n\r\n<p style=\"text-align: center;\">[latex]\\chi^2=\\sum\\dfrac{(\\text{Observed}-\\text{Expected})^2}{\\text{Expected}}[\/latex]<\/p>\r\n<\/section>\r\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">\r\n<p>[embed]https:\/\/youtu.be\/t_jfTOE44YQ[\/embed]<\/p>\r\n<\/section>\r\n<p>&nbsp;<\/p>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Complete a chi-square test of homogeneity&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:12801,&quot;3&quot;:{&quot;1&quot;:0},&quot;12&quot;:0,&quot;15&quot;:&quot;arial&quot;,&quot;16&quot;:9}\">Complete a chi-square test of homogeneity<\/span><\/li>\n<li><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Write the conclusion of a chi-square test of homogeneity in context of the problem&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:12801,&quot;3&quot;:{&quot;1&quot;:0},&quot;12&quot;:0,&quot;15&quot;:&quot;arial&quot;,&quot;16&quot;:9}\">Write the conclusion of a chi-square test of homogeneity in context of the problem<\/span><\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox recall\">A <strong>chi-square test of homogeneity<\/strong> determines if two or more populations (or subgroups of a population) have the same distribution of a single categorical variable. The formula for the chi-square test statistic:<\/p>\n<p style=\"text-align: center;\">[latex]\\chi^2=\\sum\\dfrac{(\\text{Observed}-\\text{Expected})^2}{\\text{Expected}}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Introduction to the chi-square test for homogeneity | AP Statistics | Khan Academy\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/t_jfTOE44YQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/section>\n<p>&nbsp;<\/p>\n","protected":false},"author":8,"menu_order":26,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":1388,"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\/1419"}],"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":2,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/1419\/revisions"}],"predecessor-version":[{"id":6878,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/1419\/revisions\/6878"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/parts\/1388"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/1419\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/media?parent=1419"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapter-type?post=1419"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/contributor?post=1419"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/license?post=1419"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}