[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"detail-sidebar-cat-0-en-105":3,"doc-seo-314014-105":59,"doc-detail-314014-en":131},{"code":4,"msg":5,"data":6},0,"success",[7,13,18,23,28,33,38,43,48,51,55],{"id":8,"doc_module":4,"doc_module_name":9,"category_name":10,"show_sort_weight":11,"slug":12},1,"Document","Story & Novel",90,"story-novel",{"id":14,"doc_module":4,"doc_module_name":9,"category_name":15,"show_sort_weight":16,"slug":17},2,"Literature",80,"literature",{"id":19,"doc_module":4,"doc_module_name":9,"category_name":20,"show_sort_weight":21,"slug":22},4,"Exam",70,"exam",{"id":24,"doc_module":4,"doc_module_name":9,"category_name":25,"show_sort_weight":26,"slug":27},5,"Comic",60,"comic",{"id":29,"doc_module":4,"doc_module_name":9,"category_name":30,"show_sort_weight":31,"slug":32},6,"Technology",50,"technology",{"id":34,"doc_module":4,"doc_module_name":9,"category_name":35,"show_sort_weight":36,"slug":37},7,"Healthcare",40,"healthcare",{"id":39,"doc_module":4,"doc_module_name":9,"category_name":40,"show_sort_weight":41,"slug":42},8,"Research & Report",30,"research-report",{"id":44,"doc_module":4,"doc_module_name":9,"category_name":45,"show_sort_weight":46,"slug":47},9,"Religion & Spirituality",20,"religion-spirituality",{"id":46,"doc_module":4,"doc_module_name":9,"category_name":49,"show_sort_weight":46,"slug":50},"World Cup","world-cup",{"id":52,"doc_module":4,"doc_module_name":9,"category_name":53,"show_sort_weight":52,"slug":54},10,"Lifestyle","lifestyle",{"id":56,"doc_module":4,"doc_module_name":9,"category_name":57,"show_sort_weight":24,"slug":58},19,"General","general",{"code":4,"msg":60,"data":61},"ok",{"site_id":62,"language":63,"slug":64,"title":65,"keywords":66,"description":67,"schema_data":68,"social_meta":124,"head_meta":126,"extra_data":128,"updated_unix":130},105,"en","ap-statistics-chapter-11-chi-square-goodness-of-fit-test-and-test-of-associationindependence","AP Statistics Chapter 11 - Chi-Square Goodness of Fit Test and Test of Association/Independence","","AP Statistics Chapter 11 explains chi-square inference for categorical data, covering both goodness of fit and tests of association/independence. The chapter defines how goodness of fit evaluates whether observed category proportions match hypothesized distributions, including equal and fixed (given) proportions. It specifies hypotheses, chi-square test statistics, expected counts, key conditions (random, independent, large expected counts), and degrees of freedom. It also introduces two-way tables for association between two categorical variables, with matching conditions and calculation guidance.",{"@graph":69,"@context":123},[70,84,106],{"@type":71,"itemListElement":72},"BreadcrumbList",[73,77,79,82],{"item":74,"name":75,"@type":76,"position":8},"https://docshare.wps.com","Home","ListItem",{"item":78,"name":9,"@type":76,"position":14},"https://docshare.wps.com/document/",{"item":80,"name":20,"@type":76,"position":81},"https://docshare.wps.com/document/exam/",3,{"item":83,"name":65,"@type":76,"position":19},"https://docshare.wps.com/document/ap-statistics-chapter-11-chi-square-goodness-of-fit-test-and-test-of-associationindependence/314014/",{"url":83,"name":65,"@type":85,"image":86,"author":91,"headline":65,"publisher":94,"fileFormat":97,"inLanguage":63,"description":67,"dateModified":98,"datePublished":99,"encodingFormat":97,"isAccessibleForFree":100,"interactionStatistic":101},"DigitalDocument",{"url":87,"@type":88,"width":89,"height":90},"https://docshare.wps.com/thumbnails/ap-statistics-chapter-11-chi-square-goodness-of-fit-test-and-test-of-associationindependence/314014.png","ImageObject",300,407,{"name":92,"@type":93},"Ophelia","Person",{"url":74,"name":95,"@type":96},"DocShare","Organization","application/pdf","2026-11-05","2026-09-20",true,{"@type":102,"interactionType":103,"userInteractionCount":105},"InteractionCounter",{"@type":104},"ViewAction",16,{"@type":107,"mainEntity":108},"FAQPage",[109,115,119],{"name":110,"@type":111,"acceptedAnswer":112},"What is the purpose of a chi-square goodness of fit test?","Question",{"text":113,"@type":114},"It helps determine whether a population’s observed categorical outcomes match a hypothesized distribution expressed as proportions in the categories.","Answer",{"name":116,"@type":111,"acceptedAnswer":117},"How are expected counts computed for equal versus given proportions?",{"text":118,"@type":114},"For equal proportions, expected counts are found by dividing the total count by the number of outcome categories. For fixed/given proportions, expected counts are found by multiplying the total by each given percentage.",{"name":120,"@type":111,"acceptedAnswer":121},"What conditions must be met to use the chi-square tests for categorical data?",{"text":122,"@type":114},"The data must come from a well-designed random sample or randomized experiment (random). Observations must be independent, and expected counts must be large, with all expected counts at least 5.","https://schema.org",{"og:url":83,"og:type":125,"og:title":65,"og:site_name":95,"og:description":67},"article",{"robots":127,"canonical":83},"index,follow",{"doc_id":129,"site_id":62},314014,1790184572,{"code":4,"msg":5,"data":132},{"doc_id":129,"user_id":133,"nickname":92,"user_avatar":134,"doc_module":4,"category_id":19,"category_name":20,"doc_title":65,"doc_description":67,"doc_content":135,"file_id":136,"file_url":137,"file_type":138,"file_size":139,"view_count":105,"is_deleted":4,"is_public":8,"is_downloadable":8,"audit_status":8,"page_count":14,"language":140,"language_code":63,"site_id":62,"html_lang":63,"table_of_contents":141,"faqs":142,"seo_title":143,"seo_description":67,"update_tm":144,"read_time":24,"preview_page_count":4},7971461741311,"https://ap-avatar.wpscdn.com/avatar/74000253aff267980c6?x-image-process=image/resize,m_fixed,w_180,h_180&k=1779345379180704826","AP Statistics Chapter 11: Inference for Distributions of Categorical Data  \n11.1 – Chi-Square (􀟯2 ) Goodness of Fit Test  \nGoodness of Fit  \nA goodness of fit test is used to help determine whether a population has a certain hypothesized distribution, expressed as proportions of individuals in the population falling into various outcome categories. There are two types of goodness of fit tests  \n1. Equal Proportions (all proportions are expected to be the same)  \n2. Fixed or Given Proportions (proportions are expected to follow given values)  \nHypotheses for the Goodness of Fit Test  \nHo: The stated distribution of the categorical variables in the population of interest is correct.  \nHa: The stated distribution in the population of interest is not correct.  \nThe Chi-Square Statistic  \nThe formula is  \n􀟯 2 = ∑ ~~ ~~(􀝋􀜾􀝏􀝁􀝎􀝒􀝁􀝀~~ ~~􀜿􀝁􀝋􀝔􀝑􀝌􀝊􀝁􀝐􀜿􀝐􀝀􀝁􀝔􀜿􀝌􀝋􀝁􀝑􀜿􀝊􀝐􀝐􀝁􀝀~~ ~~􀜿􀝋􀝑􀝊􀝐)2 = ∑ ~~ ~~(􀜱~~ ~~~~ ~~􀜧)2  \nExpected Counts  \n􀁸 The expected counts for the equal proportions GOF test are all the same. They are found by dividing the total of the counts by the number of outcome categories.  \n􀁸 The expected counts for the given proportions GOF test are NOT all the same. They are found by multiplying the total of the counts by each given percentage.  \nConditions for the Chi-Square Goodness of Fit Test  \n􀁸 Random – The data come from a well-designed random sample or randomized experiment.  \n􀁸 Independent – is the sample size less than 10% of the population size?  \n􀁸 Large Counts – All expected counts are at least 5.  \nDegrees of Freedom for the Goodness of Fit Test  \nThe degrees of freedom for the GOF test are n􀀐1 where n is the number of outcome categories.  \nUsing the Calculator for the 􀟯2 Test on the TI-83 Calculator *  \n􀁸 The observed counts are to be stored in L1  \n􀁸 The expected counts are to be stored in L2  \n􀁸 Let L3 = (L1-L2)2/L2  \n􀁸 The 􀟯 2 is the sum of L3 (which can be found by using 1-Var Stats)  \n*The TI-84 has a built-in test called 􀟯2 GOF-Test. Find it in the STAT-TEST menu.  \n11.2 – The 􀟯2 Test of Association/Independence for Two-Way Tables  \nTwo-Way Tables  \nWhen there are two categorical variables, data can be arranged in a row and column format, called a Two-Way Table. Here is an example:  \n\n|  | Color Choice |  |  |  |  |\n| --- | --- | --- | --- | --- | --- |\n| Grade | blue | green | red | yellow | Totals |\n| 1st | 13 | 7 | 8 | 2 | 30 |\n| 2nd | 11 | 10 | 6 | 5 | 33 |\n| Totals | 24 | 17 | 14 | 7 | 63 |\n\nHypotheses for the Test for Association between Two Categorical Variables  \nHo: There is no association between the categorical variables  \nHa: There is an association  \nOR  \nHo: The categorical variables are independent (there is no association)  \nHa: The variables are NOT independent (there is an association)  \nDegrees of Freedom for the Chi-Square Test of Association/Independence  \nThe degrees of freedom for the Test of Association/Independence test are (r 􀀐1)x(c􀀐1) where r is the number of rows and c is the number of columns in the table.  \nConditions for the Chi-Square Test of Association/Independence  \nThe conditions are the same as for the goodness of fit test  \nThe Chi-Square Statistic  \nThe formula is the same as in the goodness of fit test.  \nExpected Counts  \nThe expected counts for this test can be found as follows:  \n􀝁􀝔􀝌􀝁􀜿􀝐􀝁􀝀 􀜿􀝋􀝑􀝊􀝐 = (􀝎􀝋􀝓~~ ~~􀝐􀝋~~ ~~􀝐􀝐􀜽􀜽􀝈􀜾)􀝈(􀜿􀝁􀝋􀝐􀝈􀝋􀝑􀝐􀝉􀜽􀝈􀝊~~ ~~􀝐􀝋􀝐􀜽􀝈)  \nFor example, for the expected count for 2nd grade/green in the table above, we would use the calculation  \n(336)3(~~ ~~17) = 8.9","cbCaisa0HR5TOLjb","https://ap.wps.com/l/cbCaisa0HR5TOLjb","pdf",482198,"English","# 11.1 - Chi-Square Goodness of Fit Test\n## Goodness of Fit\n## Hypotheses for the Goodness of Fit Test\n## The Chi-Square Statistic\n## Expected Counts\n## Conditions for the Chi-Square Goodness of Fit Test\n## Degrees of Freedom for the Goodness of Fit Test\n## Using the Calculator for the χ² Test on the TI-83 Calculator\n# 11.2 - The χ² Test of Association/Independence for Two-Way Tables\n## Two-Way Tables\n## Hypotheses for the Test for Association between Two Categorical Variables\n## Degrees of Freedom for the Chi-Square Test of Association/Independence\n## Conditions for the Chi-Square Test of Association/Independence\n## The Chi-Square Statistic\n## Expected Counts","[{\"question\":\"What is the purpose of a chi-square goodness of fit test?\",\"answer\":\"It helps determine whether a population’s observed categorical outcomes match a hypothesized distribution expressed as proportions in the categories.\"},{\"question\":\"How are expected counts computed for equal versus given proportions?\",\"answer\":\"For equal proportions, expected counts are found by dividing the total count by the number of outcome categories. For fixed/given proportions, expected counts are found by multiplying the total by each given percentage.\"},{\"question\":\"What conditions must be met to use the chi-square tests for categorical data?\",\"answer\":\"The data must come from a well-designed random sample or randomized experiment (random). Observations must be independent, and expected counts must be large, with all expected counts at least 5.\"}]","AP Statistics Chapter 11 - Chi-Square Goodness of Fit Test and Test of Association/Independence | PDF",1789899847]