{"id":296,"date":"2021-07-14T15:59:09","date_gmt":"2021-07-14T15:59:09","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/goodness-of-fit-test\/"},"modified":"2023-12-05T09:42:21","modified_gmt":"2023-12-05T09:42:21","slug":"goodness-of-fit-test","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/goodness-of-fit-test\/","title":{"raw":"Chi-Square Goodness-of-Fit","rendered":"Chi-Square Goodness-of-Fit"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<section>\r\n<ul id=\"element-377\" data-bullet-style=\"bullet\">\r\n \t<li>Conduct a chi-square goodness-of-fit test and interpret the conclusion in context<\/li>\r\n<\/ul>\r\n<\/section><\/div>\r\nIn this type of hypothesis test, you determine whether the data \"fit\" a particular distribution or not. For example, you may suspect your unknown data fit a binomial distribution. You use a chi-square test (meaning the distribution for the hypothesis test is chi-square) to determine if there is a fit or not. <strong>The null and the alternative hypotheses for this test may be written in sentences or may be stated as equations or inequalities.<\/strong>\r\n\r\nThe test statistic for a goodness-of-fit test is:\r\n<p style=\"text-align: center;\">[latex]\\displaystyle{\\sum_{k}}\\frac{{({O}-{E})}^{{2}}}{{E}}[\/latex]<\/p>\r\nwhere:\r\n<ul>\r\n \t<li><em>O<\/em> = <strong>observed values<\/strong> (data)<\/li>\r\n \t<li><em>E<\/em> = <strong>expected values<\/strong> (from theory)<\/li>\r\n \t<li><em>k<\/em> = the number of different data cells or categories<\/li>\r\n<\/ul>\r\n<strong>The observed values are the data values and the expected values are the values you would expect to get if the null hypothesis were true.<\/strong> There are\u00a0<em>n <\/em>terms of the form [latex]\\displaystyle\\frac{{({O}-{E})}^{{2}}}{{E}}[\/latex].\r\n\r\nThe number of degrees of freedom is\u00a0<em>df<\/em> = (number of categories \u2013 1).\r\n\r\n<strong>The goodness-of-fit test is almost always right-tailed.<\/strong> If the observed values and the corresponding expected values are not close to each other, then the test statistic can get very large and will be way out in the right tail of the chi-square curve.\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Expected Value<\/h3>\r\nThe <strong>expected value<\/strong> is often referred to as the \"long-term\" average or mean. This means that over the long term of doing an experiment over and over, you would expect this average.\r\n\r\n<\/div>\r\n<strong>Note:\u00a0<\/strong>The expected value for each cell needs to be at least five in order for you to use this test.\r\n\r\nhttps:\/\/www.youtube.com\/embed\/2QeDRsxSF9M\r\n<div class=\"textbox exercises\">\r\n<h3>Example 1<\/h3>\r\nAbsenteeism of college students from math classes is a major concern to math instructors because missing class appears to increase the drop rate. Suppose that a study was done to determine if the actual student absenteeism rate follows faculty perception. The faculty expected that a group of 100 students would miss class according to this table.\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>Number of absences per term<\/th>\r\n<th>Expected number of students<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>0\u20132<\/td>\r\n<td>50<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>3\u20135<\/td>\r\n<td>30<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>6\u20138<\/td>\r\n<td>12<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>9\u201311<\/td>\r\n<td>6<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>12+<\/td>\r\n<td>2<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nA random survey across all mathematics courses was then done to determine the actual number\u00a0<strong>(observed)<\/strong> of absences in a course. The chart in this table displays the results of that survey.\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>Number of absences per term<\/th>\r\n<th>Actual number of students<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>0\u20132<\/td>\r\n<td>35<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>3\u20135<\/td>\r\n<td>40<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>6\u20138<\/td>\r\n<td>20<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>9\u201311<\/td>\r\n<td>1<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>12+<\/td>\r\n<td>4<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nDetermine the null and alternative hypotheses needed to conduct a goodness-of-fit test.\r\n\r\n<em>H<sub data-redactor-tag=\"sub\">0<\/sub><\/em>: Student absenteeism <strong>fits<\/strong> faculty perception.\r\n\r\nThe alternative hypothesis is the opposite of the null hypothesis.\r\n\r\n<em>H<sub data-redactor-tag=\"sub\">a<\/sub><\/em>: Student absenteeism <strong>does not fit<\/strong> faculty perception.\r\n<ol>\r\n \t<li>Can you use the information as it appears in the charts to conduct the goodness-of-fit test?<\/li>\r\n \t<li>What is the number of degrees of freedom (<em>df<\/em>)?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"833283\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"833283\"]\r\n<ol>\r\n \t<li><strong>No<\/strong><strong>. <\/strong>Notice that the expected number of absences for the \"12+\" entry is less than five (it is two). Combine that group with the \"9\u201311\" group to create new tables where the number of students for each entry is at least five. The new results are in the two tables below.\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>Number of absences per term<\/th>\r\n<th>Expected number of students<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>0\u20132<\/td>\r\n<td>50<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>3\u20135<\/td>\r\n<td>30<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>6\u20138<\/td>\r\n<td>12<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>9+<\/td>\r\n<td>8<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>Number of absences per term<\/th>\r\n<th>Actual number of students<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>0\u20132<\/td>\r\n<td>35<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>3\u20135<\/td>\r\n<td>40<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>6\u20138<\/td>\r\n<td>20<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>9+<\/td>\r\n<td>5<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n \t<li>There are four \"cells\" or categories in each of the new tables.\r\n<em>df<\/em> = number of cells \u2013 1 = 4 \u2013 1 = 3<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it 1<\/h3>\r\nA factory manager needs to understand how many products are defective versus how many are produced. The number of expected defects is listed in the table.\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>Number produced<\/th>\r\n<th>Number defective<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>0\u2013100<\/td>\r\n<td>5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>101\u2013200<\/td>\r\n<td>6<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>201\u2013300<\/td>\r\n<td>7<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>301\u2013400<\/td>\r\n<td>8<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>401\u2013500<\/td>\r\n<td>10<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nA random sample was taken to determine the actual number of defects. This table shows the results of the survey.\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>Number produced<\/th>\r\n<th>Number defective<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>0\u2013100<\/td>\r\n<td>5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>101\u2013200<\/td>\r\n<td>7<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>201\u2013300<\/td>\r\n<td>8<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>301\u2013400<\/td>\r\n<td>9<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>401\u2013500<\/td>\r\n<td>11<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nState the null and alternative hypotheses needed to conduct a goodness-of-fit test, and state the degrees of freedom.\r\n[reveal-answer q=\"473884\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"473884\"]\r\n\r\n<em>H<sub data-redactor-tag=\"sub\">0<\/sub><\/em>: The number of defaults fits expectations.\r\n\r\n<em>H<sub data-redactor-tag=\"sub\">a<\/sub><\/em>: The number of defaults does not fit expectations.\r\n\r\n<em>df<\/em> = 4\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Recall: ORDER OF OPERATIONS<\/h3>\r\n<div align=\"left\">\r\n<table style=\"border-collapse: collapse; width: 100%; height: 36px;\" border=\"1\">\r\n<tbody>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\"><strong>Please<\/strong><\/td>\r\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\"><strong>Excuse<\/strong><\/td>\r\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\"><strong>My<\/strong><\/td>\r\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\"><strong>Dear<\/strong><\/td>\r\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\"><strong>Aunt<\/strong><\/td>\r\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\"><strong>Sally<\/strong><\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\">parentheses<\/td>\r\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\">exponents<\/td>\r\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\">multiplication<\/td>\r\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\">division<\/td>\r\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\">addition<\/td>\r\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\">subtraction<\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\">\r\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\">[latex]( \\ )[\/latex]<\/td>\r\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\">[latex]x^2[\/latex]<\/td>\r\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\" colspan=\"2\">[latex]\\times \\ \\mathrm{or} \\ \\div[\/latex]<\/td>\r\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\" colspan=\"2\">[latex]+ \\ \\mathrm{or} \\ -[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h2>To calculate the test statistic for a goodness-of-fit test follow the following steps: you will end up using the formula [latex]\\frac{(O-E)^2}{E}[\/latex] over and over.<\/h2>\r\nStep 1: Pick one of the observed values and then subtract the expected value, this finds the distance between the observed value and what is expected. [latex](O-E)[\/latex]\r\n\r\nStep 2: Square this difference. [latex](O-E)^2[\/latex]\r\n\r\nStep 3: Divide step 2 by the expected value. [latex]\\frac{(O-E)^2}{E}[\/latex]\r\n\r\nStep 4: Repeat steps 1 - 3 with every observed value.\r\n\r\nStep 5: Add every value calculated in step 4, in order words find the sum of the difference between each observed value and the expected value, squared and divided by the expected value.\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example 2<\/h3>\r\nEmployers want to know which days of the week employees are absent in a five-day work week. Most employers would like to believe that employees are absent equally during the week. Suppose a random sample of 60 managers were asked on which day of the week they had the highest number of employee absences. The results were distributed as in the table below. For the population of employees, do the days for the highest number of absences occur with equal frequencies during a five-day work week? Test at a 5% significance level.\r\n\r\nDay of the Week Employees were Most Absent\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>Monday<\/th>\r\n<th>Tuesday<\/th>\r\n<th>Wednesday<\/th>\r\n<th>Thursday<\/th>\r\n<th>Friday<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>Number of Absences<\/td>\r\n<td>15<\/td>\r\n<td>12<\/td>\r\n<td>9<\/td>\r\n<td>9<\/td>\r\n<td>15<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"229341\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"229341\"]\r\n\r\nThe null and alternative hypotheses are:\r\n<ul>\r\n \t<li><em>H<sub data-redactor-tag=\"sub\">0<\/sub><\/em>: The absent days occur with equal frequencies, that is, they fit a uniform distribution.<\/li>\r\n \t<li><em>H<sub data-redactor-tag=\"sub\">a<\/sub><\/em>: The absent days occur with unequal frequencies, that is, they do not fit a uniform distribution.<\/li>\r\n<\/ul>\r\nIf the absent days occur with equal frequencies, then, out of 60 absent days (the total in the sample: 15 + 12 + 9 + 9 + 15 = 60), there would be 12 absences on Monday, 12 on Tuesday, 12 on Wednesday, 12 on Thursday, and 12 on Friday. These numbers are the\u00a0<strong>expected<\/strong> (<em>E<\/em>) values. The values in the table are the <strong>observed <\/strong>(<em>O<\/em>) values or data.\r\n\r\nThis time, calculate the\u00a0<em>\u03c7<\/em><sup>2<\/sup> test statistic by hand. Make a chart with the following headings and fill in the columns:\r\n<ul>\r\n \t<li>Expected (<em>E<\/em>) values (12, 12, 12, 12, 12)<\/li>\r\n \t<li>Observed (<em>O<\/em>) values (15, 12, 9, 9, 15)<\/li>\r\n \t<li>(<em>O<\/em> \u2013 <em>E<\/em>)<\/li>\r\n \t<li>(<em>O<\/em> \u2013 <em>E<\/em>)<sup>2<\/sup><\/li>\r\n \t<li>[latex]\\displaystyle\\frac{{({O}-{E})}^{{2}}}{{E}}[\/latex]<\/li>\r\n<\/ul>\r\nNow add (sum) the last column. The sum is three. This is the\u00a0<em>\u03c7<sup data-redactor-tag=\"sup\">2<\/sup><\/em> test statistic.\r\n\r\nTo find the\u00a0<em>p<\/em>-value, calculate <em>P<\/em>(<em>\u03c7<\/em><sup>2<\/sup> &gt; 3). This test is right-tailed. (Use a computer or calculator to find the <em>p<\/em>-value. You should get <em>p<\/em>-value = 0.5578.)\r\n\r\nThe\u00a0<em>dfs<\/em> are the number of cells \u2013 1 = 5 \u2013 1 = 4\r\n\r\n<header>\r\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR:<\/span><\/h4>\r\n<\/header>\r\n<ul>\r\n \t<li>Press\u00a0<code>2nd DISTR<\/code>.<\/li>\r\n \t<li>Arrow down to\u00a0<code>\u03c72cdf<\/code>.<\/li>\r\n \t<li>Press <code>ENTER<\/code>.<\/li>\r\n \t<li>Enter<code>(3,10^99,4)<\/code>.<\/li>\r\n<\/ul>\r\nRounded to four decimal places, you should see 0.5578, which is the p-value.\r\n\r\nNext, complete a graph like the following one with the proper labeling and shading. (You should shade the right tail.)\r\n\r\n<img class=\"aligncenter wp-image-2212 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5668\/2021\/07\/06131052\/039a6a3fbbc667b9b7ec97f2dbb5b5840ef6437d.jpeg\" alt=\"This is a blank nonsymmetrical chi-square curve for the test statistic of the days of the week absent.\" width=\"487\" height=\"229\" \/>\r\n\r\nThe decision is not to reject the null hypothesis.\r\n\r\n<strong>Conclusion: <\/strong>At a 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the absent days do not occur with equal frequencies.\r\n\r\n<strong>Using a calculator:\u00a0<\/strong>TI-83+ and some TI-84 calculators do not have a special program for the test statistic for the goodness-of-fit test. The next example has the calculator instructions.\r\n<ul>\r\n \t<li>The newer TI-84 calculators have in\u00a0<code>STAT TESTS<\/code> the test <code>Chi2 GOF<\/code>.<\/li>\r\n \t<li>To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list.<\/li>\r\n \t<li>Press <code>STAT TESTS\u00a0<\/code>and <code>Chi2 GOF<\/code>.<\/li>\r\n \t<li>Enter the list names for the Observed list and the Expected list.<\/li>\r\n \t<li>Enter the degrees of freedom and press\u00a0<code>calculate<\/code> or <code>draw<\/code>.<\/li>\r\n \t<li>Make sure you clear any lists before you start.<\/li>\r\n \t<li>To Clear Lists in the calculators: Go into\u00a0<code>STAT EDIT<\/code> and arrow up to the list name area of the particular list. Press\u00a0<code>CLEAR<\/code> and then arrow down. The list will be cleared.<\/li>\r\n \t<li>Alternatively, you can press <code>STAT<\/code> and press 4 (for\u00a0<code>ClrList<\/code>). Enter the list name and press <code>ENTER<\/code>.<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it 2<\/h3>\r\nTeachers want to know which night each week their students are doing most of their homework. Most teachers think that students do homework equally throughout the week. Suppose a random sample of 56 students were asked on which night of the week they did the most homework. The results were distributed as in the table.\r\n<table>\r\n<thead>\r\n<tr>\r\n<th><\/th>\r\n<th>Sunday<\/th>\r\n<th>Monday<\/th>\r\n<th>Tuesday<\/th>\r\n<th>Wednesday<\/th>\r\n<th>Thursday<\/th>\r\n<th>Friday<\/th>\r\n<th>Saturday<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>Number of Students<\/td>\r\n<td>11<\/td>\r\n<td>8<\/td>\r\n<td>10<\/td>\r\n<td>7<\/td>\r\n<td>10<\/td>\r\n<td>5<\/td>\r\n<td>5<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nFrom the population of students, do the nights for the highest number of students doing the majority of their homework occur with equal frequencies during a week? What type of hypothesis test should you use?\r\n[reveal-answer q=\"478005\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"478005\"]\r\n\r\n<em>df<\/em> = 6\r\n\r\n<em>p<\/em>-value = 0.6093\r\n\r\nWe decline to reject the null hypothesis. There is not enough evidence to support that students do not do the majority of their homework equally throughout the week.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example 3<\/h3>\r\nOne study indicates that the number of televisions that American families have is distributed (this is the\u00a0<strong>given<\/strong> distribution for the American population) as in the table.\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>Number of Televisions<\/th>\r\n<th>Percent<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>0<\/td>\r\n<td>10<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1<\/td>\r\n<td>16<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2<\/td>\r\n<td>55<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>3<\/td>\r\n<td>11<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>4+<\/td>\r\n<td>8<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe table contains expected (<em>E<\/em>) percents.\r\n\r\nA random sample of 600 families in the far western United States resulted in the data in this table.\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>Number of Televisions<\/th>\r\n<th>Frequency<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tfoot>\r\n<tr>\r\n<td>Total = 600<\/td>\r\n<\/tr>\r\n<\/tfoot>\r\n<tbody>\r\n<tr>\r\n<td>0<\/td>\r\n<td>66<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1<\/td>\r\n<td>119<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2<\/td>\r\n<td>340<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>3<\/td>\r\n<td>60<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>4+<\/td>\r\n<td>15<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe table contains observed (<em>O<\/em>) frequency values.\r\n\r\nAt the 1% significance level, does it appear that the distribution \"number of televisions\" of far western United States families is different from the distribution for the American population as a whole?\r\n[reveal-answer q=\"946383\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"946383\"]\r\n\r\nThis problem asks you to test whether the far western United States families' distribution fits the distribution of the American families. This test is always right-tailed.\r\n\r\nThe first table contains expected percentages. To get expected (<em>E<\/em>) frequencies, multiply the percentage by 600. The expected frequencies are shown in this table.\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>Number of Televisions<\/th>\r\n<th>Percent<\/th>\r\n<th>Expected Frequency<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>0<\/td>\r\n<td>10<\/td>\r\n<td>(0.10)(600) = 60<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1<\/td>\r\n<td>16<\/td>\r\n<td>(0.16)(600) = 96<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2<\/td>\r\n<td>55<\/td>\r\n<td>(0.55)(600) = 330<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>3<\/td>\r\n<td>11<\/td>\r\n<td>(0.11)(600) = 66<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>over 3<\/td>\r\n<td>8<\/td>\r\n<td>(0.08)(600) = 48<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nTherefore, the expected frequencies are 60, 96, 330, 66, and 48. In the TI calculators, you can let the calculator do the math. For example, instead of 60, enter 0.10*600.\r\n\r\n<em>H<sub>0<\/sub><\/em>: The \"number of televisions\" distribution of far western United States families is the same as the \"number of televisions\" distribution of the American population.\r\n\r\n<em>H<sub>a<\/sub><\/em>: The \"number of televisions\" distribution of far western United States families is different from the \"number of televisions\" distribution of the American population.\r\n\r\nDistribution for the test:\u00a0[latex]\\displaystyle\\chi^{2}_{4}[\/latex] where <i>df<\/i> = (the number of cells) \u2013 1 = 5 \u2013 1 = 4.\r\n\r\n<strong>Note<\/strong>:\u00a0[latex]df\\neq600-1[\/latex]\r\n<div id=\"example3\" class=\"example\" data-type=\"example\"><section>\r\n<div id=\"element-911\" class=\"exercise\" data-type=\"exercise\"><section>\r\n<div id=\"id6229434\" class=\"solution ui-solution-visible\" data-type=\"solution\"><section class=\"ui-body\">\r\n<p id=\"element-65\"><strong>Calculate the test statistic:\u00a0<\/strong><em data-effect=\"italics\">\u03c7<\/em><sup>2<\/sup> = 29.65<\/p>\r\n<p id=\"element-280\"><strong>Graph:<\/strong><\/p>\r\n<img class=\"aligncenter wp-image-2215 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5668\/2021\/07\/06131244\/c64bbffd1bc6d84f526b8cf8d2ba05ce360d752e.jpeg\" alt=\"This is a nonsymmetric chi-square curve with values of 0, 4, and 29.65 labeled on the horizontal axis. The value 4 coincides with the peak of the curve. A vertical upward line extends from 29.65 to the curve, and the region to the right of this line is shaded. The shaded area is equal to the p-value.\" width=\"487\" height=\"146\" \/>\r\n\r\n<strong>Probability statement:\u00a0<\/strong><em data-effect=\"italics\">p<\/em>-value = <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">\u03c7<\/em><sup>2<\/sup> &gt; 29.65) = 0.000006\r\n\r\n<strong>Compare <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value:<\/strong>\r\n<ul>\r\n \t<li><em data-effect=\"italics\">\u03b1<\/em> = 0.01<\/li>\r\n \t<li><em data-effect=\"italics\">p<\/em>-value = 0.000006<\/li>\r\n \t<li>So, <em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">\u03b1<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\"> &gt; <\/span><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">p<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">-value.<\/span><\/li>\r\n<\/ul>\r\n<strong>Make a decision:<\/strong> Since <em data-effect=\"italics\">\u03b1<\/em> &gt; <em data-effect=\"italics\">p<\/em>-value, reject <em data-effect=\"italics\">H<sub>o<\/sub><\/em>.\r\n\r\nThis means you reject the belief that the distribution for the far western states is the same as that of the American population as a whole.\r\n\r\n<strong>Conclusion:<\/strong> At the 1% significance level, from the data, there is sufficient evidence to conclude that the \"number of televisions\" distribution for the far western United States is different from the \"number of televisions\" distribution for the American population as a whole.\r\n\r\n<strong>Using a calculator:<\/strong>\r\n<ul>\r\n \t<li>Press <code>STAT<\/code> and <code>ENTER<\/code>.<\/li>\r\n \t<li>Make sure to clear lists <code>L1<\/code>, <code>L2<\/code>, and <code>L3<\/code> if they have data in them (see the note at the end of Example 2).<\/li>\r\n \t<li>Into <code>L1<\/code>, put the observed frequencies <code>66<\/code>, <code>119<\/code>,<code>349<\/code>, <code>60<\/code>, <code>15<\/code>.<\/li>\r\n \t<li>Into <code>L2<\/code>, put the expected frequencies <code>.10*600, .16*600<\/code>, <code>.55*600<\/code>,<code>.11*600<\/code>, <code>.08*600<\/code>.<\/li>\r\n \t<li>Arrow over to list <code>L3<\/code> and up to the name area <code>L3<\/code>. Enter <code>(L1-L2)^2\/L2<\/code> and <code>ENTER<\/code>.<\/li>\r\n \t<li>Press <code>2nd QUIT<\/code>. Press <code>2nd LIST<\/code> and arrow over to <code>MATH<\/code>. Press <code>5<\/code>. You should see <code>\"sum\" (Enter L3)<\/code>.<\/li>\r\n \t<li>Rounded to 2 decimal places, you should see <code>29.65<\/code>. Press <code>2nd DISTR<\/code>. Press <code>7<\/code> or Arrow down to <code>7:\u03c72cdf<\/code> and press <code>ENTER<\/code>.<\/li>\r\n \t<li>Enter <code>(29.65,1E99,4)<\/code>. Rounded to four places, you should see <code>5.77E-6 = .000006<\/code>\u00a0(rounded to six decimal places), which is the p-value.<\/li>\r\n \t<li>The newer TI-84 calculators have in <code>STAT TESTS<\/code> the test <code>Chi2 GOF<\/code>.<\/li>\r\n \t<li>To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list.<\/li>\r\n \t<li>Press <code>STAT TESTS<\/code> and <code>Chi2 GOF<\/code>. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press <code>calculate<\/code> or <code>draw<\/code>. Make sure you clear any lists before you start.<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<div id=\"example3\" class=\"example\" data-type=\"example\"><section>\r\n<div id=\"element-911\" class=\"exercise\" data-type=\"exercise\"><section>\r\n<div id=\"id6229434\" class=\"solution ui-solution-visible\" data-type=\"solution\"><section class=\"ui-body\">\r\n<div id=\"fs-idm137892176\" class=\"note statistics calculator\" data-type=\"note\" data-has-label=\"true\" data-label=\"\"><section>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it 3<\/h3>\r\n<div id=\"example3\" class=\"example\" data-type=\"example\"><\/div>\r\n<div id=\"fs-idm99666800\" class=\"note statistics try ui-has-child-title\" data-type=\"note\" data-has-label=\"true\" data-label=\"\"><section>\r\n<div id=\"eip-521\" class=\"exercise\" data-type=\"exercise\"><section>\r\n<div class=\"problem\" data-type=\"problem\">\r\n\r\nThe expected percentage of the number of pets students have in their homes is distributed (this is the given distribution for the student population of the United States) as in this table.\r\n<table id=\"fs-idm147763264\" summary=\"\">\r\n<thead>\r\n<tr>\r\n<th>Number of Pets<\/th>\r\n<th>Percent<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>0<\/td>\r\n<td>18<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1<\/td>\r\n<td>25<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2<\/td>\r\n<td>30<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>3<\/td>\r\n<td>18<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>4+<\/td>\r\n<td>9<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"eip-idp91504704\">A random sample of 1,000 students from the Eastern United States resulted in the data in the table below.<\/p>\r\n\r\n<table id=\"fs-idm64242624\" summary=\"\">\r\n<thead>\r\n<tr>\r\n<th>Number of Pets<\/th>\r\n<th>Frequency<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>0<\/td>\r\n<td>210<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1<\/td>\r\n<td>240<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2<\/td>\r\n<td>320<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>3<\/td>\r\n<td>140<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>4+<\/td>\r\n<td>90<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"eip-idp27106912\">At the 1% significance level, does it appear that the distribution \u201cnumber of pets\u201d of students in the Eastern United States is different from the distribution for the United States student population as a whole? What is the <em data-effect=\"italics\">p<\/em>-value?\r\n[reveal-answer q=\"270635\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"270635\"]<\/p>\r\n<em>p-<\/em>value\u00a0=\u00a00.0036\r\n\r\nWe reject the null hypothesis that the distributions are the same. There is sufficient evidence to conclude that the distribution \u201cnumber of pets\u201d of students in the Eastern United States is different from the distribution for the United States student population as a whole.\r\n<p id=\"eip-idp27106912\">[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-idm99666800\" class=\"note statistics try ui-has-child-title\" data-type=\"note\" data-has-label=\"true\" data-label=\"\"><section>\r\n<div id=\"eip-521\" class=\"exercise\" data-type=\"exercise\"><section>\r\n<div id=\"eip-idm61016960\" class=\"solution ui-solution-visible\" data-type=\"solution\" data-label=\"\"><section class=\"ui-body\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example 4<\/h3>\r\n<div class=\"example\" data-type=\"example\"><section>\r\n<div class=\"exercise\" data-type=\"exercise\"><section>\r\n<div id=\"id6230511\" class=\"problem\" data-type=\"problem\">\r\n\r\nSuppose you flip two coins 100 times. The results are 20 <em data-effect=\"italics\">HH<\/em>, 27 <em data-effect=\"italics\">HT<\/em>, 30 <em data-effect=\"italics\">TH<\/em>, and 23 <em data-effect=\"italics\">TT<\/em>. Are the coins fair? Test at a 5% significance level.\r\n\r\n[reveal-answer q=\"843316\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"843316\"]\r\n\r\nThis problem can be set up as a goodness-of-fit problem. The sample space for flipping two fair coins is {<em>HH, HT, TH, TT<\/em>}. Out of 100 flips, you would expect 25 <em>HH<\/em>, 25 <em>HT<\/em>, 25 <em>TH<\/em>, and 25 <em>TT<\/em>. This is the expected distribution. The question, \"Are the coins fair?\" is the same as saying, \"Does the distribution of the coins (20 <em>HH<\/em>, 27 <em>HT<\/em>, 30 <em>TH,<\/em> 23 <em>TT<\/em>) fit the expected distribution?\"\r\n\r\nRandom Variable: Let <em>X<\/em> = the number of heads in one flip of the two coins. <em>X<\/em> takes on the values 0, 1, 2. (There are 0, 1, or 2 heads in the flip of two coins.) Therefore, the <strong>number of cells is three<\/strong>. Since <em>X<\/em> = the number of heads, the observed frequencies are 20 (for two heads), 57 (for one head), and 23 (for zero heads or both tails). The expected frequencies are 25 (for two heads), 50 (for one head), and 25 (for zero heads or both tails). This test is right-tailed.\r\n\r\n<strong><em>H<sub>0<\/sub><\/em><\/strong><strong>:<\/strong> The coins are fair.\r\n<strong><em>H<sub>a<\/sub><\/em>:<\/strong> The coins are not fair.\r\n\r\n<strong>Distribution for the test:<\/strong>\u00a0[latex]\\chi^2_2[\/latex]\u00a0where df = 3 \u2013 1 = 2.\r\n\r\n<strong>Calculate the test statistic:<\/strong> \u03c72 = 2.14\r\n\r\n<strong>Graph:<\/strong>\r\n\r\n<img class=\"wp-image-1476 size-full aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5668\/2021\/07\/25000250\/Example-44.jpeg\" alt=\"This is a nonsymmetrical chi-square curve with values of 0 and 2.14 labeled on the horizontal axis. A vertical upward line extends from 2.14 to the curve and the region to the right of this line is shaded. The shaded area is equal to the p-value.\" width=\"487\" height=\"146\" \/>\r\n\r\n<strong>Probability statement:<\/strong> <em>p-<\/em>value = <em>P<\/em>(<em>\u03c7<\/em><sup>2<\/sup> &gt; 2.14) = 0.3430\r\n\r\n<strong>Compare <em>\u03b1<\/em> and the <em>p<\/em>-value:<\/strong>\r\n<ul>\r\n \t<li><em>\u03b1<\/em> = 0.05<\/li>\r\n \t<li><em>p<\/em>-value = 0.3430<\/li>\r\n \t<li><em>\u03b1<\/em> &lt; <em>p<\/em>-value.<\/li>\r\n<\/ul>\r\n<strong style=\"font-size: 1rem; orphans: 1; text-align: initial;\">Make a decision:<\/strong><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\"> Since \u03b1 &lt; p-value, do not reject H0.<\/span>\r\n\r\n<strong>Conclusion:<\/strong> There is insufficient evidence to conclude that the coins are not fair.\r\n\r\n<strong>Using a calculator:<\/strong>\r\n<ul>\r\n \t<li>Press STAT and ENTER. Make sure you clear lists L1, L2, and L3 if they have data in them.<\/li>\r\n \t<li>Into L1, put the observed frequencies 20, 57, 23.<\/li>\r\n \t<li>Into L2, put the expected frequencies 25, 50, 25.<\/li>\r\n \t<li>Arrow over to list L3 and up to the name area \"L3\". Enter(L1-L2)^2\/L2 and ENTER.<\/li>\r\n \t<li>Press 2nd QUIT. Press 2nd LIST and arrow over to MATH. Press 5. You should see \"sum\".Enter L3.<\/li>\r\n \t<li>Rounded to two decimal places, you should see 2.14. Press 2nd DISTR. Arrow down to 7:\u03c72cdf (or press 7). Press ENTER.<\/li>\r\n \t<li>Enter2.14,1E99,2). Rounded to four places, you should see .3430, which is the p-value.<\/li>\r\n \t<li>The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list.<\/li>\r\n \t<li>Press STAT TESTS and Chi2 GOF. Enter the list names for the Observed list and the Expected list.<\/li>\r\n \t<li>Enter the degrees of freedom and press calculate or draw. Make sure you clear any lists before you start.<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<div class=\"example\" data-type=\"example\"><section>\r\n<div class=\"exercise\" data-type=\"exercise\"><section>\r\n<div id=\"id6230530\" class=\"solution ui-solution-visible\" data-type=\"solution\"><section class=\"ui-body\">\r\n<div id=\"fs-idm114779376\" class=\"note statistics calculator\" data-type=\"note\" data-has-label=\"true\" data-label=\"\"><section>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it 4<\/h3>\r\n<div id=\"fs-idm153122384\" class=\"note statistics try finger ui-has-child-title\" data-type=\"note\" data-has-label=\"true\" data-label=\"\"><section>\r\n<div class=\"exercise\" data-type=\"exercise\"><section>\r\n<div class=\"problem\" data-type=\"problem\">\r\n\r\nStudents in a social studies class hypothesize that the literacy rates across the world for every region are 82%. This table\u00a0shows the actual literacy rates across the world broken down by region. What are the test statistic and the degrees of freedom?\r\n<table id=\"fs-idm84960960\" summary=\"\">\r\n<thead>\r\n<tr>\r\n<th>MDG Region<\/th>\r\n<th>Adult Literacy Rate (%)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>Developed Regions<\/td>\r\n<td>99.0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Commonwealth of Independent States<\/td>\r\n<td>99.5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Northern Africa<\/td>\r\n<td>67.3<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Sub-Saharan Africa<\/td>\r\n<td>62.5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Latin America and the Caribbean<\/td>\r\n<td>91.0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Eastern Asia<\/td>\r\n<td>93.8<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Southern Asia<\/td>\r\n<td>61.9<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>South-Eastern Asia<\/td>\r\n<td>91.9<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Western Asia<\/td>\r\n<td>84.5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Oceania<\/td>\r\n<td>66.4<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div id=\"eip-idm38888864\" class=\"solution ui-solution-visible\" data-type=\"solution\" data-label=\"\"><section class=\"ui-body\">[reveal-answer q=\"957595\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"957595\"]degrees of freedom = 9\r\n<p id=\"eip-idp94879552\">chi<sup>2<\/sup> test statistic = 26.38<\/p>\r\n<a href=\"https:\/\/courses.candelalearning.com\/masterystats1x6xmaster\/wp-content\/uploads\/sites\/419\/2015\/06\/CNX_Stats_C11_M04_tryit001.jpg\"><img class=\"size-full wp-image-738 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214758\/CNX_Stats_C11_M04_tryit001.jpg\" alt=\"This is a nonsymmetric chi-square curve with df = 9. The values 0, 9, and 26.38 are labeled on the horizontal axis. The value 9 coincides with the peak of the curve. A vertical upward line extends from 26.38 to the curve, and the region to the right of this line is shaded. The shaded area is equal to the p-value.\" width=\"487\" height=\"187\" \/><\/a>\r\n\r\n<strong>Using a calculator:<\/strong>\r\n<ul>\r\n \t<li id=\"eip-idp901200\">Press <code>STAT<\/code> and <code>ENTER<\/code>. Make sure you clear lists <code>L1, L2,<\/code> and <code>L3<\/code> if they have data in them.<\/li>\r\n \t<li>Into L1, put the observed frequencies <code>99, 99.5, 67.3, 62.5, 91, 93.8, 61.9, 91.9, 84.5, 66.4<\/code>.<\/li>\r\n \t<li>Into <code>L2<\/code>, put the expected frequencies <code>82, 82, 82, 82, 82, 82, 82, 82, 82, 82<\/code>.<\/li>\r\n \t<li>Arrow over to list <code>L3<\/code> and up to the name area <code>L3<\/code>. Enter <code>(L1-L2)^2\/L2<\/code> and<code>ENTER<\/code>.<\/li>\r\n \t<li>Press <code>2nd QUIT<\/code>. Press <code>2nd LIST,<\/code> and arrow over to <code>MATH<\/code>. Press <code>5<\/code>. You should see<code>\"sum\"<\/code>. <code>Enter L3<\/code>.<\/li>\r\n \t<li>Rounded to two decimal places, you should see <code>26.38<\/code>. Press <code>2nd DISTR<\/code>.<\/li>\r\n \t<li>Arrow down to <code>7:\u03c72cdf<\/code> (or press <code>7<\/code>). Press <code>ENTER<\/code>. Enter <code>26.38,1E99,9)<\/code>. Rounded to four places, you should see<code>.0018<\/code>, which is the <em data-effect=\"italics\">p<\/em>-value.<\/li>\r\n \t<li>The newer TI-84 calculators have in <code>STAT TESTS<\/code> the test <code>Chi2 GOF<\/code>. To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list.<\/li>\r\n \t<li>Press <code>STAT TESTS<\/code> and <code>Chi2 GOF<\/code>. Enter the list names for the Observed list and the Expected list.<\/li>\r\n \t<li>Enter the degrees of freedom and press<code>calculate<\/code> or <code>draw<\/code>. Make sure you clear any lists before you start.<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<section>\n<ul id=\"element-377\" data-bullet-style=\"bullet\">\n<li>Conduct a chi-square goodness-of-fit test and interpret the conclusion in context<\/li>\n<\/ul>\n<\/section>\n<\/div>\n<p>In this type of hypothesis test, you determine whether the data &#8220;fit&#8221; a particular distribution or not. For example, you may suspect your unknown data fit a binomial distribution. You use a chi-square test (meaning the distribution for the hypothesis test is chi-square) to determine if there is a fit or not. <strong>The null and the alternative hypotheses for this test may be written in sentences or may be stated as equations or inequalities.<\/strong><\/p>\n<p>The test statistic for a goodness-of-fit test is:<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle{\\sum_{k}}\\frac{{({O}-{E})}^{{2}}}{{E}}[\/latex]<\/p>\n<p>where:<\/p>\n<ul>\n<li><em>O<\/em> = <strong>observed values<\/strong> (data)<\/li>\n<li><em>E<\/em> = <strong>expected values<\/strong> (from theory)<\/li>\n<li><em>k<\/em> = the number of different data cells or categories<\/li>\n<\/ul>\n<p><strong>The observed values are the data values and the expected values are the values you would expect to get if the null hypothesis were true.<\/strong> There are\u00a0<em>n <\/em>terms of the form [latex]\\displaystyle\\frac{{({O}-{E})}^{{2}}}{{E}}[\/latex].<\/p>\n<p>The number of degrees of freedom is\u00a0<em>df<\/em> = (number of categories \u2013 1).<\/p>\n<p><strong>The goodness-of-fit test is almost always right-tailed.<\/strong> If the observed values and the corresponding expected values are not close to each other, then the test statistic can get very large and will be way out in the right tail of the chi-square curve.<\/p>\n<div class=\"textbox examples\">\n<h3>Recall: Expected Value<\/h3>\n<p>The <strong>expected value<\/strong> is often referred to as the &#8220;long-term&#8221; average or mean. This means that over the long term of doing an experiment over and over, you would expect this average.<\/p>\n<\/div>\n<p><strong>Note:\u00a0<\/strong>The expected value for each cell needs to be at least five in order for you to use this test.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Pearson&#39;s chi square test (goodness of fit) | Probability and Statistics | Khan Academy\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/2QeDRsxSF9M?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox exercises\">\n<h3>Example 1<\/h3>\n<p>Absenteeism of college students from math classes is a major concern to math instructors because missing class appears to increase the drop rate. Suppose that a study was done to determine if the actual student absenteeism rate follows faculty perception. The faculty expected that a group of 100 students would miss class according to this table.<\/p>\n<table>\n<thead>\n<tr>\n<th>Number of absences per term<\/th>\n<th>Expected number of students<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>0\u20132<\/td>\n<td>50<\/td>\n<\/tr>\n<tr>\n<td>3\u20135<\/td>\n<td>30<\/td>\n<\/tr>\n<tr>\n<td>6\u20138<\/td>\n<td>12<\/td>\n<\/tr>\n<tr>\n<td>9\u201311<\/td>\n<td>6<\/td>\n<\/tr>\n<tr>\n<td>12+<\/td>\n<td>2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>A random survey across all mathematics courses was then done to determine the actual number\u00a0<strong>(observed)<\/strong> of absences in a course. The chart in this table displays the results of that survey.<\/p>\n<table>\n<thead>\n<tr>\n<th>Number of absences per term<\/th>\n<th>Actual number of students<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>0\u20132<\/td>\n<td>35<\/td>\n<\/tr>\n<tr>\n<td>3\u20135<\/td>\n<td>40<\/td>\n<\/tr>\n<tr>\n<td>6\u20138<\/td>\n<td>20<\/td>\n<\/tr>\n<tr>\n<td>9\u201311<\/td>\n<td>1<\/td>\n<\/tr>\n<tr>\n<td>12+<\/td>\n<td>4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Determine the null and alternative hypotheses needed to conduct a goodness-of-fit test.<\/p>\n<p><em>H<sub data-redactor-tag=\"sub\">0<\/sub><\/em>: Student absenteeism <strong>fits<\/strong> faculty perception.<\/p>\n<p>The alternative hypothesis is the opposite of the null hypothesis.<\/p>\n<p><em>H<sub data-redactor-tag=\"sub\">a<\/sub><\/em>: Student absenteeism <strong>does not fit<\/strong> faculty perception.<\/p>\n<ol>\n<li>Can you use the information as it appears in the charts to conduct the goodness-of-fit test?<\/li>\n<li>What is the number of degrees of freedom (<em>df<\/em>)?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q833283\">Show Answer<\/span><\/p>\n<div id=\"q833283\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li><strong>No<\/strong><strong>. <\/strong>Notice that the expected number of absences for the &#8220;12+&#8221; entry is less than five (it is two). Combine that group with the &#8220;9\u201311&#8221; group to create new tables where the number of students for each entry is at least five. The new results are in the two tables below.<br \/>\n<table>\n<thead>\n<tr>\n<th>Number of absences per term<\/th>\n<th>Expected number of students<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>0\u20132<\/td>\n<td>50<\/td>\n<\/tr>\n<tr>\n<td>3\u20135<\/td>\n<td>30<\/td>\n<\/tr>\n<tr>\n<td>6\u20138<\/td>\n<td>12<\/td>\n<\/tr>\n<tr>\n<td>9+<\/td>\n<td>8<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table>\n<thead>\n<tr>\n<th>Number of absences per term<\/th>\n<th>Actual number of students<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>0\u20132<\/td>\n<td>35<\/td>\n<\/tr>\n<tr>\n<td>3\u20135<\/td>\n<td>40<\/td>\n<\/tr>\n<tr>\n<td>6\u20138<\/td>\n<td>20<\/td>\n<\/tr>\n<tr>\n<td>9+<\/td>\n<td>5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<li>There are four &#8220;cells&#8221; or categories in each of the new tables.<br \/>\n<em>df<\/em> = number of cells \u2013 1 = 4 \u2013 1 = 3<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it 1<\/h3>\n<p>A factory manager needs to understand how many products are defective versus how many are produced. The number of expected defects is listed in the table.<\/p>\n<table>\n<thead>\n<tr>\n<th>Number produced<\/th>\n<th>Number defective<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>0\u2013100<\/td>\n<td>5<\/td>\n<\/tr>\n<tr>\n<td>101\u2013200<\/td>\n<td>6<\/td>\n<\/tr>\n<tr>\n<td>201\u2013300<\/td>\n<td>7<\/td>\n<\/tr>\n<tr>\n<td>301\u2013400<\/td>\n<td>8<\/td>\n<\/tr>\n<tr>\n<td>401\u2013500<\/td>\n<td>10<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>A random sample was taken to determine the actual number of defects. This table shows the results of the survey.<\/p>\n<table>\n<thead>\n<tr>\n<th>Number produced<\/th>\n<th>Number defective<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>0\u2013100<\/td>\n<td>5<\/td>\n<\/tr>\n<tr>\n<td>101\u2013200<\/td>\n<td>7<\/td>\n<\/tr>\n<tr>\n<td>201\u2013300<\/td>\n<td>8<\/td>\n<\/tr>\n<tr>\n<td>301\u2013400<\/td>\n<td>9<\/td>\n<\/tr>\n<tr>\n<td>401\u2013500<\/td>\n<td>11<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>State the null and alternative hypotheses needed to conduct a goodness-of-fit test, and state the degrees of freedom.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q473884\">Show Answer<\/span><\/p>\n<div id=\"q473884\" class=\"hidden-answer\" style=\"display: none\">\n<p><em>H<sub data-redactor-tag=\"sub\">0<\/sub><\/em>: The number of defaults fits expectations.<\/p>\n<p><em>H<sub data-redactor-tag=\"sub\">a<\/sub><\/em>: The number of defaults does not fit expectations.<\/p>\n<p><em>df<\/em> = 4<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Recall: ORDER OF OPERATIONS<\/h3>\n<div style=\"text-align: left;\">\n<table style=\"border-collapse: collapse; width: 100%; height: 36px;\">\n<tbody>\n<tr style=\"height: 12px;\">\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\"><strong>Please<\/strong><\/td>\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\"><strong>Excuse<\/strong><\/td>\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\"><strong>My<\/strong><\/td>\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\"><strong>Dear<\/strong><\/td>\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\"><strong>Aunt<\/strong><\/td>\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\"><strong>Sally<\/strong><\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\">parentheses<\/td>\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\">exponents<\/td>\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\">multiplication<\/td>\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\">division<\/td>\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\">addition<\/td>\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\">subtraction<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\">[latex]( \\ )[\/latex]<\/td>\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\">[latex]x^2[\/latex]<\/td>\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\" colspan=\"2\">[latex]\\times \\ \\mathrm{or} \\ \\div[\/latex]<\/td>\n<td style=\"width: 16.6667%; height: 12px; text-align: center;\" colspan=\"2\">[latex]+ \\ \\mathrm{or} \\ -[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2>To calculate the test statistic for a goodness-of-fit test follow the following steps: you will end up using the formula [latex]\\frac{(O-E)^2}{E}[\/latex] over and over.<\/h2>\n<p>Step 1: Pick one of the observed values and then subtract the expected value, this finds the distance between the observed value and what is expected. [latex](O-E)[\/latex]<\/p>\n<p>Step 2: Square this difference. [latex](O-E)^2[\/latex]<\/p>\n<p>Step 3: Divide step 2 by the expected value. [latex]\\frac{(O-E)^2}{E}[\/latex]<\/p>\n<p>Step 4: Repeat steps 1 &#8211; 3 with every observed value.<\/p>\n<p>Step 5: Add every value calculated in step 4, in order words find the sum of the difference between each observed value and the expected value, squared and divided by the expected value.<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example 2<\/h3>\n<p>Employers want to know which days of the week employees are absent in a five-day work week. Most employers would like to believe that employees are absent equally during the week. Suppose a random sample of 60 managers were asked on which day of the week they had the highest number of employee absences. The results were distributed as in the table below. For the population of employees, do the days for the highest number of absences occur with equal frequencies during a five-day work week? Test at a 5% significance level.<\/p>\n<p>Day of the Week Employees were Most Absent<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>Monday<\/th>\n<th>Tuesday<\/th>\n<th>Wednesday<\/th>\n<th>Thursday<\/th>\n<th>Friday<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Number of Absences<\/td>\n<td>15<\/td>\n<td>12<\/td>\n<td>9<\/td>\n<td>9<\/td>\n<td>15<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q229341\">Show Answer<\/span><\/p>\n<div id=\"q229341\" class=\"hidden-answer\" style=\"display: none\">\n<p>The null and alternative hypotheses are:<\/p>\n<ul>\n<li><em>H<sub data-redactor-tag=\"sub\">0<\/sub><\/em>: The absent days occur with equal frequencies, that is, they fit a uniform distribution.<\/li>\n<li><em>H<sub data-redactor-tag=\"sub\">a<\/sub><\/em>: The absent days occur with unequal frequencies, that is, they do not fit a uniform distribution.<\/li>\n<\/ul>\n<p>If the absent days occur with equal frequencies, then, out of 60 absent days (the total in the sample: 15 + 12 + 9 + 9 + 15 = 60), there would be 12 absences on Monday, 12 on Tuesday, 12 on Wednesday, 12 on Thursday, and 12 on Friday. These numbers are the\u00a0<strong>expected<\/strong> (<em>E<\/em>) values. The values in the table are the <strong>observed <\/strong>(<em>O<\/em>) values or data.<\/p>\n<p>This time, calculate the\u00a0<em>\u03c7<\/em><sup>2<\/sup> test statistic by hand. Make a chart with the following headings and fill in the columns:<\/p>\n<ul>\n<li>Expected (<em>E<\/em>) values (12, 12, 12, 12, 12)<\/li>\n<li>Observed (<em>O<\/em>) values (15, 12, 9, 9, 15)<\/li>\n<li>(<em>O<\/em> \u2013 <em>E<\/em>)<\/li>\n<li>(<em>O<\/em> \u2013 <em>E<\/em>)<sup>2<\/sup><\/li>\n<li>[latex]\\displaystyle\\frac{{({O}-{E})}^{{2}}}{{E}}[\/latex]<\/li>\n<\/ul>\n<p>Now add (sum) the last column. The sum is three. This is the\u00a0<em>\u03c7<sup data-redactor-tag=\"sup\">2<\/sup><\/em> test statistic.<\/p>\n<p>To find the\u00a0<em>p<\/em>-value, calculate <em>P<\/em>(<em>\u03c7<\/em><sup>2<\/sup> &gt; 3). This test is right-tailed. (Use a computer or calculator to find the <em>p<\/em>-value. You should get <em>p<\/em>-value = 0.5578.)<\/p>\n<p>The\u00a0<em>dfs<\/em> are the number of cells \u2013 1 = 5 \u2013 1 = 4<\/p>\n<header>\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR:<\/span><\/h4>\n<\/header>\n<ul>\n<li>Press\u00a0<code>2nd DISTR<\/code>.<\/li>\n<li>Arrow down to\u00a0<code>\u03c72cdf<\/code>.<\/li>\n<li>Press <code>ENTER<\/code>.<\/li>\n<li>Enter<code>(3,10^99,4)<\/code>.<\/li>\n<\/ul>\n<p>Rounded to four decimal places, you should see 0.5578, which is the p-value.<\/p>\n<p>Next, complete a graph like the following one with the proper labeling and shading. (You should shade the right tail.)<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2212 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5668\/2021\/07\/06131052\/039a6a3fbbc667b9b7ec97f2dbb5b5840ef6437d.jpeg\" alt=\"This is a blank nonsymmetrical chi-square curve for the test statistic of the days of the week absent.\" width=\"487\" height=\"229\" \/><\/p>\n<p>The decision is not to reject the null hypothesis.<\/p>\n<p><strong>Conclusion: <\/strong>At a 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the absent days do not occur with equal frequencies.<\/p>\n<p><strong>Using a calculator:\u00a0<\/strong>TI-83+ and some TI-84 calculators do not have a special program for the test statistic for the goodness-of-fit test. The next example has the calculator instructions.<\/p>\n<ul>\n<li>The newer TI-84 calculators have in\u00a0<code>STAT TESTS<\/code> the test <code>Chi2 GOF<\/code>.<\/li>\n<li>To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list.<\/li>\n<li>Press <code>STAT TESTS\u00a0<\/code>and <code>Chi2 GOF<\/code>.<\/li>\n<li>Enter the list names for the Observed list and the Expected list.<\/li>\n<li>Enter the degrees of freedom and press\u00a0<code>calculate<\/code> or <code>draw<\/code>.<\/li>\n<li>Make sure you clear any lists before you start.<\/li>\n<li>To Clear Lists in the calculators: Go into\u00a0<code>STAT EDIT<\/code> and arrow up to the list name area of the particular list. Press\u00a0<code>CLEAR<\/code> and then arrow down. The list will be cleared.<\/li>\n<li>Alternatively, you can press <code>STAT<\/code> and press 4 (for\u00a0<code>ClrList<\/code>). Enter the list name and press <code>ENTER<\/code>.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it 2<\/h3>\n<p>Teachers want to know which night each week their students are doing most of their homework. Most teachers think that students do homework equally throughout the week. Suppose a random sample of 56 students were asked on which night of the week they did the most homework. The results were distributed as in the table.<\/p>\n<table>\n<thead>\n<tr>\n<th><\/th>\n<th>Sunday<\/th>\n<th>Monday<\/th>\n<th>Tuesday<\/th>\n<th>Wednesday<\/th>\n<th>Thursday<\/th>\n<th>Friday<\/th>\n<th>Saturday<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Number of Students<\/td>\n<td>11<\/td>\n<td>8<\/td>\n<td>10<\/td>\n<td>7<\/td>\n<td>10<\/td>\n<td>5<\/td>\n<td>5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>From the population of students, do the nights for the highest number of students doing the majority of their homework occur with equal frequencies during a week? What type of hypothesis test should you use?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q478005\">Show Answer<\/span><\/p>\n<div id=\"q478005\" class=\"hidden-answer\" style=\"display: none\">\n<p><em>df<\/em> = 6<\/p>\n<p><em>p<\/em>-value = 0.6093<\/p>\n<p>We decline to reject the null hypothesis. There is not enough evidence to support that students do not do the majority of their homework equally throughout the week.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example 3<\/h3>\n<p>One study indicates that the number of televisions that American families have is distributed (this is the\u00a0<strong>given<\/strong> distribution for the American population) as in the table.<\/p>\n<table>\n<thead>\n<tr>\n<th>Number of Televisions<\/th>\n<th>Percent<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>0<\/td>\n<td>10<\/td>\n<\/tr>\n<tr>\n<td>1<\/td>\n<td>16<\/td>\n<\/tr>\n<tr>\n<td>2<\/td>\n<td>55<\/td>\n<\/tr>\n<tr>\n<td>3<\/td>\n<td>11<\/td>\n<\/tr>\n<tr>\n<td>4+<\/td>\n<td>8<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The table contains expected (<em>E<\/em>) percents.<\/p>\n<p>A random sample of 600 families in the far western United States resulted in the data in this table.<\/p>\n<table>\n<thead>\n<tr>\n<th>Number of Televisions<\/th>\n<th>Frequency<\/th>\n<\/tr>\n<\/thead>\n<tfoot>\n<tr>\n<td>Total = 600<\/td>\n<\/tr>\n<\/tfoot>\n<tbody>\n<tr>\n<td>0<\/td>\n<td>66<\/td>\n<\/tr>\n<tr>\n<td>1<\/td>\n<td>119<\/td>\n<\/tr>\n<tr>\n<td>2<\/td>\n<td>340<\/td>\n<\/tr>\n<tr>\n<td>3<\/td>\n<td>60<\/td>\n<\/tr>\n<tr>\n<td>4+<\/td>\n<td>15<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The table contains observed (<em>O<\/em>) frequency values.<\/p>\n<p>At the 1% significance level, does it appear that the distribution &#8220;number of televisions&#8221; of far western United States families is different from the distribution for the American population as a whole?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q946383\">Show Answer<\/span><\/p>\n<div id=\"q946383\" class=\"hidden-answer\" style=\"display: none\">\n<p>This problem asks you to test whether the far western United States families&#8217; distribution fits the distribution of the American families. This test is always right-tailed.<\/p>\n<p>The first table contains expected percentages. To get expected (<em>E<\/em>) frequencies, multiply the percentage by 600. The expected frequencies are shown in this table.<\/p>\n<table>\n<thead>\n<tr>\n<th>Number of Televisions<\/th>\n<th>Percent<\/th>\n<th>Expected Frequency<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>0<\/td>\n<td>10<\/td>\n<td>(0.10)(600) = 60<\/td>\n<\/tr>\n<tr>\n<td>1<\/td>\n<td>16<\/td>\n<td>(0.16)(600) = 96<\/td>\n<\/tr>\n<tr>\n<td>2<\/td>\n<td>55<\/td>\n<td>(0.55)(600) = 330<\/td>\n<\/tr>\n<tr>\n<td>3<\/td>\n<td>11<\/td>\n<td>(0.11)(600) = 66<\/td>\n<\/tr>\n<tr>\n<td>over 3<\/td>\n<td>8<\/td>\n<td>(0.08)(600) = 48<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Therefore, the expected frequencies are 60, 96, 330, 66, and 48. In the TI calculators, you can let the calculator do the math. For example, instead of 60, enter 0.10*600.<\/p>\n<p><em>H<sub>0<\/sub><\/em>: The &#8220;number of televisions&#8221; distribution of far western United States families is the same as the &#8220;number of televisions&#8221; distribution of the American population.<\/p>\n<p><em>H<sub>a<\/sub><\/em>: The &#8220;number of televisions&#8221; distribution of far western United States families is different from the &#8220;number of televisions&#8221; distribution of the American population.<\/p>\n<p>Distribution for the test:\u00a0[latex]\\displaystyle\\chi^{2}_{4}[\/latex] where <i>df<\/i> = (the number of cells) \u2013 1 = 5 \u2013 1 = 4.<\/p>\n<p><strong>Note<\/strong>:\u00a0[latex]df\\neq600-1[\/latex]<\/p>\n<div id=\"example3\" class=\"example\" data-type=\"example\">\n<section>\n<div id=\"element-911\" class=\"exercise\" data-type=\"exercise\">\n<section>\n<div id=\"id6229434\" class=\"solution ui-solution-visible\" data-type=\"solution\">\n<section class=\"ui-body\">\n<p id=\"element-65\"><strong>Calculate the test statistic:\u00a0<\/strong><em data-effect=\"italics\">\u03c7<\/em><sup>2<\/sup> = 29.65<\/p>\n<p id=\"element-280\"><strong>Graph:<\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-2215 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5668\/2021\/07\/06131244\/c64bbffd1bc6d84f526b8cf8d2ba05ce360d752e.jpeg\" alt=\"This is a nonsymmetric chi-square curve with values of 0, 4, and 29.65 labeled on the horizontal axis. The value 4 coincides with the peak of the curve. A vertical upward line extends from 29.65 to the curve, and the region to the right of this line is shaded. The shaded area is equal to the p-value.\" width=\"487\" height=\"146\" \/><\/p>\n<p><strong>Probability statement:\u00a0<\/strong><em data-effect=\"italics\">p<\/em>-value = <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">\u03c7<\/em><sup>2<\/sup> &gt; 29.65) = 0.000006<\/p>\n<p><strong>Compare <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value:<\/strong><\/p>\n<ul>\n<li><em data-effect=\"italics\">\u03b1<\/em> = 0.01<\/li>\n<li><em data-effect=\"italics\">p<\/em>-value = 0.000006<\/li>\n<li>So, <em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">\u03b1<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\"> &gt; <\/span><em style=\"font-size: 1rem; orphans: 1; text-align: initial;\" data-effect=\"italics\">p<\/em><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">-value.<\/span><\/li>\n<\/ul>\n<p><strong>Make a decision:<\/strong> Since <em data-effect=\"italics\">\u03b1<\/em> &gt; <em data-effect=\"italics\">p<\/em>-value, reject <em data-effect=\"italics\">H<sub>o<\/sub><\/em>.<\/p>\n<p>This means you reject the belief that the distribution for the far western states is the same as that of the American population as a whole.<\/p>\n<p><strong>Conclusion:<\/strong> At the 1% significance level, from the data, there is sufficient evidence to conclude that the &#8220;number of televisions&#8221; distribution for the far western United States is different from the &#8220;number of televisions&#8221; distribution for the American population as a whole.<\/p>\n<p><strong>Using a calculator:<\/strong><\/p>\n<ul>\n<li>Press <code>STAT<\/code> and <code>ENTER<\/code>.<\/li>\n<li>Make sure to clear lists <code>L1<\/code>, <code>L2<\/code>, and <code>L3<\/code> if they have data in them (see the note at the end of Example 2).<\/li>\n<li>Into <code>L1<\/code>, put the observed frequencies <code>66<\/code>, <code>119<\/code>,<code>349<\/code>, <code>60<\/code>, <code>15<\/code>.<\/li>\n<li>Into <code>L2<\/code>, put the expected frequencies <code>.10*600, .16*600<\/code>, <code>.55*600<\/code>,<code>.11*600<\/code>, <code>.08*600<\/code>.<\/li>\n<li>Arrow over to list <code>L3<\/code> and up to the name area <code>L3<\/code>. Enter <code>(L1-L2)^2\/L2<\/code> and <code>ENTER<\/code>.<\/li>\n<li>Press <code>2nd QUIT<\/code>. Press <code>2nd LIST<\/code> and arrow over to <code>MATH<\/code>. Press <code>5<\/code>. You should see <code>\"sum\" (Enter L3)<\/code>.<\/li>\n<li>Rounded to 2 decimal places, you should see <code>29.65<\/code>. Press <code>2nd DISTR<\/code>. Press <code>7<\/code> or Arrow down to <code>7:\u03c72cdf<\/code> and press <code>ENTER<\/code>.<\/li>\n<li>Enter <code>(29.65,1E99,4)<\/code>. Rounded to four places, you should see <code>5.77E-6 = .000006<\/code>\u00a0(rounded to six decimal places), which is the p-value.<\/li>\n<li>The newer TI-84 calculators have in <code>STAT TESTS<\/code> the test <code>Chi2 GOF<\/code>.<\/li>\n<li>To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list.<\/li>\n<li>Press <code>STAT TESTS<\/code> and <code>Chi2 GOF<\/code>. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press <code>calculate<\/code> or <code>draw<\/code>. Make sure you clear any lists before you start.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<div id=\"example3\" class=\"example\" data-type=\"example\">\n<section>\n<div id=\"element-911\" class=\"exercise\" data-type=\"exercise\">\n<section>\n<div id=\"id6229434\" class=\"solution ui-solution-visible\" data-type=\"solution\">\n<section class=\"ui-body\">\n<div id=\"fs-idm137892176\" class=\"note statistics calculator\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<section>\n<div class=\"textbox key-takeaways\">\n<h3>try it 3<\/h3>\n<div id=\"example3\" class=\"example\" data-type=\"example\"><\/div>\n<div id=\"fs-idm99666800\" class=\"note statistics try ui-has-child-title\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<section>\n<div id=\"eip-521\" class=\"exercise\" data-type=\"exercise\">\n<section>\n<div class=\"problem\" data-type=\"problem\">\n<p>The expected percentage of the number of pets students have in their homes is distributed (this is the given distribution for the student population of the United States) as in this table.<\/p>\n<table id=\"fs-idm147763264\" summary=\"\">\n<thead>\n<tr>\n<th>Number of Pets<\/th>\n<th>Percent<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>0<\/td>\n<td>18<\/td>\n<\/tr>\n<tr>\n<td>1<\/td>\n<td>25<\/td>\n<\/tr>\n<tr>\n<td>2<\/td>\n<td>30<\/td>\n<\/tr>\n<tr>\n<td>3<\/td>\n<td>18<\/td>\n<\/tr>\n<tr>\n<td>4+<\/td>\n<td>9<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"eip-idp91504704\">A random sample of 1,000 students from the Eastern United States resulted in the data in the table below.<\/p>\n<table id=\"fs-idm64242624\" summary=\"\">\n<thead>\n<tr>\n<th>Number of Pets<\/th>\n<th>Frequency<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>0<\/td>\n<td>210<\/td>\n<\/tr>\n<tr>\n<td>1<\/td>\n<td>240<\/td>\n<\/tr>\n<tr>\n<td>2<\/td>\n<td>320<\/td>\n<\/tr>\n<tr>\n<td>3<\/td>\n<td>140<\/td>\n<\/tr>\n<tr>\n<td>4+<\/td>\n<td>90<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"eip-idp27106912\">At the 1% significance level, does it appear that the distribution \u201cnumber of pets\u201d of students in the Eastern United States is different from the distribution for the United States student population as a whole? What is the <em data-effect=\"italics\">p<\/em>-value?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q270635\">Show Answer<\/span><\/p>\n<div id=\"q270635\" class=\"hidden-answer\" style=\"display: none\">\n<p><em>p-<\/em>value\u00a0=\u00a00.0036<\/p>\n<p>We reject the null hypothesis that the distributions are the same. There is sufficient evidence to conclude that the distribution \u201cnumber of pets\u201d of students in the Eastern United States is different from the distribution for the United States student population as a whole.<\/p>\n<p id=\"eip-idp27106912\"><\/div>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-idm99666800\" class=\"note statistics try ui-has-child-title\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<section>\n<div id=\"eip-521\" class=\"exercise\" data-type=\"exercise\">\n<section>\n<div id=\"eip-idm61016960\" class=\"solution ui-solution-visible\" data-type=\"solution\" data-label=\"\">\n<section class=\"ui-body\">\n<div class=\"textbox exercises\">\n<h3>Example 4<\/h3>\n<div class=\"example\" data-type=\"example\">\n<section>\n<div class=\"exercise\" data-type=\"exercise\">\n<section>\n<div id=\"id6230511\" class=\"problem\" data-type=\"problem\">\n<p>Suppose you flip two coins 100 times. The results are 20 <em data-effect=\"italics\">HH<\/em>, 27 <em data-effect=\"italics\">HT<\/em>, 30 <em data-effect=\"italics\">TH<\/em>, and 23 <em data-effect=\"italics\">TT<\/em>. Are the coins fair? Test at a 5% significance level.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q843316\">Show Answer<\/span><\/p>\n<div id=\"q843316\" class=\"hidden-answer\" style=\"display: none\">\n<p>This problem can be set up as a goodness-of-fit problem. The sample space for flipping two fair coins is {<em>HH, HT, TH, TT<\/em>}. Out of 100 flips, you would expect 25 <em>HH<\/em>, 25 <em>HT<\/em>, 25 <em>TH<\/em>, and 25 <em>TT<\/em>. This is the expected distribution. The question, &#8220;Are the coins fair?&#8221; is the same as saying, &#8220;Does the distribution of the coins (20 <em>HH<\/em>, 27 <em>HT<\/em>, 30 <em>TH,<\/em> 23 <em>TT<\/em>) fit the expected distribution?&#8221;<\/p>\n<p>Random Variable: Let <em>X<\/em> = the number of heads in one flip of the two coins. <em>X<\/em> takes on the values 0, 1, 2. (There are 0, 1, or 2 heads in the flip of two coins.) Therefore, the <strong>number of cells is three<\/strong>. Since <em>X<\/em> = the number of heads, the observed frequencies are 20 (for two heads), 57 (for one head), and 23 (for zero heads or both tails). The expected frequencies are 25 (for two heads), 50 (for one head), and 25 (for zero heads or both tails). This test is right-tailed.<\/p>\n<p><strong><em>H<sub>0<\/sub><\/em><\/strong><strong>:<\/strong> The coins are fair.<br \/>\n<strong><em>H<sub>a<\/sub><\/em>:<\/strong> The coins are not fair.<\/p>\n<p><strong>Distribution for the test:<\/strong>\u00a0[latex]\\chi^2_2[\/latex]\u00a0where df = 3 \u2013 1 = 2.<\/p>\n<p><strong>Calculate the test statistic:<\/strong> \u03c72 = 2.14<\/p>\n<p><strong>Graph:<\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1476 size-full aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5668\/2021\/07\/25000250\/Example-44.jpeg\" alt=\"This is a nonsymmetrical chi-square curve with values of 0 and 2.14 labeled on the horizontal axis. A vertical upward line extends from 2.14 to the curve and the region to the right of this line is shaded. The shaded area is equal to the p-value.\" width=\"487\" height=\"146\" \/><\/p>\n<p><strong>Probability statement:<\/strong> <em>p-<\/em>value = <em>P<\/em>(<em>\u03c7<\/em><sup>2<\/sup> &gt; 2.14) = 0.3430<\/p>\n<p><strong>Compare <em>\u03b1<\/em> and the <em>p<\/em>-value:<\/strong><\/p>\n<ul>\n<li><em>\u03b1<\/em> = 0.05<\/li>\n<li><em>p<\/em>-value = 0.3430<\/li>\n<li><em>\u03b1<\/em> &lt; <em>p<\/em>-value.<\/li>\n<\/ul>\n<p><strong style=\"font-size: 1rem; orphans: 1; text-align: initial;\">Make a decision:<\/strong><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\"> Since \u03b1 &lt; p-value, do not reject H0.<\/span><\/p>\n<p><strong>Conclusion:<\/strong> There is insufficient evidence to conclude that the coins are not fair.<\/p>\n<p><strong>Using a calculator:<\/strong><\/p>\n<ul>\n<li>Press STAT and ENTER. Make sure you clear lists L1, L2, and L3 if they have data in them.<\/li>\n<li>Into L1, put the observed frequencies 20, 57, 23.<\/li>\n<li>Into L2, put the expected frequencies 25, 50, 25.<\/li>\n<li>Arrow over to list L3 and up to the name area &#8220;L3&#8221;. Enter(L1-L2)^2\/L2 and ENTER.<\/li>\n<li>Press 2nd QUIT. Press 2nd LIST and arrow over to MATH. Press 5. You should see &#8220;sum&#8221;.Enter L3.<\/li>\n<li>Rounded to two decimal places, you should see 2.14. Press 2nd DISTR. Arrow down to 7:\u03c72cdf (or press 7). Press ENTER.<\/li>\n<li>Enter2.14,1E99,2). Rounded to four places, you should see .3430, which is the p-value.<\/li>\n<li>The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list.<\/li>\n<li>Press STAT TESTS and Chi2 GOF. Enter the list names for the Observed list and the Expected list.<\/li>\n<li>Enter the degrees of freedom and press calculate or draw. Make sure you clear any lists before you start.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<div class=\"example\" data-type=\"example\">\n<section>\n<div class=\"exercise\" data-type=\"exercise\">\n<section>\n<div id=\"id6230530\" class=\"solution ui-solution-visible\" data-type=\"solution\">\n<section class=\"ui-body\">\n<div id=\"fs-idm114779376\" class=\"note statistics calculator\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<section>\n<div class=\"textbox key-takeaways\">\n<h3>try it 4<\/h3>\n<div id=\"fs-idm153122384\" class=\"note statistics try finger ui-has-child-title\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<section>\n<div class=\"exercise\" data-type=\"exercise\">\n<section>\n<div class=\"problem\" data-type=\"problem\">\n<p>Students in a social studies class hypothesize that the literacy rates across the world for every region are 82%. This table\u00a0shows the actual literacy rates across the world broken down by region. What are the test statistic and the degrees of freedom?<\/p>\n<table id=\"fs-idm84960960\" summary=\"\">\n<thead>\n<tr>\n<th>MDG Region<\/th>\n<th>Adult Literacy Rate (%)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Developed Regions<\/td>\n<td>99.0<\/td>\n<\/tr>\n<tr>\n<td>Commonwealth of Independent States<\/td>\n<td>99.5<\/td>\n<\/tr>\n<tr>\n<td>Northern Africa<\/td>\n<td>67.3<\/td>\n<\/tr>\n<tr>\n<td>Sub-Saharan Africa<\/td>\n<td>62.5<\/td>\n<\/tr>\n<tr>\n<td>Latin America and the Caribbean<\/td>\n<td>91.0<\/td>\n<\/tr>\n<tr>\n<td>Eastern Asia<\/td>\n<td>93.8<\/td>\n<\/tr>\n<tr>\n<td>Southern Asia<\/td>\n<td>61.9<\/td>\n<\/tr>\n<tr>\n<td>South-Eastern Asia<\/td>\n<td>91.9<\/td>\n<\/tr>\n<tr>\n<td>Western Asia<\/td>\n<td>84.5<\/td>\n<\/tr>\n<tr>\n<td>Oceania<\/td>\n<td>66.4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div id=\"eip-idm38888864\" class=\"solution ui-solution-visible\" data-type=\"solution\" data-label=\"\">\n<section class=\"ui-body\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q957595\">Show Answer<\/span><\/p>\n<div id=\"q957595\" class=\"hidden-answer\" style=\"display: none\">degrees of freedom = 9<\/p>\n<p id=\"eip-idp94879552\">chi<sup>2<\/sup> test statistic = 26.38<\/p>\n<p><a href=\"https:\/\/courses.candelalearning.com\/masterystats1x6xmaster\/wp-content\/uploads\/sites\/419\/2015\/06\/CNX_Stats_C11_M04_tryit001.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-738 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214758\/CNX_Stats_C11_M04_tryit001.jpg\" alt=\"This is a nonsymmetric chi-square curve with df = 9. The values 0, 9, and 26.38 are labeled on the horizontal axis. The value 9 coincides with the peak of the curve. A vertical upward line extends from 26.38 to the curve, and the region to the right of this line is shaded. The shaded area is equal to the p-value.\" width=\"487\" height=\"187\" \/><\/a><\/p>\n<p><strong>Using a calculator:<\/strong><\/p>\n<ul>\n<li id=\"eip-idp901200\">Press <code>STAT<\/code> and <code>ENTER<\/code>. Make sure you clear lists <code>L1, L2,<\/code> and <code>L3<\/code> if they have data in them.<\/li>\n<li>Into L1, put the observed frequencies <code>99, 99.5, 67.3, 62.5, 91, 93.8, 61.9, 91.9, 84.5, 66.4<\/code>.<\/li>\n<li>Into <code>L2<\/code>, put the expected frequencies <code>82, 82, 82, 82, 82, 82, 82, 82, 82, 82<\/code>.<\/li>\n<li>Arrow over to list <code>L3<\/code> and up to the name area <code>L3<\/code>. Enter <code>(L1-L2)^2\/L2<\/code> and<code>ENTER<\/code>.<\/li>\n<li>Press <code>2nd QUIT<\/code>. Press <code>2nd LIST,<\/code> and arrow over to <code>MATH<\/code>. Press <code>5<\/code>. You should see<code>\"sum\"<\/code>. <code>Enter L3<\/code>.<\/li>\n<li>Rounded to two decimal places, you should see <code>26.38<\/code>. Press <code>2nd DISTR<\/code>.<\/li>\n<li>Arrow down to <code>7:\u03c72cdf<\/code> (or press <code>7<\/code>). Press <code>ENTER<\/code>. Enter <code>26.38,1E99,9)<\/code>. Rounded to four places, you should see<code>.0018<\/code>, which is the <em data-effect=\"italics\">p<\/em>-value.<\/li>\n<li>The newer TI-84 calculators have in <code>STAT TESTS<\/code> the test <code>Chi2 GOF<\/code>. To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list.<\/li>\n<li>Press <code>STAT TESTS<\/code> and <code>Chi2 GOF<\/code>. Enter the list names for the Observed list and the Expected list.<\/li>\n<li>Enter the degrees of freedom and press<code>calculate<\/code> or <code>draw<\/code>. Make sure you clear any lists before you start.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\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-296\">\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>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>Pearson&#039;s chi-square test (goodness of fit) | Probability and Statistics | Khan Academy . <strong>Authored by<\/strong>: Khan Academy. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.youtube.com\/embed\/2QeDRsxSF9M\">https:\/\/www.youtube.com\/embed\/2QeDRsxSF9M<\/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":10,"template":"","meta":{"_candela_citation":"[{\"type\":\"copyrighted_video\",\"description\":\"Pearson\\'s chi-square test (goodness of fit) | Probability and Statistics | Khan Academy \",\"author\":\"Khan Academy\",\"organization\":\"\",\"url\":\"https:\/\/www.youtube.com\/embed\/2QeDRsxSF9M\",\"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-296","chapter","type-chapter","status-publish","hentry"],"part":293,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapters\/296","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":23,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapters\/296\/revisions"}],"predecessor-version":[{"id":4057,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapters\/296\/revisions\/4057"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/parts\/293"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapters\/296\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/wp\/v2\/media?parent=296"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapter-type?post=296"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/wp\/v2\/contributor?post=296"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/wp\/v2\/license?post=296"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}