{"id":441,"date":"2016-04-21T22:43:38","date_gmt":"2016-04-21T22:43:38","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstats1xmaster\/?post_type=chapter&#038;p=441"},"modified":"2021-06-24T20:32:13","modified_gmt":"2021-06-24T20:32:13","slug":"test-of-independence","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ntcc-introstats1\/chapter\/test-of-independence\/","title":{"raw":"11.3: Test of Independence","rendered":"11.3: Test of Independence"},"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 and interpret chi-square test of independence hypothesis tests<\/li>\r\n<\/ul>\r\n<\/section><\/div>\r\nTests of independence involve using a\u00a0<strong>contingency table<\/strong> of observed (data) values.\r\n\r\nThe test statistic for a\u00a0<strong>test of independence<\/strong> is similar to that of a goodness-of-fit test:\r\n\r\n[latex]\\displaystyle{\\sum_{(i\\cdot{j})}}\\frac{{({O}-{E})}^{{2}}}{{E}}[\/latex]\r\n\r\nwhere:\r\n<ul>\r\n \t<li><em>O<\/em> = observed values<\/li>\r\n \t<li><em>E<\/em> = expected values<\/li>\r\n \t<li><em>i<\/em> = the number of rows in the table<\/li>\r\n \t<li><em>j<\/em> = the number of columns in the table<\/li>\r\n<\/ul>\r\nThere are\u00a0[latex]\\displaystyle{i}\\cdot{j}[\/latex] terms of the form [latex]\\frac{{({O}-{E})}^{{2}}}{{E}}[\/latex].\r\n\r\n<strong>A test of independence determines whether two factors are independent or not. <\/strong>\r\n\r\n<strong>Note: <\/strong>The expected value for each cell needs to be at least five in order for you to use this test.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSuppose\u00a0<em>A<\/em> = a speeding violation in the last year and <em>B<\/em> = a cell phone user while driving. If <em>A<\/em> and <em>B<\/em> are independent then <em>P<\/em>(<em>A<\/em> AND <em>B<\/em>) = <em>P<\/em>(<em>A<\/em>)<em>P<\/em>(<em>B<\/em>). <em>A <\/em>AND <em>B<\/em> is the event that a driver received a speeding violation last year and also used a cell phone while driving. Suppose, in a study of drivers who received speeding violations in the last year, and who used cell phone while driving, that 755 people were surveyed. Out of the 755, 70 had a speeding violation and 685 did not; 305 used cell phones while driving and 450 did not.\r\n\r\nLet\u00a0<em>y<\/em> = expected number of drivers who used a cell phone while driving and received speeding violations.\r\n\r\nIf\u00a0<em>A<\/em> and <em>B<\/em> are independent, then <em>P<\/em>(<em>A<\/em> AND <em>B<\/em>) = <em>P<\/em>(<em>A<\/em>)<em>P<\/em>(<em>B<\/em>). By substitution,\r\n\r\n[latex]\\displaystyle\\frac{{y}}{{755}}={(\\frac{{70}}{{755}})}{(\\frac{{305}}{{755}})}[\/latex]\r\n\r\nSolve for\r\n<em>y<\/em>: [latex]\\displaystyle{y}=\\frac{{{({70})}{({305})}}}{{755}}={28.3}[\/latex]\r\n\r\nAbout 28 people from the sample are expected to use cell phones while driving and to receive speeding violations.\r\n\r\nIn a test of independence, we state the null and alternative hypotheses in words. Since the contingency table consists of\u00a0<strong>two factors<\/strong>, the null hypothesis states that the factors are <strong>independent<\/strong> and the alternative hypothesis states that they are <strong>not independent (dependent)<\/strong>. If we do a test of independence using the example, then the null hypothesis is:\r\n\r\n<em>H<sub data-redactor-tag=\"sub\">0<\/sub><\/em>: Being a cell phone user while driving and receiving a speeding violation are independent events.\r\n\r\nIf the null hypothesis were true, we would expect about 28 people to use cell phones while driving and to receive a speeding violation.\r\n\r\n<strong>The test of independence is always right-tailed<\/strong> because of the calculation of the test statistic. If the expected and observed values are not close together, then the test statistic is very large and way out in the right tail of the chi-square curve, as it is in a goodness-of-fit.\r\n\r\nThe number of degrees of freedom for the test of independence is:\r\n\r\n<em>df<\/em> = (number of columns \u2013 1)(number of rows \u2013 1)\r\n\r\nThe following formula calculates the\u00a0<strong>expected number<\/strong> (<em>E<\/em>):\r\n\r\n[latex]\\displaystyle{E}=\\frac{{{(\\text{row total})}{(\\text{column total})}}}{\\text{total number surveyed}}[\/latex]\r\n\r\n<\/div>\r\n<h3><\/h3>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nA sample of 300 students is taken. Of the students surveyed, 50 were music students, while 250 were not. Ninety-seven were on the honor roll, while 203 were not. If we assume being a music student and being on the honor roll are independent events, what is the expected number of music students who are also on the honor roll?\r\n\r\nAbout 16 students are expected to be music students and on the honor roll.\r\n\r\n<\/div>\r\n<h3><\/h3>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIn a volunteer group, adults 21 and older volunteer from one to nine hours each week to spend time with a disabled senior citizen. The program recruits among community college students, four-year college students, and nonstudents. The table below is a sample of the adult volunteers and the number of hours they volunteer per week.\r\n\r\nNumber of Hours Worked Per Week by Volunteer Type (Observed)The table contains\u00a0<strong>observed (O)<\/strong> values (data).\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>Type of Volunteer<\/th>\r\n<th>1\u20133 Hours<\/th>\r\n<th>4\u20136 Hours<\/th>\r\n<th>7\u20139 Hours<\/th>\r\n<th>Row Total<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>Community College Students<\/td>\r\n<td>111<\/td>\r\n<td>96<\/td>\r\n<td>48<\/td>\r\n<td>255<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Four-Year College Students<\/td>\r\n<td>96<\/td>\r\n<td>133<\/td>\r\n<td>61<\/td>\r\n<td>290<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Nonstudents<\/td>\r\n<td>91<\/td>\r\n<td>150<\/td>\r\n<td>53<\/td>\r\n<td>294<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Column Total<\/td>\r\n<td>298<\/td>\r\n<td>379<\/td>\r\n<td>162<\/td>\r\n<td>839<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nIs the number of hours4 volunteered\u00a0<strong>independent<\/strong> of the type of volunteer?\r\n\r\nSolution:\r\n\r\nThe\u00a0<strong>observed table<\/strong> and the question at the end of the problem, \"Is the number of hours volunteered independent of the type of volunteer?\" tell you this is a test of independence. The two factors are <strong>number of hours volunteered<\/strong> and <strong>type of volunteer<\/strong>. This test is always right-tailed.\r\n\r\n<em>H<sub data-redactor-tag=\"sub\">0<\/sub><\/em>: The number of hours volunteered is <strong>independent<\/strong> of the type of volunteer.\r\n\r\n<em>H<sub data-redactor-tag=\"sub\">a<\/sub><\/em>: The number of hours volunteered is <strong>dependent<\/strong> on the type of volunteer.\r\n\r\nThe expected result are in the table below.\r\n<table>\r\n<thead>\r\n<tr>\r\n<th colspan=\"4\">Number of Hours Worked Per Week by Volunteer Type (Expected)\r\nThe table contains expected (<em>E<\/em>) values (data)<\/th>\r\n<\/tr>\r\n<tr>\r\n<th>Type of Volunteer<\/th>\r\n<th>1\u20133 Hours<\/th>\r\n<th>4\u20136 Hours<\/th>\r\n<th>7\u20139 Hours<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>Community College Students<\/td>\r\n<td>90.57<\/td>\r\n<td>115.19<\/td>\r\n<td>49.24<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Four-Year College Students<\/td>\r\n<td>103.00<\/td>\r\n<td>131.00<\/td>\r\n<td>56.00<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Nonstudents<\/td>\r\n<td>104.42<\/td>\r\n<td>132.81<\/td>\r\n<td>56.77<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nFor example, the calculation for the expected frequency for the top left cell is\r\n\r\n[latex]\\displaystyle{E}=\\frac{{{(\\text{row total})}{(\\text{column total})}}}{\\text{total number surveyed}}=\\frac{{{({255})}{({298})}}}{{839}}={90.57}[\/latex]\r\n\r\n<strong>Calculate the test statistic:<\/strong><em>\u03c7<\/em><sup>2<\/sup> = 12.99 (calculator or computer)\r\n\r\n<strong>Distribution for the test:<\/strong> [latex]\\displaystyle\\chi^{2}_{4}[\/latex]\r\n<div class=\"example\" data-type=\"example\"><section>\r\n<div id=\"exercise102\" class=\"exercise\" data-type=\"exercise\"><section>\r\n<div id=\"id8698856\" class=\"solution ui-solution-visible\" data-type=\"solution\"><section class=\"ui-body\"><strong>Graph:<\/strong><a href=\"https:\/\/courses.candelalearning.com\/masterystats1x6xmaster\/wp-content\/uploads\/sites\/419\/2015\/06\/fig-ch11_05_01N.jpg\"><img class=\"alignnone size-full wp-image-744\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214800\/fig-ch11_05_01N.jpg\" alt=\"Nonsymmetrical chi-square curve with values of 0 and 12.99 on the x-axis representing the test statistic of number of hours worked by volunteers of different types. A vertical upward line extends from 12.99 to the curve and the area to the right of this is equal to the p-value.\" width=\"487\" height=\"146\" \/><\/a>\r\n<p id=\"element-117\"><strong>Probability statement:<\/strong><em data-effect=\"italics\">p<\/em>-value=<em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">\u03c7<sup>2<\/sup><\/em> &gt; 12.99) = 0.0113<\/p>\r\n<p id=\"element-347\"><strong>Compare <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value:<\/strong> Since no <em data-effect=\"italics\">\u03b1<\/em> is given, assume <em data-effect=\"italics\">\u03b1<\/em> = 0.05. <em data-effect=\"italics\">p<\/em>-value = 0.0113. <em data-effect=\"italics\">\u03b1<\/em> &gt; <em data-effect=\"italics\">p<\/em>-value.<\/p>\r\n<p id=\"fs-idm62402592\"><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>0<\/sub><\/em>. This means that the factors are not independent.<\/p>\r\n<strong>Conclusion:<\/strong> At a 5% level of significance, from the data, there is sufficient evidence to conclude that the number of hours volunteered and the type of volunteer are dependent on one another.\r\n\r\nFor the example in the table titled \"Number of Hours Worked Per Week by Volunteer Type (Expected),\" if there had been another type of volunteer, teenagers, what would the degrees of freedom be?\r\n<div id=\"fs-idp30286464\" class=\"note statistics calculator\" data-type=\"note\" data-has-label=\"true\" data-label=\"\"><section>\r\n<p id=\"element-854\">Press the <code>MATRX<\/code> key and arrow over to <code>EDIT<\/code>. Press <code>1:[A]<\/code>. Press <code>3 ENTER 3 ENTER<\/code>. Enter the table values by row from the table. Press <code>ENTER<\/code> after each. Press <code>2nd QUIT<\/code>. Press<code>STAT<\/code> and arrow over to <code>TESTS<\/code>. Arrow down to <code>C:\u03c72-TEST<\/code>. Press <code>ENTER<\/code>. You should see <code>Observed:[A] and Expected:[B]<\/code>. Arrow down to <code>Calculate<\/code>. Press <code>ENTER<\/code>. The test statistic is 12.9909 and the <em data-effect=\"italics\">p<\/em>-value = 0.0113. Do the procedure a second time, but arrow down to <code>Draw<\/code> instead of <code>calculate<\/code>.<\/p>\r\n\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n&nbsp;\r\n<div id=\"fs-idm24618832\" class=\"note statistics try finger ui-has-child-title\" data-type=\"note\" data-has-label=\"true\" data-label=\"\"><header>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<section>\r\n<div id=\"eip-685\" class=\"exercise\" data-type=\"exercise\"><section>\r\n<div id=\"eip-903\" class=\"problem\" data-type=\"problem\">\r\n<p id=\"eip-845\">The Bureau of Labor Statistics gathers data about employment in the United States. A sample is taken to calculate the number of U.S. citizens working in one of several industry sectors over time. The table below\u00a0shows the results:<\/p>\r\n\r\n<table id=\"fs-idp61742592\" summary=\"\">\r\n<thead>\r\n<tr>\r\n<th>Industry Sector<\/th>\r\n<th>2000<\/th>\r\n<th>2010<\/th>\r\n<th>2020<\/th>\r\n<th>Total<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>Nonagriculture wage and salary<\/td>\r\n<td>13,243<\/td>\r\n<td>13,044<\/td>\r\n<td>15,018<\/td>\r\n<td>41,305<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Goods-producing, excluding agriculture<\/td>\r\n<td>2,457<\/td>\r\n<td>1,771<\/td>\r\n<td>1,950<\/td>\r\n<td>6,178<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Services-providing<\/td>\r\n<td>10,786<\/td>\r\n<td>11,273<\/td>\r\n<td>13,068<\/td>\r\n<td>35,127<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Agriculture, forestry, fishing, and hunting<\/td>\r\n<td>240<\/td>\r\n<td>214<\/td>\r\n<td>201<\/td>\r\n<td>655<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Nonagriculture self-employed and unpaid family worker<\/td>\r\n<td>931<\/td>\r\n<td>894<\/td>\r\n<td>972<\/td>\r\n<td>2,797<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Secondary wage and salary jobs in agriculture and private household industries<\/td>\r\n<td>14<\/td>\r\n<td>11<\/td>\r\n<td>11<\/td>\r\n<td>36<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Secondary jobs as a self-employed or unpaid family worker<\/td>\r\n<td>196<\/td>\r\n<td>144<\/td>\r\n<td>152<\/td>\r\n<td>492<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Total<\/td>\r\n<td>27,867<\/td>\r\n<td>27,351<\/td>\r\n<td>31,372<\/td>\r\n<td>86,590<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"eip-idp2741344\">We want to know if the change in the number of jobs is independent of the change in years. State the null and alternative hypotheses and the degrees of freedom.<\/p>\r\n\r\n<\/div>\r\n<div id=\"eip-idm75764640\" class=\"solution ui-solution-visible\" data-type=\"solution\" data-label=\"\"><section class=\"ui-body\"><em data-effect=\"italics\">H<\/em><sub>0<\/sub>: The number of jobs is independent of the year.<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: The number of jobs is dependent on the year.<em data-effect=\"italics\">df<\/em> = 12<a href=\"https:\/\/courses.candelalearning.com\/masterystats1x6xmaster\/wp-content\/uploads\/sites\/419\/2015\/06\/CNX_Stats_C11_M05_tryit001annoN.jpg\"><img class=\"alignnone size-full wp-image-746\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214802\/CNX_Stats_C11_M05_tryit001annoN.jpg\" alt=\"Nonsymmetrical chi-square curve with values of 0, 12, and 227.73 on the x-axis. A vertical upward line extends from 227.73 to the curve and the area to the right of this is equal to the p-value. p-value = almost zero.\" width=\"487\" height=\"187\" \/><\/a><\/section><section class=\"ui-body\">\r\n<p id=\"eip-idm36315600\">Press the <code>MATRX<\/code> key and arrow over to <code>EDIT<\/code>. Press <code>1:[A]<\/code>. Press <code>3 ENTER 3 ENTER<\/code>. Enter the table values by row. Press <code>ENTER<\/code> after each. Press <code>2nd QUIT<\/code>. Press <code>STAT<\/code> and arrow over to <code>TESTS<\/code>. Arrow down to <code>C:\u03c72-TEST<\/code>. Press <code>ENTER<\/code>. You should see <code>Observed:[A] and Expected:[B]<\/code>. Arrow down to <code>Calculate<\/code>. Press <code>ENTER<\/code>. The test statistic is 227.73 and the<em data-effect=\"italics\">p<\/em>\u2212value = 5.90E - 42 = 0. Do the procedure a second time but arrow down to <code>Draw<\/code> instead of<code>calculate<\/code>.<\/p>\r\n\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n&nbsp;\r\n\r\n<\/header><\/div>\r\n<div class=\"example\" data-type=\"example\"><section>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nDe Anza College is interested in the relationship between anxiety level and the need to succeed in school. A random sample of 400 students took a test that measured anxiety level and need to succeed in school. This table\u00a0shows the results. De Anza College wants to know if anxiety level and need to succeed in school are independent events.\r\n<table summary=\"This table presents the need to succeed in school in the first column and the various anxiety levels (high to low) and the row total in the second to seventh columns. The first row is for high need, the second row is for medium need, third row is for low need, and the column total is in the fourth row.\">\r\n<thead valign=\"top\">\r\n<tr>\r\n<th colspan=\"7\">Need to Succeed in School vs. Anxiety Level<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<th>Need to Succeed in School<\/th>\r\n<th>High Anxiety<\/th>\r\n<th>Med-high Anxiety<\/th>\r\n<th>Medium Anxiety<\/th>\r\n<th>Med-low Anxiety<\/th>\r\n<th>Low Anxiety<\/th>\r\n<th>Row Total<\/th>\r\n<\/tr>\r\n<tr>\r\n<td>High Need<\/td>\r\n<td>35<\/td>\r\n<td>42<\/td>\r\n<td>53<\/td>\r\n<td>15<\/td>\r\n<td>10<\/td>\r\n<td>155<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Medium Need<\/td>\r\n<td>18<\/td>\r\n<td>48<\/td>\r\n<td>63<\/td>\r\n<td>33<\/td>\r\n<td>31<\/td>\r\n<td>193<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Low Need<\/td>\r\n<td>4<\/td>\r\n<td>5<\/td>\r\n<td>11<\/td>\r\n<td>15<\/td>\r\n<td>17<\/td>\r\n<td>52<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Column Total<\/td>\r\n<td>57<\/td>\r\n<td>95<\/td>\r\n<td>127<\/td>\r\n<td>63<\/td>\r\n<td>58<\/td>\r\n<td>400<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div class=\"exercise\" data-type=\"exercise\"><section>\r\n<div id=\"id9399628\" class=\"problem\" data-type=\"problem\">\r\n<ol>\r\n \t<li id=\"element-671\">How many high anxiety level students are expected to have a high need to succeed in school?<\/li>\r\n \t<li>If the two variables are independent, how many students do you expect to have a low need to succeed in school and a med-low level of anxiety?<\/li>\r\n \t<li>[latex]\\displaystyle{E}=\\frac{(\\text{row total})(\\text{column total})}{\\text{total surveyed}}[\/latex]\u00a0= ________<\/li>\r\n \t<li>The expected number of students who have a med-low anxiety level and a low need to succeed in school is about ________.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"id9399648\" class=\"solution ui-solution-visible\" data-type=\"solution\">\r\n<p class=\"ui-toggle-wrapper\">Solution:<\/p>\r\n\r\n<section class=\"ui-body\">\r\n<ol>\r\n \t<li>The column total for a high anxiety level is 57. The row total for high need to succeed in school is 155. The sample size or total surveyed is 400.\r\n[latex]\\displaystyle{E}=\\frac{(\\text{row total})(\\text{column total})}{\\text{total surveyed}}=\\frac{155\\cdot57}{400}=22.09[\/latex]\r\nThe expected number of students who have a high anxiety level and a high need to succeed in school is about 22.<\/li>\r\n \t<li>The column total for a med-low anxiety level is 63. The row total for a low need to succeed in school is 52. The sample size or total surveyed is 400.<\/li>\r\n \t<li>[latex]\\displaystyle{E}=\\frac{(\\text{row total})(\\text{column total})}{\\text{total surveyed}} = 8.19[\/latex]<\/li>\r\n \t<li>8<\/li>\r\n<\/ol>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/section><\/div>\r\n<div id=\"fs-idm15185536\" class=\"note statistics try ui-has-child-title\" data-type=\"note\" data-has-label=\"true\" data-label=\"\"><header>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<section>\r\n<div id=\"eip-358\" class=\"exercise\" data-type=\"exercise\"><section>\r\n<div id=\"eip-928\" class=\"problem\" data-type=\"problem\">\r\n<p id=\"eip-757\">Refer back to the information in the Try It about the Bureau of Labor Statistics. How many service providing jobs are there expected to be in 2020? How many nonagriculture wage and salary jobs are there expected to be in 2020?<\/p>\r\n\r\n<\/div>\r\n<div id=\"eip-idm119976240\" class=\"solution ui-solution-visible\" data-type=\"solution\" data-label=\"\">\r\n<div class=\"ui-toggle-wrapper\">\u00a012,727, 14,965<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/header><\/div>\r\nhttps:\/\/youtu.be\/xEiQn6sGM20","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 and interpret chi-square test of independence hypothesis tests<\/li>\n<\/ul>\n<\/section>\n<\/div>\n<p>Tests of independence involve using a\u00a0<strong>contingency table<\/strong> of observed (data) values.<\/p>\n<p>The test statistic for a\u00a0<strong>test of independence<\/strong> is similar to that of a goodness-of-fit test:<\/p>\n<p>[latex]\\displaystyle{\\sum_{(i\\cdot{j})}}\\frac{{({O}-{E})}^{{2}}}{{E}}[\/latex]<\/p>\n<p>where:<\/p>\n<ul>\n<li><em>O<\/em> = observed values<\/li>\n<li><em>E<\/em> = expected values<\/li>\n<li><em>i<\/em> = the number of rows in the table<\/li>\n<li><em>j<\/em> = the number of columns in the table<\/li>\n<\/ul>\n<p>There are\u00a0[latex]\\displaystyle{i}\\cdot{j}[\/latex] terms of the form [latex]\\frac{{({O}-{E})}^{{2}}}{{E}}[\/latex].<\/p>\n<p><strong>A test of independence determines whether two factors are independent or not. <\/strong><\/p>\n<p><strong>Note: <\/strong>The expected value for each cell needs to be at least five in order for you to use this test.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Suppose\u00a0<em>A<\/em> = a speeding violation in the last year and <em>B<\/em> = a cell phone user while driving. If <em>A<\/em> and <em>B<\/em> are independent then <em>P<\/em>(<em>A<\/em> AND <em>B<\/em>) = <em>P<\/em>(<em>A<\/em>)<em>P<\/em>(<em>B<\/em>). <em>A <\/em>AND <em>B<\/em> is the event that a driver received a speeding violation last year and also used a cell phone while driving. Suppose, in a study of drivers who received speeding violations in the last year, and who used cell phone while driving, that 755 people were surveyed. Out of the 755, 70 had a speeding violation and 685 did not; 305 used cell phones while driving and 450 did not.<\/p>\n<p>Let\u00a0<em>y<\/em> = expected number of drivers who used a cell phone while driving and received speeding violations.<\/p>\n<p>If\u00a0<em>A<\/em> and <em>B<\/em> are independent, then <em>P<\/em>(<em>A<\/em> AND <em>B<\/em>) = <em>P<\/em>(<em>A<\/em>)<em>P<\/em>(<em>B<\/em>). By substitution,<\/p>\n<p>[latex]\\displaystyle\\frac{{y}}{{755}}={(\\frac{{70}}{{755}})}{(\\frac{{305}}{{755}})}[\/latex]<\/p>\n<p>Solve for<br \/>\n<em>y<\/em>: [latex]\\displaystyle{y}=\\frac{{{({70})}{({305})}}}{{755}}={28.3}[\/latex]<\/p>\n<p>About 28 people from the sample are expected to use cell phones while driving and to receive speeding violations.<\/p>\n<p>In a test of independence, we state the null and alternative hypotheses in words. Since the contingency table consists of\u00a0<strong>two factors<\/strong>, the null hypothesis states that the factors are <strong>independent<\/strong> and the alternative hypothesis states that they are <strong>not independent (dependent)<\/strong>. If we do a test of independence using the example, then the null hypothesis is:<\/p>\n<p><em>H<sub data-redactor-tag=\"sub\">0<\/sub><\/em>: Being a cell phone user while driving and receiving a speeding violation are independent events.<\/p>\n<p>If the null hypothesis were true, we would expect about 28 people to use cell phones while driving and to receive a speeding violation.<\/p>\n<p><strong>The test of independence is always right-tailed<\/strong> because of the calculation of the test statistic. If the expected and observed values are not close together, then the test statistic is very large and way out in the right tail of the chi-square curve, as it is in a goodness-of-fit.<\/p>\n<p>The number of degrees of freedom for the test of independence is:<\/p>\n<p><em>df<\/em> = (number of columns \u2013 1)(number of rows \u2013 1)<\/p>\n<p>The following formula calculates the\u00a0<strong>expected number<\/strong> (<em>E<\/em>):<\/p>\n<p>[latex]\\displaystyle{E}=\\frac{{{(\\text{row total})}{(\\text{column total})}}}{\\text{total number surveyed}}[\/latex]<\/p>\n<\/div>\n<h3><\/h3>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>A sample of 300 students is taken. Of the students surveyed, 50 were music students, while 250 were not. Ninety-seven were on the honor roll, while 203 were not. If we assume being a music student and being on the honor roll are independent events, what is the expected number of music students who are also on the honor roll?<\/p>\n<p>About 16 students are expected to be music students and on the honor roll.<\/p>\n<\/div>\n<h3><\/h3>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>In a volunteer group, adults 21 and older volunteer from one to nine hours each week to spend time with a disabled senior citizen. The program recruits among community college students, four-year college students, and nonstudents. The table below is a sample of the adult volunteers and the number of hours they volunteer per week.<\/p>\n<p>Number of Hours Worked Per Week by Volunteer Type (Observed)The table contains\u00a0<strong>observed (O)<\/strong> values (data).<\/p>\n<table>\n<thead>\n<tr>\n<th>Type of Volunteer<\/th>\n<th>1\u20133 Hours<\/th>\n<th>4\u20136 Hours<\/th>\n<th>7\u20139 Hours<\/th>\n<th>Row Total<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Community College Students<\/td>\n<td>111<\/td>\n<td>96<\/td>\n<td>48<\/td>\n<td>255<\/td>\n<\/tr>\n<tr>\n<td>Four-Year College Students<\/td>\n<td>96<\/td>\n<td>133<\/td>\n<td>61<\/td>\n<td>290<\/td>\n<\/tr>\n<tr>\n<td>Nonstudents<\/td>\n<td>91<\/td>\n<td>150<\/td>\n<td>53<\/td>\n<td>294<\/td>\n<\/tr>\n<tr>\n<td>Column Total<\/td>\n<td>298<\/td>\n<td>379<\/td>\n<td>162<\/td>\n<td>839<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Is the number of hours4 volunteered\u00a0<strong>independent<\/strong> of the type of volunteer?<\/p>\n<p>Solution:<\/p>\n<p>The\u00a0<strong>observed table<\/strong> and the question at the end of the problem, &#8220;Is the number of hours volunteered independent of the type of volunteer?&#8221; tell you this is a test of independence. The two factors are <strong>number of hours volunteered<\/strong> and <strong>type of volunteer<\/strong>. This test is always right-tailed.<\/p>\n<p><em>H<sub data-redactor-tag=\"sub\">0<\/sub><\/em>: The number of hours volunteered is <strong>independent<\/strong> of the type of volunteer.<\/p>\n<p><em>H<sub data-redactor-tag=\"sub\">a<\/sub><\/em>: The number of hours volunteered is <strong>dependent<\/strong> on the type of volunteer.<\/p>\n<p>The expected result are in the table below.<\/p>\n<table>\n<thead>\n<tr>\n<th colspan=\"4\">Number of Hours Worked Per Week by Volunteer Type (Expected)<br \/>\nThe table contains expected (<em>E<\/em>) values (data)<\/th>\n<\/tr>\n<tr>\n<th>Type of Volunteer<\/th>\n<th>1\u20133 Hours<\/th>\n<th>4\u20136 Hours<\/th>\n<th>7\u20139 Hours<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Community College Students<\/td>\n<td>90.57<\/td>\n<td>115.19<\/td>\n<td>49.24<\/td>\n<\/tr>\n<tr>\n<td>Four-Year College Students<\/td>\n<td>103.00<\/td>\n<td>131.00<\/td>\n<td>56.00<\/td>\n<\/tr>\n<tr>\n<td>Nonstudents<\/td>\n<td>104.42<\/td>\n<td>132.81<\/td>\n<td>56.77<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>For example, the calculation for the expected frequency for the top left cell is<\/p>\n<p>[latex]\\displaystyle{E}=\\frac{{{(\\text{row total})}{(\\text{column total})}}}{\\text{total number surveyed}}=\\frac{{{({255})}{({298})}}}{{839}}={90.57}[\/latex]<\/p>\n<p><strong>Calculate the test statistic:<\/strong><em>\u03c7<\/em><sup>2<\/sup> = 12.99 (calculator or computer)<\/p>\n<p><strong>Distribution for the test:<\/strong> [latex]\\displaystyle\\chi^{2}_{4}[\/latex]<\/p>\n<div class=\"example\" data-type=\"example\">\n<section>\n<div id=\"exercise102\" class=\"exercise\" data-type=\"exercise\">\n<section>\n<div id=\"id8698856\" class=\"solution ui-solution-visible\" data-type=\"solution\">\n<section class=\"ui-body\"><strong>Graph:<\/strong><a href=\"https:\/\/courses.candelalearning.com\/masterystats1x6xmaster\/wp-content\/uploads\/sites\/419\/2015\/06\/fig-ch11_05_01N.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-744\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214800\/fig-ch11_05_01N.jpg\" alt=\"Nonsymmetrical chi-square curve with values of 0 and 12.99 on the x-axis representing the test statistic of number of hours worked by volunteers of different types. A vertical upward line extends from 12.99 to the curve and the area to the right of this is equal to the p-value.\" width=\"487\" height=\"146\" \/><\/a><\/p>\n<p id=\"element-117\"><strong>Probability statement:<\/strong><em data-effect=\"italics\">p<\/em>-value=<em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">\u03c7<sup>2<\/sup><\/em> &gt; 12.99) = 0.0113<\/p>\n<p id=\"element-347\"><strong>Compare <em data-effect=\"italics\">\u03b1<\/em> and the <em data-effect=\"italics\">p<\/em>-value:<\/strong> Since no <em data-effect=\"italics\">\u03b1<\/em> is given, assume <em data-effect=\"italics\">\u03b1<\/em> = 0.05. <em data-effect=\"italics\">p<\/em>-value = 0.0113. <em data-effect=\"italics\">\u03b1<\/em> &gt; <em data-effect=\"italics\">p<\/em>-value.<\/p>\n<p id=\"fs-idm62402592\"><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>0<\/sub><\/em>. This means that the factors are not independent.<\/p>\n<p><strong>Conclusion:<\/strong> At a 5% level of significance, from the data, there is sufficient evidence to conclude that the number of hours volunteered and the type of volunteer are dependent on one another.<\/p>\n<p>For the example in the table titled &#8220;Number of Hours Worked Per Week by Volunteer Type (Expected),&#8221; if there had been another type of volunteer, teenagers, what would the degrees of freedom be?<\/p>\n<div id=\"fs-idp30286464\" class=\"note statistics calculator\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<section>\n<p id=\"element-854\">Press the <code>MATRX<\/code> key and arrow over to <code>EDIT<\/code>. Press <code>1:[A]<\/code>. Press <code>3 ENTER 3 ENTER<\/code>. Enter the table values by row from the table. Press <code>ENTER<\/code> after each. Press <code>2nd QUIT<\/code>. Press<code>STAT<\/code> and arrow over to <code>TESTS<\/code>. Arrow down to <code>C:\u03c72-TEST<\/code>. Press <code>ENTER<\/code>. You should see <code>Observed:[A] and Expected:[B]<\/code>. Arrow down to <code>Calculate<\/code>. Press <code>ENTER<\/code>. The test statistic is 12.9909 and the <em data-effect=\"italics\">p<\/em>-value = 0.0113. Do the procedure a second time, but arrow down to <code>Draw<\/code> instead of <code>calculate<\/code>.<\/p>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div id=\"fs-idm24618832\" class=\"note statistics try finger ui-has-child-title\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<header>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<section>\n<div id=\"eip-685\" class=\"exercise\" data-type=\"exercise\">\n<section>\n<div id=\"eip-903\" class=\"problem\" data-type=\"problem\">\n<p id=\"eip-845\">The Bureau of Labor Statistics gathers data about employment in the United States. A sample is taken to calculate the number of U.S. citizens working in one of several industry sectors over time. The table below\u00a0shows the results:<\/p>\n<table id=\"fs-idp61742592\" summary=\"\">\n<thead>\n<tr>\n<th>Industry Sector<\/th>\n<th>2000<\/th>\n<th>2010<\/th>\n<th>2020<\/th>\n<th>Total<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Nonagriculture wage and salary<\/td>\n<td>13,243<\/td>\n<td>13,044<\/td>\n<td>15,018<\/td>\n<td>41,305<\/td>\n<\/tr>\n<tr>\n<td>Goods-producing, excluding agriculture<\/td>\n<td>2,457<\/td>\n<td>1,771<\/td>\n<td>1,950<\/td>\n<td>6,178<\/td>\n<\/tr>\n<tr>\n<td>Services-providing<\/td>\n<td>10,786<\/td>\n<td>11,273<\/td>\n<td>13,068<\/td>\n<td>35,127<\/td>\n<\/tr>\n<tr>\n<td>Agriculture, forestry, fishing, and hunting<\/td>\n<td>240<\/td>\n<td>214<\/td>\n<td>201<\/td>\n<td>655<\/td>\n<\/tr>\n<tr>\n<td>Nonagriculture self-employed and unpaid family worker<\/td>\n<td>931<\/td>\n<td>894<\/td>\n<td>972<\/td>\n<td>2,797<\/td>\n<\/tr>\n<tr>\n<td>Secondary wage and salary jobs in agriculture and private household industries<\/td>\n<td>14<\/td>\n<td>11<\/td>\n<td>11<\/td>\n<td>36<\/td>\n<\/tr>\n<tr>\n<td>Secondary jobs as a self-employed or unpaid family worker<\/td>\n<td>196<\/td>\n<td>144<\/td>\n<td>152<\/td>\n<td>492<\/td>\n<\/tr>\n<tr>\n<td>Total<\/td>\n<td>27,867<\/td>\n<td>27,351<\/td>\n<td>31,372<\/td>\n<td>86,590<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"eip-idp2741344\">We want to know if the change in the number of jobs is independent of the change in years. State the null and alternative hypotheses and the degrees of freedom.<\/p>\n<\/div>\n<div id=\"eip-idm75764640\" class=\"solution ui-solution-visible\" data-type=\"solution\" data-label=\"\">\n<section class=\"ui-body\"><em data-effect=\"italics\">H<\/em><sub>0<\/sub>: The number of jobs is independent of the year.<em data-effect=\"italics\">H<sub>a<\/sub><\/em>: The number of jobs is dependent on the year.<em data-effect=\"italics\">df<\/em> = 12<a href=\"https:\/\/courses.candelalearning.com\/masterystats1x6xmaster\/wp-content\/uploads\/sites\/419\/2015\/06\/CNX_Stats_C11_M05_tryit001annoN.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-746\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214802\/CNX_Stats_C11_M05_tryit001annoN.jpg\" alt=\"Nonsymmetrical chi-square curve with values of 0, 12, and 227.73 on the x-axis. A vertical upward line extends from 227.73 to the curve and the area to the right of this is equal to the p-value. p-value = almost zero.\" width=\"487\" height=\"187\" \/><\/a><\/section>\n<section class=\"ui-body\">\n<p id=\"eip-idm36315600\">Press the <code>MATRX<\/code> key and arrow over to <code>EDIT<\/code>. Press <code>1:[A]<\/code>. Press <code>3 ENTER 3 ENTER<\/code>. Enter the table values by row. Press <code>ENTER<\/code> after each. Press <code>2nd QUIT<\/code>. Press <code>STAT<\/code> and arrow over to <code>TESTS<\/code>. Arrow down to <code>C:\u03c72-TEST<\/code>. Press <code>ENTER<\/code>. You should see <code>Observed:[A] and Expected:[B]<\/code>. Arrow down to <code>Calculate<\/code>. Press <code>ENTER<\/code>. The test statistic is 227.73 and the<em data-effect=\"italics\">p<\/em>\u2212value = 5.90E &#8211; 42 = 0. Do the procedure a second time but arrow down to <code>Draw<\/code> instead of<code>calculate<\/code>.<\/p>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<p>&nbsp;<\/p>\n<\/header>\n<\/div>\n<div class=\"example\" data-type=\"example\">\n<section>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>De Anza College is interested in the relationship between anxiety level and the need to succeed in school. A random sample of 400 students took a test that measured anxiety level and need to succeed in school. This table\u00a0shows the results. De Anza College wants to know if anxiety level and need to succeed in school are independent events.<\/p>\n<table summary=\"This table presents the need to succeed in school in the first column and the various anxiety levels (high to low) and the row total in the second to seventh columns. The first row is for high need, the second row is for medium need, third row is for low need, and the column total is in the fourth row.\">\n<thead valign=\"top\">\n<tr>\n<th colspan=\"7\">Need to Succeed in School vs. Anxiety Level<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<th>Need to Succeed in School<\/th>\n<th>High Anxiety<\/th>\n<th>Med-high Anxiety<\/th>\n<th>Medium Anxiety<\/th>\n<th>Med-low Anxiety<\/th>\n<th>Low Anxiety<\/th>\n<th>Row Total<\/th>\n<\/tr>\n<tr>\n<td>High Need<\/td>\n<td>35<\/td>\n<td>42<\/td>\n<td>53<\/td>\n<td>15<\/td>\n<td>10<\/td>\n<td>155<\/td>\n<\/tr>\n<tr>\n<td>Medium Need<\/td>\n<td>18<\/td>\n<td>48<\/td>\n<td>63<\/td>\n<td>33<\/td>\n<td>31<\/td>\n<td>193<\/td>\n<\/tr>\n<tr>\n<td>Low Need<\/td>\n<td>4<\/td>\n<td>5<\/td>\n<td>11<\/td>\n<td>15<\/td>\n<td>17<\/td>\n<td>52<\/td>\n<\/tr>\n<tr>\n<td>Column Total<\/td>\n<td>57<\/td>\n<td>95<\/td>\n<td>127<\/td>\n<td>63<\/td>\n<td>58<\/td>\n<td>400<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"exercise\" data-type=\"exercise\">\n<section>\n<div id=\"id9399628\" class=\"problem\" data-type=\"problem\">\n<ol>\n<li id=\"element-671\">How many high anxiety level students are expected to have a high need to succeed in school?<\/li>\n<li>If the two variables are independent, how many students do you expect to have a low need to succeed in school and a med-low level of anxiety?<\/li>\n<li>[latex]\\displaystyle{E}=\\frac{(\\text{row total})(\\text{column total})}{\\text{total surveyed}}[\/latex]\u00a0= ________<\/li>\n<li>The expected number of students who have a med-low anxiety level and a low need to succeed in school is about ________.<\/li>\n<\/ol>\n<\/div>\n<div id=\"id9399648\" class=\"solution ui-solution-visible\" data-type=\"solution\">\n<p class=\"ui-toggle-wrapper\">Solution:<\/p>\n<section class=\"ui-body\">\n<ol>\n<li>The column total for a high anxiety level is 57. The row total for high need to succeed in school is 155. The sample size or total surveyed is 400.<br \/>\n[latex]\\displaystyle{E}=\\frac{(\\text{row total})(\\text{column total})}{\\text{total surveyed}}=\\frac{155\\cdot57}{400}=22.09[\/latex]<br \/>\nThe expected number of students who have a high anxiety level and a high need to succeed in school is about 22.<\/li>\n<li>The column total for a med-low anxiety level is 63. The row total for a low need to succeed in school is 52. The sample size or total surveyed is 400.<\/li>\n<li>[latex]\\displaystyle{E}=\\frac{(\\text{row total})(\\text{column total})}{\\text{total surveyed}} = 8.19[\/latex]<\/li>\n<li>8<\/li>\n<\/ol>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/section>\n<\/div>\n<div id=\"fs-idm15185536\" class=\"note statistics try ui-has-child-title\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<header>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<section>\n<div id=\"eip-358\" class=\"exercise\" data-type=\"exercise\">\n<section>\n<div id=\"eip-928\" class=\"problem\" data-type=\"problem\">\n<p id=\"eip-757\">Refer back to the information in the Try It about the Bureau of Labor Statistics. How many service providing jobs are there expected to be in 2020? How many nonagriculture wage and salary jobs are there expected to be in 2020?<\/p>\n<\/div>\n<div id=\"eip-idm119976240\" class=\"solution ui-solution-visible\" data-type=\"solution\" data-label=\"\">\n<div class=\"ui-toggle-wrapper\">\u00a012,727, 14,965<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<\/header>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"How to calculate Chi Square Test for Independence (two way)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/xEiQn6sGM20?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\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-441\">\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 Illowski, Susan Dean. <strong>Provided by<\/strong>: Open Stax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.44\">http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.44<\/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>: Download for free at http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.44<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">All rights reserved content<\/div><ul class=\"citation-list\"><li>How to calculate Chi Square Test for Independence (two way). <strong>Authored by<\/strong>: statisticsfun. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/xEiQn6sGM20\">https:\/\/youtu.be\/xEiQn6sGM20<\/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":21,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Introductory Statistics \",\"author\":\"Barbara Illowski, Susan Dean\",\"organization\":\"Open Stax\",\"url\":\"http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.44\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.44\"},{\"type\":\"copyrighted_video\",\"description\":\"How to calculate Chi Square Test for Independence (two way)\",\"author\":\"statisticsfun\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/xEiQn6sGM20\",\"project\":\"\",\"license\":\"arr\",\"license_terms\":\"Standard YouTube License\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-441","chapter","type-chapter","status-publish","hentry"],"part":411,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/ntcc-introstats1\/wp-json\/pressbooks\/v2\/chapters\/441","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/ntcc-introstats1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/ntcc-introstats1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ntcc-introstats1\/wp-json\/wp\/v2\/users\/21"}],"version-history":[{"count":7,"href":"https:\/\/courses.lumenlearning.com\/ntcc-introstats1\/wp-json\/pressbooks\/v2\/chapters\/441\/revisions"}],"predecessor-version":[{"id":2032,"href":"https:\/\/courses.lumenlearning.com\/ntcc-introstats1\/wp-json\/pressbooks\/v2\/chapters\/441\/revisions\/2032"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/ntcc-introstats1\/wp-json\/pressbooks\/v2\/parts\/411"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/ntcc-introstats1\/wp-json\/pressbooks\/v2\/chapters\/441\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/ntcc-introstats1\/wp-json\/wp\/v2\/media?parent=441"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ntcc-introstats1\/wp-json\/pressbooks\/v2\/chapter-type?post=441"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ntcc-introstats1\/wp-json\/wp\/v2\/contributor?post=441"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ntcc-introstats1\/wp-json\/wp\/v2\/license?post=441"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}