{"id":1170,"date":"2021-06-25T12:12:39","date_gmt":"2021-06-25T16:12:39","guid":{"rendered":"https:\/\/openbooks.macewan.ca\/rcommander\/?post_type=chapter&p=1170"},"modified":"2025-05-07T18:09:17","modified_gmt":"2025-05-07T22:09:17","slug":"11-4-chi-square-independence-test","status":"publish","type":"chapter","link":"https:\/\/openbooks.macewan.ca\/introstats\/chapter\/11-4-chi-square-independence-test\/","title":{"raw":"11.4 Chi-Square Independence Test","rendered":"11.4 Chi-Square Independence Test"},"content":{"raw":"The chi-square independence test is used to test for an association between two categorical variables of a population.\r\n

11.4.1 Terminologies Used for a Contingency Table<\/h2>\r\nRecall that a contingency table summarizes the counts of two categorical variables. For example, the following contingency table groups 200 females according to their breast cancer status and smoking status:\r\n

Table 11.6<\/strong>: Contingency Table of Cancer Status (row) and Smoking Status (column)<\/p>\r\n\r\n

\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
<\/td>\r\n\r\n
Smoker [latex]\\color{blue}{(S_1)}[\/latex]<\/span><\/div><\/th>\r\n
\r\n
Non-smoker [latex]\\color{blue}{(S_2)}[\/latex]<\/span><\/div><\/th>\r\n
Total<\/strong><\/th>\r\n<\/tr>\r\n<\/thead>\r\n
Breast Cancer [latex]\\color{red}{(C_1)}[\/latex]<\/th>\r\n10<\/strong> [latex]( C_1 \\: \\& \\: S_1 )[\/latex]<\/td>\r\n30<\/strong>\u00a0[latex]( C_1 \\: \\& \\: S_2 )[\/latex]<\/td>\r\n40\u00a0 <\/strong><\/td>\r\n<\/tr>\r\n
Cancer-free [latex]\\color{red}{(C_2)}[\/latex]<\/th>\r\n20<\/strong>\u00a0 [latex]( C_2 \\: \\& \\: S_1 )[\/latex]<\/td>\r\n140<\/strong>\u00a0 [latex]( C_2 \\: \\& \\: S_2 )[\/latex]<\/td>\r\n160\u00a0 <\/strong><\/td>\r\n<\/tr>\r\n
Total<\/strong><\/th>\r\n30<\/strong><\/span><\/td>\r\n170<\/strong><\/span><\/td>\r\n200<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\nSuppose we randomly select an individual from this sample. Define the events:\r\n\r\n[latex]\\begin{align*}\r\nS_1 &= \\text{the subject is a smoker}; \\\\\r\nS_2 &= \\text{the subject is a non-smoker}; \\\\\r\nC_1 &= \\text{the subject has breast cancer}; \\\\\r\nC_2 &= \\text{the subject does not have breast cancer.}\r\n\\end{align*}[\/latex]\r\n\r\nThe joint events are:\r\n\r\n[latex]\\begin{align*}\r\nC_1 \\: \\& \\: S_1 &= \\text{the subject has cancer and is a smoker}; \\\\\r\nC_1 \\: \\& \\: S_2 &= \\text{the subject has cancer and is a non-smoker}; \\\\\r\nC_2 \\: \\& \\: S_1 &= \\text{the subject does not have cancer and is a smoker}; \\\\\r\nC_2 \\: \\& \\: S_2 &= \\text{the subject does not have cancer and is a non-smoker.}\r\n\\end{align*}[\/latex]\r\n\r\nThe variable \u201cCancer Status\u201d is called the row variable, and it has two possible values\u2014cancer or cancer-free<\/span>. The variable \u201cSmoking Status\u201d is the column variable,<\/span> and it has two values\u2014smoker and non-smoker. The two numbers in the last column (40 and 160) are the row totals<\/span> and the two in the last row (30, 170) are the column totals<\/span>. The sample size is also called the grand total<\/em>. The four numbers in bold are the joint frequencies. The boxes that contain the joint frequencies are referred to as cells.\r\n\r\nBased on the [latex]\\frac{f}{N}[\/latex]\u00a0rule, the marginal distribution<\/strong> of the row<\/span> (column<\/span>) variable equals the row<\/span> (column<\/span>) totals divided by [latex]n[\/latex]. The joint distribution<\/strong> is given by the joint frequencies divided by [latex]n[\/latex]. The following table shows the marginal distribution of \u201cCancer Status\u201d in the last column, the marginal distribution of \u201cSmoking Status\u201d in the last row, and the joint distribution of the four cells.\r\n

Table 11.7<\/strong>: Marginal and Joint Probability Distributions of Cancer Status and Smoking Status<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
<\/td>\r\n\r\n
Smoker [latex]\\color{blue}{(S_1)}[\/latex]<\/span><\/div><\/th>\r\n
\r\n
Non-smoker [latex]\\color{blue}{(S_2)}[\/latex]<\/span><\/div><\/th>\r\n
\r\n
Total<\/strong><\/div><\/th>\r\n<\/tr>\r\n<\/thead>\r\n
\r\n
Breast Cancer [latex]\\color{red}{(C_1)}[\/latex]<\/div><\/th>\r\n
[latex]P(C_1 \\: \\& \\: S_1) = \\frac{10}{200} = 0.05[\/latex]<\/td>\r\n[latex]P(C_1 \\: \\& \\: S_2) = \\frac{30}{200} = 0.15[\/latex]<\/td>\r\n[latex]\\color{red}{P(C_1) = \\frac{40}{200} = 0.2}[\/latex]<\/span><\/td>\r\n<\/tr>\r\n
\r\n
Cancer-free [latex]\\color{red}{(C_2)}[\/latex]<\/span><\/div><\/th>\r\n
[latex]P(C_2 \\: \\& \\: S_1) =\\frac{20}{200} = 0.1[\/latex]<\/td>\r\n[latex]P(C_2 \\: \\& \\: S_2) = \\frac{140}{200} = 0.7[\/latex]<\/td>\r\n[latex]\\color{red}{P(C_1) = \\frac{160}{200} = 0.8}[\/latex]<\/span><\/td>\r\n<\/tr>\r\n
\r\n
Total<\/strong><\/div><\/th>\r\n
\u00a0[latex]\\color{blue}{P(S_1) = \\frac{30}{200} = 0.15}[\/latex]<\/span><\/td>\r\n [latex]\\color{blue}{P(S_2) = \\frac{170}{200} = 0.85}[\/latex]<\/span><\/td>\r\n\r\n
1<\/div><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe want to test for an association between the two variables in a contingency table. Two variables are said to be associated if they are NOT independent. If two variables are associated, then differences exist among the conditional distributions of one variable, given different values of the other variable. For example, the conditional distributions of \u201cCancer Status\u201d given \u201cSmoking Status\u201d are given in the following table. Notice that the conditional distributions are simply the relative frequencies of \u201cCancer\u201d within smoker and non-smoker groups.\r\n

Table 11.8<\/strong>: Conditional Probability Distribution of Cancer Status Given Smoking Status<\/p>\r\n\r\n

\r\n\r\n<\/thead>\r\n\r\n\r\n\r\n\r\n\r\n\r\n
\r\n
<\/div><\/td>\r\n
\r\n
Smoker [latex]\\color{blue}{(S_1)}[\/latex]<\/span><\/div><\/th>\r\n
\r\n
Non-smoker [latex]\\color{blue}{(S_2)}[\/latex]<\/span><\/div><\/th>\r\n
\r\n
Marginal Distribution<\/strong>\r\nOf Cancer Status<\/strong><\/div><\/th>\r\n<\/tr>\r\n<\/thead>\r\n
\r\n
Breast Cancer [latex]\\color{red}{(C_1)}[\/latex]<\/span><\/div><\/th>\r\n
[latex]P(C_1 | S_1) = \\frac{10}{30} = 0.333[\/latex]<\/td>\r\n[latex]P(C_1 | S_2) = \\frac{30}{170} = 0.176[\/latex]<\/td>\r\n[latex]\\color{red}{P(C_1) = \\frac{40}{200} = 0.2}[\/latex]<\/span><\/td>\r\n<\/tr>\r\n
\r\n
Cancer-free [latex]\\color{red}{(C_2)}[\/latex]<\/span><\/div><\/th>\r\n
[latex]P(C_2 | S_1) = \\frac{20}{30} = 0.677[\/latex]<\/td>\r\n[latex]P(C_2 | S_2) = \\frac{140}{170} = 0.824[\/latex]<\/td>\r\n[latex]\\color{red}{P(C_2) = \\frac{160}{200} = 0.8}[\/latex]<\/span><\/td>\r\n<\/tr>\r\n
Total<\/strong><\/th>\r\n\r\n
1<\/strong><\/div><\/td>\r\n
\r\n
1<\/div><\/td>\r\n
\r\n
1<\/div><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\nA segmented bar graph helps us visualize conditional distributions and the concept of association. The figure below is the segmented bar graph that displays the conditional distributions of \u201cCancer Status\u201d for smokers and non-smokers and the marginal distribution of \"Cancer Status\". The three bars should be identical if \u201cCancer Status\u201d and \u201cSmoking Status\u201d are independent. That is, the conditional probabilities should equal the unconditional probabilities:\r\n

[latex]P(C_1 | S_1) = P(C_1 | S_2) = P(C_1); P(C_2 | S_1) = P(C_2 | S_2) = P(C_2).[\/latex]<\/a><\/p>\r\n\r\n\r\n\r\n\r\n
[caption id=\"attachment_1173\" align=\"aligncenter\" width=\"291\"]\"A<\/a> Figure 11.2<\/strong>: Segment Bar Chart. [Image Description (See Appendix D Figure 11.2)<\/a>] Click on the image to enlarge it.[\/caption]<\/td>\r\nInterpretation:\r\n\r\nThe proportion or percentage of females with breast cancer (the green bar) is higher among the smokers than the non-smokers. Therefore, \u201cCancer Status\u201d and \u201cSmoking Status\u201d might be associated; we can test this by a chi-square independence test.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

11.4.2 Main Idea Behind Chi-Square Independence Test<\/h2>\r\nThe null hypothesis is that the two variables are independent; the alternative is that they are associated. The test statistic is the same as that from the chi-square goodness-of-fit test; for each cell, compute the difference between the observed frequency (O)\u00a0and the expected frequency ([latex]E[\/latex]), square it, and divide\u00a0by the expected frequency. The expected frequency is the number we expect to observe if the null is true. A large chi-square statistic means the observed and the expected frequencies are significantly different, which provides evidence against the null hypothesis. Therefore, we should reject the null if the observed chi-square statistic is sufficiently large. More specifically, given the significance level [latex]\\alpha[\/latex], reject [latex]H_0[\/latex]\u00a0if the P-value [latex]\\leq \\alpha[\/latex], where the\u00a0P-value is the area to the right<\/strong> of the observed test statistic under the chi-square curve.\r\n\r\nThe test procedure is straightforward\u2014the key is calculating each cell's expected frequency. Recall that two events, A and B , are independent if [latex]P(A \\: \\& B) = P(A) \\times P(B)[\/latex]. For example, if the events \u201cBreast Cancer\u201d and \u201cSmoker\u201d are independent, then [latex]P(\\text{Breast Cancer \\& Smoker}) = P(\\text{Breast Cancer}) \\times P(\\text{Smoker})[\/latex] where [latex]P(\\text{Breast Cancer})[\/latex] and [latex]P(\\text{Smoker})[\/latex] are given by the marginal distribution of \u201cCancer Status\u201d and \u201cSmoking Status\u201d respectively. That is,\r\n[latex] P(\\text{Breast Cancer}) = \\frac{40}{200} = 0.2; P(\\text{Smoker}) = \\frac{30}{200} = 0.15. [\/latex]\r\nIf [latex]H_0[\/latex] (the two variables are independent) is true, the expected frequency for the cell \u201cCancer and Smoker\u201d is\r\n[latex] E = n P(\\text{Cancer and Smoker}) = n P(\\text{Cancer}) P(\\text{Smoker})= 200 \\times \\frac{40}{200} \\times \\frac{30}{200} = \\frac{40 \\times 30}{200} = 6. [\/latex]\r\nIn general,\r\n[latex] \\text{Expected frequency of the cell in rth row and cth column} = \\frac{\\text{rth row total} \\times \\text{cth column total}}{n}. [\/latex]\r\nApplying the above formula to each cell yields the following expected frequencies:\r\n