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Chapter 11 The Chi-Square Test of Association/Independence. Target Goal: I can perform a chi-square test for association/independence to determine whether there is convincing evidence of an association between two categorical variables. 11.2b h.w : pg. 728: 49, 51, 53 - 58.

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## Chapter 11 The Chi-Square Test of Association/Independence

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**Chapter 11The Chi-Square Test of Association/Independence**Target Goal: I can perform a chi-square test for association/independence to determine whether there is convincing evidence of an association between two categorical variables. 11.2b h.w: pg. 728: 49, 51, 53 - 58**The chi-square test can also be used to show evidence that**there is a relationship between two categorical variables. • Use this if you have independent SRS’s from several populations where one variable is categorical and the other is the sample number. • Or, if you have a single SRS with each individual classified according to two categorical variables. • Or, if you have an entire population with each individual classified according to two categorical variables.**Ex: Smoking and SES**An example that classifies observations from a single population in two ways: by smoking habits and SES. • In a study of heart disease in male federal employees, researchers classified 356 volunteer subjects according to their socioeconomic status (SES) and their smoking status.**Observed Counts for smoking and SES**SES Smoking High Middle Low Total Current 51 22 43 116 Former 92 21 28 141 Never 68 9 22 99 Total 211 52 93 356 • This is a 3x3 table with added margin totals. Even though this example is different than comparing several proportions, we can still apply the chi-square test because the row and column variables are not related to each other.**The Chi-Square Test of Association/Independence**Use the chi-square test of association/independence to test the null hypothesis, Ho: there is no relationship between two categorical variables when you have a two way table from a single SRS, with each individual is classified according to both oftwo categorical variables.**SES cont.**• SES is the explanatory variable therefore we need to compare the column percents that give the conditional distribution of smoking within each SES category.**Calculate Column Percents:**• 51/211 = 0.242 about 24.2% of the high-SES group are current smokers. • Fill in the rest of the table.**Column percents for Smoking and SES**SES Smoking High Middle Low Current 24.2 42.3 46.2 Former 43.6 40.4 30.1 Never 32.2 17.3 23.7 Total 100.0 100.0 100.0 What do the column percents suggest?**There is a negative association between smoking and SES.**• The lower the SES, the more likely to smoke.**Computing Expected Cell Counts**• 116 x 211 = 68.75 356**Expected Count for Smoking and SES**SES Smoking High Middle Low Total Current 68.75 16.94 30.30 115.99 Former 83.57 20.60 36.83 141.00 Never 58.68 14.46 25.86 99.00 Total 211 52 92.99 355.99**Chi-square Test for Association/Independence**Step 1: State - We want to perform a test of Ho: There is no association between smoking and SES. Ha: There is an association between smoking and SES.**Step 2: Plan**If conditions are met, we should carry out a chi-square test of association/independence. Random: The subjects were volunteers, we may not be able to generalize our results. Large Sample Size: • To use chi-square we must check all expected counts. • We did this and all counts ≥ 1 and no more than 20% < 5.**Independence:**• Because we are sampling without replacement, we need to check the 10% condition.It is safe to assume that the total number of male federal employees is at least 10(356) = 3560. • Thus, knowing the values of both variables for one person gives us no meaningful information about the variables for another person. So, individual observations are independent.**Step 3: Carry out the inference procedure.**• The test statistic • Calculate by hand with df = (r-1)(c-1) = • Or with calculator, need to enter observed counts into matrix table A. • Note: the calculator will calculate the expected counts for you when you execute the X2 test.**Note: if doing by hand, could write calculator program to**do “expected counts” or must do by hand. • Enter observed values in matrix A, • Then STAT:TESTS: -Test • The calculator enters expected values in matrix B. • P-value = .00098 Note: the association does not mean that SES causes smoking behavior.**Step 4: Conclude –Interpret the results in context.**• With a p-value this low, we reject the null hypothesis at the alpha = .01 level and conclude that there is strong evidence of an association between smoking and SES in the population of male federal employees.**Follow-up Analysis**Inference for Relationships Start by examining which cells in the two-way table show large deviations between the observed and expected counts. Then look at the individual components to see which terms contribute most to the chi-square statistic. Minitab output for the wine and music study displays the individual components that contribute to the chi-square statistic.**Follow-up Analysis**Inference for Relationships Looking at the output, we see that just two of the nine components that make up the chi-square statistic contribute about 14 (almost 77%) of the total χ2 = 18.28. We are led to a specific conclusion: sales of Italian wine are strongly affected by Italian and French music.

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