Chi-Square With Base Hypothesis That

¶ … Chi-Square with base hypothesis that all political groups contribute equal amounts. b) Chi-square with base hypothesis appropriate to the attitude question asked. c) T-test. d) Chi-square with base hypothesis of equal average salaries between regions. Perform a chi-square test of the following data: a) Regulation is the best way to ensure a safe workplace Managers Blue-collar workers Agree Disagree Totals One potential hypothesis for this data is that managers and blue-collar workers do not have very different opinions about regulation.

Some regulations may make the workplace safer for managers as well as blue-collar workers. Many blue-collar workers are politically conservative, which would influence them against over-regulation of the workplace. So, the expectations for my chi-square test will be: Managers Blue-collar Workers Agree Disagree No Opinion Total Expected Obtained (Obs-Exp)^2/Exp Man-agree 62 58 0.258064516 Man-disag 29 34 0.862068966 Man-NA 9 8 0. Work-agree 62 66 0.258064516 Work-disag 29 24 0.862068966 Work-NA 9 10 0. 2 = 2.46 This does not exceed the critical value of 3.84 for a chi square test with ? = 0.05 and df = 1.

Therefore, we can assume that these data are not significantly different from the expected values, and do not violate the null hypothesis. b) Ownership of residence Male Female Yes 25 16 No 7 8 Totals 32 24 In this case, we are testing current numbers against the null hypothesis that home ownership has equalized across genders. The expected value table is: Male Female Yes 23.4 17.6 No 8.6 6.4 Totals 32 24 Expected Obtained (Obs-Exp)^2/Exp M-Yes 23.4 25 0.105400697 M-No 8.6 7 0.288095238 F-Yes 17.6 16 0.140534262 F-No 6.4 8 0.384126984 In this case as well, we fail to violate the null hypothesis, ?2 = 0.92.

c) Age of shopper Store a Store B 20-34 27 73 35-54 31 82 55+ 11 93 Totals 69 The null hypothesis in this case is that both stores draw equally from the same age groups, although Store B. draws more customers overall. The expected value table is: Store a Store B 20-34 21.8 78.2 35-54 24.6 88.4 55+ 22.6 81.4 Totals 69 Expected Obtained (Obs-Exp)^2/Exp 20-34 21.8 27 1.25830064 35-54 24.6 31 1.66726748 55+ 22.6 11 5.98240235 20-34 78.23343849 73 0.35009171 35-54 88.40378549 82 0.46387684 55+ 81.36277603 93 1.66445872 2 = 11.39 This value does exceed the critical ?2 value for df = 2 at ? = 0.05. Therefore, we can assume that one of the observed values is significantly different from the expected value for that group.

Without post-hoc pairwise tests it is impossible to say exactly which group is different. We can make an educated guess, however, that the proportion of 55+ shoppers in store a is statistically different from what would be expected by chance. 3. Collapse the response categories in the following table so that it meets the assumption of the Chi-square test, then perform the test. Ownership (Collapsed) Education Owners Non-owners Some High School or Below 5 17 High School graduate 30 25 22 26 Post-Baccalaureate 5 7 Total 62 75 2 = 6.49.

This does not exceed the critical ?2 value for df = 3, so we cannot assume that there is any significant difference between the observed counts of home ownership by educational level and those expected by chance. 4. A ?2 test to determine whether the sample is significantly different from the expected distribution would be most appropriate. The data in this case yield ?2 = 2.51, below the critical value cutoff for ? = 0.05. We can assume.