Question: Conduct the chi-square test for independence using the aggregated results for Example 7.18 using a level of significance of 0.05. Data from Example 7.18 Violations

Conduct the chi-square test for independence using the aggregated results for Example 7.18 using a level of significance of 0.05.


Data from Example 7.18

Violations of Chi-Square Assumptions 

A survey of 100 students at a university queried their beverage preferences at a local coffee shop. The results are shown in the table below. 

Female Male Total Brewed coffee 6 16 22 Iced coffee 10 5


The expected frequencies are shown next.

15 Espresso 2 8 10 Cappuccino 4 2 6 Latter 7 3


If we were to conduct a chi-square test of independence, we would see that of the 16 cells, five, or over 30%, have frequencies smaller than 5. Four of them are in the Cappuccino, Latte, and Mocha columns; these can be aggregated into one column called Hot Specialty beverages 

10 Mocha 9 2 11 Iced blended Tea 10 7 17 8


Now only 2 of 12 cells have an expected frequency less than 5; this now meets the assumptions of the chi-square test.

1 9 Total 56 44 100

Female Male Total Brewed coffee 6 16 22 Iced coffee 10 5 15 Espresso 2 8 10 Cappuccino 4 2 6 Latter 7 3 10 Mocha 9 2 11 Iced blended Tea 10 7 17 8 1 9 Total 56 44 100

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