Conduct the chi-square test for independence using the aggregated results for Example 7.18 using a level of
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 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.
The expected frequencies are shown next.
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
Now only 2 of 12 cells have an expected frequency less than 5; this now meets the assumptions of the chi-square test.
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