In a vending machine study with four drink types, the investigator asks if four types are equally favored. Which test is appropriate?

In a vending machine study with four drink types, the investigator asks if four types are equally favored. Which test is appropriate?

• A. Chi-square goodness-of-fit
• B. Chi-square test of independence
• C. t-test for proportions
• D. ANOVA

Answer: (A)

This question is testing whether observed counts across several categories match an expected distribution. With four drink types and the question of equal popularity, the natural null is that each type is chosen with the same probability, so the expected counts are equal for all four types. The appropriate method is the chi-square goodness-of-fit test. It compares what you actually observed (how many times each drink type was chosen) to what you would expect if all four types were equally favored (equal counts). The test statistic, sum of (observed minus expected) squared over the expected count for each category, follows a chi-square distribution with degrees of freedom equal to the number of categories minus one (three in this case). A small p-value indicates the observed distribution deviates beyond what randomness would explain, leading to the conclusion that the four types are not equally favored. Why not the other tests? The chi-square test of independence looks at relationships between two categorical variables, which isn’t the scenario here. A t-test for proportions applies when comparing one or two proportions, not when you have four categories; it’s not suited for testing equality across multiple categories. ANOVA compares means of a continuous outcome across groups, which isn’t about categorical counts.

This question is testing whether observed counts across several categories match an expected distribution. With four drink types and the question of equal popularity, the natural null is that each type is chosen with the same probability, so the expected counts are equal for all four types.

The appropriate method is the chi-square goodness-of-fit test. It compares what you actually observed (how many times each drink type was chosen) to what you would expect if all four types were equally favored (equal counts). The test statistic, sum of (observed minus expected) squared over the expected count for each category, follows a chi-square distribution with degrees of freedom equal to the number of categories minus one (three in this case). A small p-value indicates the observed distribution deviates beyond what randomness would explain, leading to the conclusion that the four types are not equally favored.

Why not the other tests? The chi-square test of independence looks at relationships between two categorical variables, which isn’t the scenario here. A t-test for proportions applies when comparing one or two proportions, not when you have four categories; it’s not suited for testing equality across multiple categories. ANOVA compares means of a continuous outcome across groups, which isn’t about categorical counts.