A manufacturing plant investigates defect types from three vendors to see if the distribution of defect types is the same across vendors. Which test is used?

A manufacturing plant investigates defect types from three vendors to see if the distribution of defect types is the same across vendors. Which test is used?

• A. Chi-square goodness-of-fit
• B. Chi-square test of homogeneity
• C. Chi-square test of independence
• D. ANOVA

Answer: (B)

When you want to know if a categorical outcome is distributed the same across multiple groups, you use a chi-square test of homogeneity. Here, defect type is the categorical outcome and the groups are the three vendors. You'd arrange the data in a contingency table with vendors as rows and defect types as columns, and compare the observed counts in each cell to the counts you’d expect if every vendor had the same distribution of defect types. Under the null hypothesis, no matter which vendor you look at, the proportion of each defect type is the same across all vendors. The alternative is that at least one vendor has a different distribution of defect types. You compute the chi-square statistic from (observed − expected)²/expected for all cells, where expected counts come from the product of the corresponding row total and column total divided by the grand total. This test is distinct from the chi-square goodness-of-fit, which checks a single population against a specific expected distribution, and the chi-square test of independence, which asks whether two categorical variables are associated within one population. ANOVA is for comparing means of a continuous variable across groups, not for categorical outcomes. Since the goal here is to compare distributions of a defect-type category across multiple vendors, the chi-square test of homogeneity is the appropriate choice.

When you want to know if a categorical outcome is distributed the same across multiple groups, you use a chi-square test of homogeneity. Here, defect type is the categorical outcome and the groups are the three vendors. You'd arrange the data in a contingency table with vendors as rows and defect types as columns, and compare the observed counts in each cell to the counts you’d expect if every vendor had the same distribution of defect types.

Under the null hypothesis, no matter which vendor you look at, the proportion of each defect type is the same across all vendors. The alternative is that at least one vendor has a different distribution of defect types. You compute the chi-square statistic from (observed − expected)²/expected for all cells, where expected counts come from the product of the corresponding row total and column total divided by the grand total.

This test is distinct from the chi-square goodness-of-fit, which checks a single population against a specific expected distribution, and the chi-square test of independence, which asks whether two categorical variables are associated within one population. ANOVA is for comparing means of a continuous variable across groups, not for categorical outcomes. Since the goal here is to compare distributions of a defect-type category across multiple vendors, the chi-square test of homogeneity is the appropriate choice.