For a bag of peanut M&M's, observed counts for Brown 39, Yellow 44, Red 36, Blue 78, Orange 73, and Green 48. Test whether the bag's color distribution matches the official distribution.

For a bag of peanut M&M's, observed counts for Brown 39, Yellow 44, Red 36, Blue 78, Orange 73, and Green 48. Test whether the bag's color distribution matches the official distribution.

• A. Chi-square independence test with 9 degrees of freedom
• B. Chi-square goodness of fit with 5 degrees of freedom
• C. One-way ANOVA
• D. T-test for proportions

Answer: (B)

Think of this as a chi-square goodness-of-fit situation: you have observed counts in several categories (the six M&M colors) and you want to know if they follow a specified distribution (the official color proportions). The test compares what you actually saw to what you would expect if the bag exactly matched the official mix. There are six color categories, and the official distribution fixes the expected proportions, so the degrees of freedom are six minus one, giving five. You’d compute the total number of observed candies, which is 318, then multiply the official proportions by 318 to get the expected counts for each color. The chi-square statistic sums (observed − expected)² divided by the expected across all colors, and you compare that value to a chi-square distribution with five degrees of freedom to assess the fit. This approach is the right one because it specifically tests whether a single categorical distribution matches a specified model. Other options don’t fit: the independence test is for two categorical variables in a contingency table, not a single categorical distribution; ANOVA is for comparing means of continuous outcomes across groups; and a t-test for proportions handles one or two proportions, not goodness-of-fit across multiple categories.

Think of this as a chi-square goodness-of-fit situation: you have observed counts in several categories (the six M&M colors) and you want to know if they follow a specified distribution (the official color proportions). The test compares what you actually saw to what you would expect if the bag exactly matched the official mix.

There are six color categories, and the official distribution fixes the expected proportions, so the degrees of freedom are six minus one, giving five. You’d compute the total number of observed candies, which is 318, then multiply the official proportions by 318 to get the expected counts for each color. The chi-square statistic sums (observed − expected)² divided by the expected across all colors, and you compare that value to a chi-square distribution with five degrees of freedom to assess the fit.

This approach is the right one because it specifically tests whether a single categorical distribution matches a specified model. Other options don’t fit: the independence test is for two categorical variables in a contingency table, not a single categorical distribution; ANOVA is for comparing means of continuous outcomes across groups; and a t-test for proportions handles one or two proportions, not goodness-of-fit across multiple categories.