 Solution

Use the Pearson chi-squared test of homogeneity for two multinomial samples. Under equal preference distributions, estimate the common probabilities from pooled counts, giving $(1/4,1/4,1/2)$. With row totals forty and sixty, the expected table isThe Pearson chi-squared statistic for contingency tables isThere are $(2-1)(3-1)=2$ degrees of freedom, after fitting the common category probabilities. All expected counts exceed five. ThusAt five percent, do not reject the common preference distribution. These data are consistent with that hypothesis; failure to reject does not prove equality. This conclusion differs from part a because a common nonuniform distribution, especially an overall preference for chocolate, is allowed here.