Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 3 / 8C (Statistics)

Under the null hypothesis the three preference probabilities for girls are each $1/3$. Conditional on their total of sixty, the counts have a multinomial distribution, giving expected counts $(20,20,20)$. The Pearson chi-squared goodness-of-fit test statistic isNo probabilities have been fitted, so the asymptotic null distribution is chi-squared distribution with $3-1=2$ degrees of freedom. All expected counts are comfortably large for this approximation. For two degrees of freedom its upper-tail probability is $e^{-X^2/2}$, soAt the conventional five-percent or one-percent significance level, reject equal attractiveness: there is substantial evidence against the null distribution. No level was specified, so reporting the p-value makes the decision rule explicit. The test assumes independent responses and a sampling interpretation appropriate to the children surveyed.

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.