 Solution

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.