= 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 is
$$
\boxed{X^2=\frac{(13-20)^2+(14-20)^2+(33-20)^2}{20}=12.7.}
$$
No 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 $e^{-X^2/2}$, so
$$
\boxed{p=e^{-6.35}\simeq0.00175.}
$$
At the conventional five-percent or one-percent significance level, \b[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.
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