下表给出了 250 窝幼犬中每窝雄性幼犬的数量，每窝均有 5 只幼犬。

Jeff 认为二项分布会是一个合适的模型。

(a) 求这些每窝 5 只的幼犬中雄性所占的比例。

Jeff 计算得到如下期望频数，保留到 2 位小数。

(b) 求 $r$ 的值。

对于表中除 $r$ 以外给出的数据，$\sum\frac{(O-E)^2}{E}=5.70$，保留到 2 位小数。

(c) 在 5% 的显著性水平下，检验二项分布是否适合作为这 250 窝幼犬中每窝雄性幼犬数量的模型。 你应写出原假设、自由度和所用临界值。

先用“雄性数量 × 对应窝数”求出雄性小狗总数，再除以 $250\times5$ 只小狗的总数，所得样本比例就是二项模型中 $p$ 的估计值。

The total number of male puppies is

The total number of puppies is $250\times5=1250$. Therefore, the proportion of male puppies is

各格期望频数之和必须等于总窝数 250。按照题表中已经保留两位小数的其他五格，从 250 中扣除它们即可得到 $r$。

The expected frequencies must sum to 250. Hence

Equivalently, using $\widehat p=0.488$ from part (a),

题目已经给出其余五格的卡方贡献之和，所以只需计算“0 只雄性”这一格的贡献并加上 5.70。六格中估计了一个参数 $p$，因此自由度为 $6-1-1=4$，再与 $5\%$ 临界值比较。

The hypotheses are

For the cell corresponding to no male puppies, the contribution to the test statistic is

Therefore,

There are 6 cells and one parameter has been estimated, so

At the $5\%$ significance level, the critical value is $9.488$. Since

$H_0$ is rejected. There is significant evidence that a binomial distribution is not a suitable model for the number of male puppies in these litters.

将每一格的 $(O-E)^2/E$ 展开后求和，利用 $\sum O=\sum E=250$，可把统计量改写为 $\sum O^2/E-250$。直接使用六格期望频数计算，避免逐格求差和平方；最后仍按四个自由度作检验。

The hypotheses are

Since $\sum O=\sum E=250$,

Using $r=8.79$ and the other expected frequencies in the table,

There are 6 cells and one estimated parameter, so the number of degrees of freedom is $6-1-1=4$. The critical value at the $5\%$ significance level is $9.488$.

Since $10.95>9.488$, reject $H_0$. There is significant evidence that a binomial distribution is not a suitable model for the number of male puppies in these litters.