记录了帮助台每小时收到的邮件数量。随机抽取的 80 个一小时区间的结果如下表。

(a) 证明样本中每小时邮件数的平均值为 3。

经理认为，每小时收到的邮件数量可以用泊松分布来建模。

下表给出了一些期望频数。

(b) 求 $r$、$s$ 和 $t$ 的值，答案保留 3 位小数。

(c) 在 10% 显著性水平下，检验泊松模型是否合理。 你应清楚写出原假设与备择假设、检验统计量和所用的临界值。

用各邮件数乘以相应频数，所得总和除以样本量 $80$，即可求得样本均值。

The sample mean is

以样本均值估计泊松分布参数，所以 $\lambda=3$。各单点的期望频数为 $80\operatorname{P}(X=x)$；最后一组是 $X\geqslant6$，可用总期望频数为 $80$ 求补集。

Let $X\sim\operatorname{Po}(3)$. For $x=0,1,2,\ldots$,

Hence

and

Using $r=3.983$ and $s=8.066$ to 3 decimal places, the remaining expected frequency is chosen so that the total is $80$:

先提出泊松模型是否合理的假设。由于 $X=0$ 的期望频数小于 $5$，把 $0$ 与 $1$ 合并后进行卡方拟合优度检验。合并后有 $6$ 组，并从样本估计了一个参数，因此自由度为 $6-1-1=4$。

The hypotheses are

Since the expected frequency for $X=0$ is less than $5$, the classes $X=0$ and $X=1$ are combined.

The test statistic is

There are $6$ classes after combining, and one parameter has been estimated, so

At the $10\%$ significance level, the critical value is

Since

the test statistic lies in the critical region. Therefore, $H_0$ is rejected. There is sufficient evidence at the $10\%$ significance level to conclude that a Poisson distribution is not a reasonable model for the number of emails received per hour.

也可对合并后的六组使用 $\chi^2=\sum\frac{O^2}{E}-80$。合并是因为第一组期望频数不足 $5$，并非改变泊松分布参数的估计；因此仍须扣除由样本估计的一个参数，自由度为 $4$。

We test whether a Poisson distribution is a reasonable model:

Since the expected frequency for zero emails is below $5$, combine the zero- and one-email classes. Their observed frequency is $1+10=11$ and their expected frequency is $3.983+11.949=15.932$. Using the six combined classes in Method 1,

There are $6-1-1=4$ degrees of freedom because there are six combined classes and one estimated parameter. The $10\%$ critical value is $7.779$. Since $7.900587\ldots>7.779$, we reject $H_0$: there is sufficient evidence that a Poisson distribution is not a reasonable model for the number of emails received per hour.