一名统计学家认为，某咨询中心在 10 分钟区间内接到的电话数量可以用 Poisson 分布 $\operatorname{Po}(1.9)$ 建模。在 100 天中，每天随机选取一个 10 分钟区间并记录接到的电话数量。结果总结在下表中，同时列出了一些与分布 $\operatorname{Po}(1.9)$ 对应的期望频数。

(a) 完成表格。

(b) 在 10% 显著性水平下进行拟合优度检验，以判断该统计学家的看法是否合理。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

用 $100$ 乘泊松分布的概率。分别代入缺失三格的次数。

For $N\sim\operatorname{Poisson}(1.9)$, the expected frequency at $j$ is $100\mathrm e^{-1.9}1.9^j/j!$. For $j=3,4,5$ these are respectively $17.098$, $8.122$ and $3.086$ to three decimal places.

最后两格合计的期望频数仍小于 $5$，因此合并 $4$ 次及以上。均值已给定，无须为参数估计再减自由度。

Test $H_0$ that $\operatorname{Poisson}(1.9)$ fits the number of calls per ten minutes against $H_1$ that it does not fit. Combine the last three categories. The resulting observed counts are $10,18,35,21,16$, with expected counts $14.957,28.418,26.997,17.098,12.530$ shown to three decimal places. All pooled expected counts exceed $5$.

Using unrounded expected counts,

There are $5-1=4$ degrees of freedom. The upper $10\%$ critical value is $7.779$. Since $9.68582\ldots>7.779$, we reject $H_0$. The data give sufficient evidence at the $10\%$ level that $\operatorname{Poisson}(1.9)$ does not fit the call counts.