一个停车场的经理声称，进入停车场的汽车数量服从均值为 $2.8$ 的 Poisson 分布。在一个工作日中，连续每 5 分钟记录进入停车场的汽车数量。下表给出了观察频数，以及大部分期望频数和它们对 $\chi^2$ 检验统计量的贡献。

(a) 求 $p$ 和 $q$ 的值。

(b) 在 5% 显著性水平下进行拟合优度检验，以调查该经理的声称。

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。

先求泊松分布取 $3$ 的概率，再乘以记录时段总数。用观测频数与期望频数求这一格的贡献。

The observed frequencies sum to $100$. Under $X\sim\operatorname{Poisson}(2.8)$,

Thus $q=(29-p)^2/p=2.048888\ldots$. To three significant figures, $p=22.2$ and $q=2.05$.

题目给定泊松均值，无须从数据估计。检查期望频数后，将七格贡献相加并用六个自由度。

The null hypothesis is that $\operatorname{Poisson}(2.8)$ fits the number of cars entering per five-minute period. The alternative is that it does not fit. All expected frequencies exceed $5$, so no cells need to be combined. The statistic is

With $7-1=6$ degrees of freedom, the upper $5\%$ critical value is $12.592$. Since $11.4<12.592$, we fail to reject the null hypothesis. At the $5\%$ level there is insufficient evidence that the proposed Poisson model fails to fit the car counts.