题目

Problem

The number of jobs sent to a printer per hour in a small office is recorded for 120 hours. The results are summarised in the following table.

Number of jobs	0	1	2	3	4	5
Frequency	24	34	28	21	8	5

(a) Show that the mean number of jobs sent to the printer per hour for these data is 1.75

(1)

The office manager believes that the number of jobs sent to the printer per hour can be modelled using a Poisson distribution. The office manager uses the mean given in part (a) to calculate the expected frequencies for this model. Some of the results are given in the following table.

Number of jobs	0	1	2	3	4	5 or more
Expected frequency	20.85	36.49	31.93	r	s	3.95

(b) Show that the value of s is 8.15 to 2 decimal places.

(1)

The value of $\sum \frac{(O_i-E_i)^2}{E_i}$ for the first four frequencies in the table is 1.43

(d) Test, at the 5% level of significance, whether or not the number of jobs sent to the printer per hour can be modelled using a Poisson distribution. Show your working clearly, stating your hypotheses, test statistic and critical value.

(7)

(Total for Question 4 is 10 marks)

题目中文翻译

某小型办公室记录了 120 小时内每小时送往打印机的作业数量。结果汇总如下表。

作业数	0	1	2	3	4	5
频数	24	34	28	21	8	5

(a) 证明这些数据中每小时送往打印机的作业平均数为 1.75。

办公室经理认为，每小时送往打印机的作业数量可以用泊松分布建模。经理使用 (a) 小题中的平均数计算该模型的期望频数。部分结果如下表。

作业数	0	1	2	3	4	5 或以上
期望频数	20.85	36.49	31.93	r	s	3.95

(b) 证明 s 的值为 8.15，保留到 2 位小数。

表中前四个频数对应的 $\sum \frac{(O-E)^2}{E}$ 的值为 1.43。

(d) 在 5% 显著性水平下检验每小时送往打印机的作业数量是否可以用泊松分布建模。请清楚写出你的原假设、检验统计量和临界值。

解答

(a)

解法一

思路

展开

用每小时作业数乘以对应频数，求出 120 小时内的作业总数，再除以 120。

答题过程

展开

The sample mean is

\begin{align*} \overline x =&\,\frac{0(24)+1(34)+2(28)}{120}\\ &\qquad{}+\frac{3(21)+4(8)+5(5)}{120}\\ =&\,\frac{210}{120}\\ =&\,1.75, \end{align*}

as required.

(b)

解法一

思路

展开

在泊松模型中，参数取 (a) 的均值 $\lambda=1.75$ 。 $s$ 是恰有 4 个作业的期望频数，所以用 $120P(X=4)$ 计算并自然得到题目要求的数值。

答题过程

展开

Let $X\sim\operatorname{Po}(1.75)$ . Then

\begin{align*} s =&\,120P(X=4)\\ =&\,120\left( \frac{e^{-1.75}(1.75)^4}{4!} \right)\\ =&\,8.149\ldots\\ =&\,8.15\quad\text{(2 d.p.)}, \end{align*}

as required.

(c)

解法一

思路

展开

$r$ 是恰有 3 个作业的期望频数，计算 $120P(X=3)$ 即可。

答题过程

展开

\begin{align*} r =&\,120P(X=3)\\ =&\,120\left( \frac{e^{-1.75}(1.75)^3}{3!} \right)\\ =&\,18.626\ldots\\ =&\,\boxed{18.63}\quad\text{(2 d.p.)}. \end{align*}

(d)

解法一

思路

展开

“5 个或以上”一格的期望频数 $3.95<5$ ，必须与“4 个”一格合并。题目已给前四格的卡方贡献为 1.43，所以只需加入合并格的贡献。合并后有 5 格，并估计了一个泊松参数，故自由度为 3。

答题过程

展开

The hypotheses are

\begin{aligned} H_0:&\ \text{A Poisson distribution is a suitable model.}\\ H_1:&\ \text{A Poisson distribution is not a suitable model.} \end{aligned}

The cells for 4 jobs and 5 or more jobs are combined because the latter has expected frequency less than 5. For the combined cell,

\begin{align*} O=&\,8+5=13,\\ E=&\,8.15+3.95=12.10. \end{align*}

Therefore,

\begin{align*} \chi^2 =&\,1.43+\frac{(13-12.10)^2}{12.10}\\ =&\,1.4969\ldots\\ =&\,1.50\quad\text{(3 s.f.)}. \end{align*}

There are 5 cells after combining, and one parameter has been estimated. Hence

\text{degrees of freedom}=5-1-1=3.

At the $5\%$ significance level, the critical value is $7.815$ . Since

1.50<7.815,

$H_0$ is not rejected. There is insufficient evidence that a Poisson distribution is not suitable; the number of jobs sent to the printer per hour is consistent with a Poisson model.

IAL 2024 Jan S3 Q4

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程

(d)

解法一

思路

答题过程