题目

Problem

The number of emails per hour received by a helpdesk were recorded. The results for a random sample of 80 one-hour periods are shown in the table.

Number of emails per hour	0	1	2	3	4	5	6
Frequencies	1	10	23	15	19	9	3

(a) Show that the mean number of emails per hour in the sample is 3

(1)

The manager believes that the number of emails per hour received could be modelled by a Poisson distribution.

The following table shows some of the expected frequencies.

Number of emails per hour	Expected Frequencies
0	r
1	11.949
2	17.923
3	17.923
4	13.443
5	s
6+	t

(b) Find the values of r, s and t, giving your answers to 3 decimal places.

(4)

(c) Using a 10% significance level, test whether or not a Poisson model is reasonable. You should clearly state your hypotheses, test statistic and the critical value used.

(7)

(Total 12 marks)

题目中文翻译

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。

每小时邮件数	0	1	2	3	4	5	6
频数	1	10	23	15	19	9	3

(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。

每小时邮件数	期望频数
0	r
1	11.949
2	17.923
3	17.923
4	13.443
5	s
6+	t

(b) 求 r、s 和 t 的值，答案保留 3 位小数。

解答

(a)

解法一

思路

展开

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。

答题过程

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The sample mean is

\begin{align*} \bar{x} =&\,\frac{0(1)+1(10)+2(23)+3(15)+4(19)+5(9)+6(3)}{80}\\ =&\,\frac{240}{80}\\ =&\,\boxed{3}. \end{align*}

(b)

解法一

思路

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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。

答题过程

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

\operatorname{P}(X=x)=\frac{e^{-3}3^x}{x!}.

Hence

\begin{align*} r =&\,80\operatorname{P}(X=0)\\ =&\,80e^{-3}\\ =&\,3.982965\ldots\\ =&\,\boxed{3.983}, \end{align*}

and

\begin{align*} s =&\,80\operatorname{P}(X=5)\\ =&\,80\left(\frac{e^{-3}3^5}{5!}\right)\\ =&\,8.065505\ldots\\ =&\,\boxed{8.066}. \end{align*}

Using the fact that the expected frequencies sum to $80$ ,

\begin{align*} t =&\,80-(r+11.949+17.923+17.923+13.443+s)\\ =&\,6.713\ldots\\ =&\,\boxed{6.713}. \end{align*}

(c)

解法一

思路

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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 $6-1-1=4$ 。

答题过程

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The hypotheses are

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

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

Number of emails	Observed frequency, $O$	Expected frequency, $E$
$\leqslant 1$	11	15.932
$2$	23	17.923
$3$	15	17.923
$4$	19	13.443
$5$	9	8.066
$\geqslant 6$	3	6.713

The test statistic is

\begin{align*} \chi^2 =&\,\sum\frac{(O-E)^2}{E}\\ =&\,\frac{(11-15.932)^2}{15.932} +\frac{(23-17.923)^2}{17.923}\\ &+\frac{(15-17.923)^2}{17.923} +\frac{(19-13.443)^2}{13.443}\\ &+\frac{(9-8.066)^2}{8.066} +\frac{(3-6.713)^2}{6.713}\\ =&\,7.901\ldots \end{align*}

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

\nu=6-1-1=4.

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

\chi_{4}^{2}(0.10)=7.779.

Since

7.901>7.779,

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.

IAL 2022 Jan S3 UNUSED Q6

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