题目

Problem

Jeff records the number of births announced in his local weekly newspaper each week for $n$ consecutive weeks, where $n$ is an integer. He decides the resulting data can be modelled as independent samples from a Poisson distribution with mean 2.8 and calculates the expected frequencies using this model.

The following table shows the observed frequencies and the expected frequencies, to 2 decimal places.

The observed and expected frequencies for 5 births are not given.

Number of births	Observed frequency ( $O$ )	Expected frequency ( $E$ )
0	8	5.84
1	27	16.35
2	25	22.88
3	16	21.36
4	12	14.95
5	$a$	$b$
6	2	3.91
$\geqslant 7$	1	2.34

(a) Find the value of $a$ and the value of $b$

(2)

The value of $\sum \dfrac{(O - E)^2}{E}$ for the given values for the number of births 0, 1, 2, 3 and 4 is 9.86

(b) Using a 5% significance level, test whether or not this Poisson model is suitable. Show your working clearly, stating your hypotheses, test statistic and critical value.

(7)

(Total for Question 5 is 9 marks)

题目中文翻译

Jeff 连续 n 周记录当地周报上每周公布的新生婴儿数，其中 n 为整数。他认为这些数据可以用均值为 2.8 的泊松分布的独立样本来建模，并据此计算期望频数。

下表给出了观测频数和期望频数，期望频数保留到 2 位小数。

出生数为 5 的观测频数和期望频数没有给出。

出生数	观测频数 (O)	期望频数 (E)
0	8	5.84
1	27	16.35
2	25	22.88
3	16	21.36
4	12	14.95
5	a	b
6	2	3.91
≥ 7	1	2.34

(a) 求 a 和 b 的值。

对于出生数 0、1、2、3 和 4，∑((O - E)^2 / E) = 9.86。

(b) 在 5% 的显著性水平下检验该泊松模型是否适合。请清楚写出你的过程，说明原假设、检验统计量和临界值。

解答

(a)

解法一

思路

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若每周出生数 $X\sim\operatorname{Po}(2.8)$ ，则 0 次出生的概率为 $e^{-2.8}$ ，所以 0 次出生的期望频数为 $ne^{-2.8}$ 。结合题目给出的两位小数及 $n$ 必须为整数，可确定 $n=96$ ；然后分别用观测频数总和及泊松概率求 $a$ 、 $b$ 。

答题过程

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Let $X$ be the number of births announced in a week. Under the proposed model,

X\sim\operatorname{Po}(2.8).

The expected frequency for zero births is

nP(X=0)=ne^{-2.8}.

Since the displayed expected frequency is $5.84$ to 2 decimal places and $n$ is an integer,

96e^{-2.8}=5.837766\ldots\approx5.84,

so $n=96$ .

The known observed frequencies total

8+27+25+16+12+2+1=91.

Hence,

a=96-91=\boxed{5}.

For 5 births, the expected frequency is

\begin{align*} b =&\,96P(X=5)\\ =&\,96\left(e^{-2.8}\frac{2.8^5}{5!}\right)\\ =&\,8.3725\ldots\\ =&\,\boxed{8.37}\quad\text{(2 d.p.)}. \end{align*}

(b)

解法一

思路

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这是卡方拟合优度检验。由于“6 次”和“至少 7 次”的期望频数都小于 5，先合并为一个尾部组，使其观测频数为 $3$ 、期望频数为 $6.25$ 。再把题目已给的前五组贡献 $9.86$ 与其余两组贡献相加。模型参数 2.8 是预先给定而非由样本估计，因此 7 个合并后类别给出 $7-1=6$ 个自由度。

答题过程

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Let $H_0$ be the hypothesis that a Poisson distribution with mean 2.8 is a suitable model.

Let $H_1$ be the hypothesis that a Poisson distribution with mean 2.8 is not a suitable model.

The final two classes must be combined because their individual expected frequencies are less than 5. For the combined class,

O=2+1=3, \qquad E=3.91+2.34=6.25.

Using the contribution $9.86$ given for 0 to 4 births,

\begin{align*} \chi^2 =&\,9.86+\frac{(5-8.37)^2}{8.37}\\ +&\,\frac{(3-6.25)^2}{6.25}\\ =&\,12.91\ldots\\ =&\,12.9\quad\text{(3 s.f.)}. \end{align*}

After combining, there are 7 classes. Since the mean 2.8 was specified and was not estimated from the data,

\nu=7-1=6.

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

\chi^2_{6,0.05}=12.592.

Since

12.9>12.592,

the result is significant and $H_0$ is rejected. The data are not consistent with a Poisson distribution with mean 2.8, so the proposed model is not suitable.

IAL 2026 Jan S3 A Q5

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程