题目

Problem

The following table shows the number of male puppies born in 250 dog litters of size 5.

Number of males	0	1	2	3	4	5
Number of litters	2	40	90	85	30	3

Jeff believes that a binomial distribution would be a suitable model for these data.

(a) Find the proportion of male puppies born in these litters of size 5.

(2)

Jeff calculates expected frequencies, to 2 decimal places, as follows.

Number of males	0	1	2	3	4	5
Expected frequencies	r	41.92	79.91	76.16	36.30	6.92

(b) Find the value of r.

(2)

The value of ∑((O - E)^2 / E) for the given values in the table excluding r is 5.70 to 2 decimal places.

(c) Using a 5% significance level, test whether or not a binomial distribution is a suitable model for the number of male puppies born in these 250 litters. You should state the hypotheses, the degrees of freedom and the critical value used.

(7)

(Total for Question 5 is 11 marks)

题目中文翻译

下表给出了 250 个大小为 5 的狗窝中出生的雄性小狗数量。

雄性小狗数量	0	1	2	3	4	5
狗窝数	2	40	90	85	30	3

Jeff 认为二项分布会是一个合适的模型。

(a) 求这些大小为 5 的狗窝中出生雄性小狗所占的比例。

Jeff 计算得到如下期望频数，保留到 2 位小数。

雄性小狗数量	0	1	2	3	4	5
期望频数	r	41.92	79.91	76.16	36.30	6.92

(b) 求 r 的值。

对于表中除 r 以外给出的数据，∑((O - E)^2 / E) = 5.70，保留到 2 位小数。

解答

(a)

解法一

思路

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先用“雄性数量 × 对应窝数”求出雄性小狗总数，再除以 $250\times5$ 只小狗的总数，所得样本比例就是二项模型中 $p$ 的估计值。

答题过程

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The total number of male puppies is

\begin{align*} &\,0(2)+1(40)+2(90)\\ &\qquad{}+3(85)+4(30)+5(3)\\ =&\,610. \end{align*}

The total number of puppies is $250\times5=1250$ . Therefore, the proportion of male puppies is

\begin{align*} \widehat p =&\,\frac{610}{1250}\\ =&\,\boxed{0.488}. \end{align*}

(b)

解法一

思路

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各格期望频数之和必须等于总窝数 250。按照题表中已经保留两位小数的其他五格，从 250 中扣除它们即可得到 $r$ ；这是官方评分方案采用的计算路线。

答题过程

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The expected frequencies must sum to 250. Hence

\begin{align*} r =&\,250-(41.92+79.91+76.16\\ &\qquad{}+36.30+6.92)\\ =&\,\boxed{8.79}. \end{align*}

(c)

解法一

思路

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题目已经给出其余五格的卡方贡献之和，所以只需计算“0 只雄性”这一格的贡献并加上 5.70。六格中估计了一个参数 $p$ ，因此自由度为 $6-1-1=4$ ，再与 $5\%$ 临界值比较。

答题过程

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The hypotheses are

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

For the cell corresponding to no male puppies, the contribution to the test statistic is

\begin{align*} \frac{(O-E)^2}{E} =&\,\frac{(2-8.79)^2}{8.79}\\ =&\,5.245\ldots. \end{align*}

Therefore,

\begin{align*} \chi^2 =&\,5.70+5.245\ldots\\ =&\,10.945\ldots\\ =&\,10.95\quad\text{(2 d.p.)}. \end{align*}

There are 6 cells and one parameter has been estimated, so

\text{degrees of freedom}=6-1-1=4.

At the $5\%$ significance level, the critical value is $9.488$ . Since

10.95>9.488,

$H_0$ is rejected. There is significant evidence that a binomial distribution is not a suitable model for the number of male puppies in these litters.

IAL 2025 Jun S3 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

(c)

解法一

思路

答题过程