题目

Problem

A supermarket sells apples in trays. Each tray contains 10 apples.

Trays are inspected for rotten apples daily. The following table shows the numbers of rotten apples in a random sample of 150 trays on one day.

Number of rotten apples	0	1	2	3	4	5	6
Frequency	25	47	37	24	8	6	3

The manager believes that these data can be modelled by a binomial distribution B(10, p).

(a) Use these data to estimate the value of the parameter p for this model. Give your answer to 2 decimal places.

(2)

The manager uses the estimated value of p to 2 decimal places and calculates expected frequencies to 2 decimal places as follows.

Number of rotten apples	0	1	2	3	4	5	$\geqslant 6$
Expected frequency	20.62	45.26	44.71	26.17	10.05	r	s

(b) Show that the value of r is 2.65 to 2 decimal places.

(1)

The manager says that the cells for 4, 5 and $\geqslant 6$ need to be combined into a single cell.

(d) Explain why it is necessary for the manager to do this.

(1)

The value of χ^2 for the first 4 frequencies given in each table is 2.51.

(e) Using a 10% significance level, test whether or not a binomial distribution is a suitable model. You must state your hypotheses, test statistic and the critical value used.

(7)

(Total for Question 4 is 12 marks)

题目中文翻译

一家超市出售装有 10 个苹果的苹果筐。

每天都会检查这些苹果筐中的腐烂苹果数。下表给出了某一天随机抽取的 150 个苹果筐中腐烂苹果的数量。

腐烂苹果数	0	1	2	3	4	5	$\geqslant 6$
频数	25	47	37	24	8	6	3

经理认为这些数据可以用二项分布 B(10, p) 来建模。

(a) 用这些数据估计参数 p 的值，答案保留 2 位小数。

经理将 p 估计到 2 位小数后，计算出如下期望频数。

腐烂苹果数	0	1	2	3	4	5	6
期望频数	20.62	45.26	44.71	26.17	10.05	r	s

(b) 证明 r 的值为 2.65，保留到 2 位小数。

经理说必须把 4、5 和至少 6 的格子合并成一个格子。

(d) 解释为什么有必要这样做。

对两张表中前 4 个频数计算得到的 χ^2 值为 2.51。

(e) 在 10% 的显著性水平下，检验二项分布是否适合作为模型。你必须写出原假设、检验统计量和所用临界值。

解答

(a)

解法一

思路

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先由频数表求样本中腐烂苹果的总数，再除以苹果总数 $150\times10$ ，所得样本比例就是 $p$ 的估计值。

答题过程

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The total number of rotten apples is

\begin{align*} &\,0(25)+1(47)+2(37)+3(24)\\ &\qquad{}+4(8)+5(6)+6(3)\\ =&\,273. \end{align*}

There are $150\times10=1500$ apples in total. Therefore,

\begin{align*} \widehat p =&\,\frac{273}{1500}\\ =&\,0.182\\ \approx&\,\boxed{0.18}\quad\text{(2 d.p.)}. \end{align*}

(b)

解法一

思路

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使用题目指定的两位小数估计值 $p=0.18$ 。每筐恰有 5 个腐烂苹果的概率由二项分布给出，再乘以 150 筐便得到期望频数 $r$ 。

答题过程

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Let $X$ be the number of rotten apples in a tray. Using the estimated value of $p$ ,

X\sim\operatorname{B}(10,0.18).

Hence

\begin{align*} r =&\,150P(X=5)\\ =&\,150\binom{10}{5}(0.18)^5(0.82)^5\\ =&\,2.647\ldots\\ =&\,\boxed{2.65}\quad\text{(2 d.p.)}. \end{align*}

(c)

解法一

思路

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各格期望频数之和应等于总频数 150。按照题目表格已经保留两位小数的数值，从 150 中减去其余各格即可得到 $s$ ，这也与官方评分方案采用的数值一致。

答题过程

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The expected frequencies must sum to 150. Therefore, using the expected frequencies in the table,

\begin{align*} s =&\,150-(20.62+45.26+44.71\\ &\qquad{}+26.17+10.05+2.65)\\ =&\,\boxed{0.54}. \end{align*}

(d)

解法一

思路

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卡方拟合优度检验要求每一格的期望频数至少为 5，而腐烂苹果数为 5 和至少为 6 的两格都不满足此条件，所以必须与相邻格合并。

答题过程

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The expected frequencies for 5 and at least 6 rotten apples are both less than 5. The cells must therefore be combined with the cell for 4 rotten apples so that the expected frequency in the combined cell is at least 5.

(e)

解法一

思路

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合并后三格的观察频数为 17、期望频数为 13.24。把这一格的卡方贡献加到题目已给出的 2.51，再根据合并后的 5 格及已估计的一个参数确定自由度为 3，最后与 $10\%$ 临界值比较。

答题过程

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Let

\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 combined cell,

\begin{align*} O=&\,8+6+3=17,\\ E=&\,10.05+2.65+0.54=13.24. \end{align*}

Therefore, the test statistic is

\begin{align*} \chi^2 =&\,2.51+\frac{(17-13.24)^2}{13.24}\\ =&\,3.577\ldots\\ =&\,3.58. \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 $10\%$ significance level, the critical value is $6.251$ . Since

3.58<6.251,

$H_0$ is not rejected. There is insufficient evidence that the binomial distribution is not a suitable model, so a binomial distribution is suitable for these data.

IAL 2026 Jan S3 Q4

题目

解答

(a)

解法一

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(b)

解法一

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(c)

解法一

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(d)

解法一

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(e)

解法一

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