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CIE 9709 2026 June Paper 65 Q5

A Level / CIE / S2

CIE 9709 2026 June Paper 65 Paper · Question 5

题目

Problem

The manufacturer of Beta cars claims that the proportion of adults in a certain town who own a Beta car is 15\frac{1}{5}. Raju suspects that the true proportion is less than 15\frac{1}{5}. He plans to carry out a hypothesis test using a random sample of 4040 adults from the town. If fewer than 44 of these adults own a Beta car, he will conclude that the true proportion is less than 15\frac{1}{5}.

(a) Use a binomial distribution to find the smallest significance level of the test. Give your answer correct to 33 significant figures.

(3)

(b) State the probability of a Type I error.

(1)

It is now given that the true proportion is 120\frac{1}{20}.

(c) Find the probability of a Type II error.

(3)
题目中文翻译

Beta 汽车制造商声称，某镇成年人中拥有 Beta 汽车者的比例为 15\frac{1}{5}。Raju 怀疑真实比例小于 15\frac{1}{5}。他计划从该镇随机抽取 4040 名成年人进行假设检验。如果这些人中拥有 Beta 汽车的少于 44 人，他就判定真实比例小于 15\frac{1}{5}。

(a) 用二项分布求该检验的最小显著性水平。答案保留 33 位有效数字。

(3)

(b) 写出发生第一类错误的概率。

(1)

现已知真实比例为 120\frac{1}{20}。

(c) 求发生第二类错误的概率。

(3)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

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显著性水平由原假设真实时落入既定拒绝域的概率决定。“少于四人”包括 0,1,2,30,1,2,3，不能漏掉三人。

答题过程

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Let KK be the number of Beta-car owners in the sample. Under the manufacturer’s claim, K∼B(40,0.2)K\sim\mathrm B(40,0.2), and the proposed critical region is K<4K<4.

Its probability under the null hypothesis is

P(K<4)= (0.8)40+40(0.2)(0.8)39 +(402)(0.2)2(0.8)38 +(403)(0.2)3(0.8)37≈ 0.0284621.\begin{align*} \mathrm P(K<4) =&\,(0.8)^{40}+40(0.2)(0.8)^{39}\\[4mm] &\,\hspace{2pt}+\binom{40}{2}(0.2)^2(0.8)^{38}\\[4mm] &\,\hspace{4pt}+\binom{40}{3}(0.2)^3(0.8)^{37}\\[4mm] \approx&\,0.0284621. \end{align*}

Therefore the smallest significance level for this rejection rule is

2.85%(3 significant figures).\boxed{2.85\%}\quad\text{(3 significant figures)}.

(b)

解法一

思路

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第一类错误是在原假设成立时拒绝它，与(a)所算的拒绝域概率相同。

答题过程

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Using part (a),

P(Type I error)= P(K<4∣p=0.2)≈ 0.0285.\begin{align*} \mathrm P(\text{Type I error}) =&\,\mathrm P(K<4\mid p=0.2)\\[4mm] \approx&\,\boxed{0.0285}. \end{align*}

(c)

解法一

思路

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第二类错误使用真实比例 1/201/20，但仍保留原来的拒绝规则；未拒绝对应样本车主数至少为四。

答题过程

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When the true proportion is 1/20=0.051/20=0.05,

K∼B(40,0.05).K\sim\mathrm B(40,0.05).

A Type II error occurs when K≥4K\ge4. Hence

 P(Type II error)= 1−∑r=03(40r)(0.05)r(0.95)40−r= 1−(0.95)40−40(0.05)(0.95)39 −(402)(0.05)2(0.95)38 −(403)(0.05)3(0.95)37≈ 0.138.\begin{align*} &\,\mathrm P(\text{Type II error})\\[4mm] =&\,1-\sum_{r=0}^3\binom{40}{r}(0.05)^r(0.95)^{40-r}\\[4mm] =&\,1-(0.95)^{40}-40(0.05)(0.95)^{39}\\[4mm] &\,-\binom{40}{2}(0.05)^2(0.95)^{38}\\[4mm] &\,\hspace{2pt}-\binom{40}{3}(0.05)^3(0.95)^{37}\\[4mm] \approx&\,\boxed{0.138}. \end{align*}