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CIE 9709 2025 June Paper 62 Q8

A Level / CIE / S2

CIE 9709 2025 June Paper 62 Paper · Question 8

题目

Problem

Birgitte has a six-sided dice. She suspects that the dice is biased so that the probability, pp, that it will show a six on one throw is less than 16\frac16. She throws the dice 3030 times and finds that it shows a six on exactly 22 throws.

(a) Use a binomial distribution with a 5%5\% significance level to test Birgitte’s suspicion.

(5)

Later, Birgitte carries out a similar test at the 5%5\% significance level, using another 3030 throws of the dice.

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

(2)

(c) Given that the value of pp is actually 0.020.02, calculate the probability of a Type II error.

(3)
题目中文翻译

Birgitte 有一枚六面骰子。她怀疑骰子不均匀,掷一次出现六点的概率 pp 小于 16\frac16。她掷了 3030 次,发现恰有 22 次出现六点。

(a) 用二项分布在 5%5\% 的显著性水平下检验 Birgitte 的猜测。

(5)

后来,Birgitte 又掷了 3030 次,并在 5%5\% 的显著性水平下进行类似的检验。

(b) 计算发生第一类错误的概率。

(2)

(c) 已知 pp 的实际值为 0.020.02,计算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

展开

怀疑出现六点的概率偏小,所以用左尾二项检验;把零、一、二次都纳入尾概率。

答题过程

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Let KK be the number of sixes in 3030 throws. Test

H0:p=16,H1:p<16.H_0:p=\frac16,\qquad H_1:p<\frac16.

Under H0H_0, K∼B(30,1/6)K\sim\mathrm B(30,1/6). For the observed K=2K=2,

P(K≤2)= (56)30 +30(16)(56)29 +(302)(16)2(56)28≈ 0.103.\begin{align*} \mathrm P(K\le2) =&\,\bigg(\frac56\bigg)^{30}\\[4mm] &\,+30\bigg(\frac16\bigg)\bigg(\frac56\bigg)^{29}\\[4mm] &\,\hspace{2pt}+\binom{30}{2}\bigg(\frac16\bigg)^2\bigg(\frac56\bigg)^{28}\\[4mm] \approx&\,0.103. \end{align*}

Since 0.103>0.050.103>0.05, do not reject H0H_0. There is insufficient evidence at the 5%5\% level that the probability of a six is less than 1/61/6, so Birgitte’s suspicion is not supported.

(b)

解法一

思路

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实际第一类错误概率来自可达到的整数拒绝域,不必恰等于名义的 5%5\%;结合(a)排除边界二,再检查边界一。

答题过程

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Under H0H_0,

P(K≤1)= (56)30 +30(16)(56)29≈ 0.029489<0.05.\begin{align*} \mathrm P(K\le1) =&\,\bigg(\frac56\bigg)^{30}\\[4mm] &\,+30\bigg(\frac16\bigg)\bigg(\frac56\bigg)^{29}\\[4mm] \approx&\,0.029489<0.05. \end{align*}

Part (a) gives P(K≤2)≈0.103>0.05\mathrm P(K\le2)\approx0.103>0.05, so the critical region is K≤1K\le1. Therefore

P(Type I error)≈0.0295.\boxed{\mathrm P(\text{Type I error})\approx0.0295}.

(c)

解法一

思路

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真实概率变为 0.020.02 时,检验仍只在至多一次六点时拒绝。第二类错误为至少两次六点。

答题过程

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When p=0.02p=0.02, K∼B(30,0.02)K\sim\mathrm B(30,0.02). A Type II error occurs when K≥2K\ge2, so

 P(Type II error)= 1−P(K≤1∣p=0.02)= 1−[(0.98)30+30(0.02)(0.98)29]≈ 0.121.\begin{align*} &\,\mathrm P(\text{Type II error})\\[4mm] =&\,1-\mathrm P(K\le1\mid p=0.02)\\[4mm] =&\,1-\big[(0.98)^{30}+30(0.02)(0.98)^{29}\big]\\[4mm] \approx&\,\boxed{0.121}. \end{align*}