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CIE 9709 2023 June Paper 63 Q3

A Level / CIE / S2

CIE 9709 2023 June Paper 63 Paper · Question 3

题目

Problem

In a random sample of 100100 students at Luciana’s college, xx students said that they liked exams. Luciana used this result to find an approximate 90%90\% confidence interval for the proportion, pp, of all students at her college who liked exams. Her confidence interval had width 0.157 920.157\,92.

(a) Find the two possible values of xx.

(4)

Suzma independently took another random sample and found another approximate 90%90\% confidence interval for pp.

(b) Find the probability that neither of the two confidence intervals contains the true value of pp.

(1)
题目中文翻译

从 Luciana 所在学院随机抽取的 100100 名学生中，有 xx 人表示喜欢考试。Luciana 用这个结果求出该学院所有学生中喜欢考试的人所占比例 pp 的近似 90%90\% 置信区间。她求得的置信区间宽度为 0.157 920.157\,92。

(a) 求 xx 的两个可能值。

(4)

Suzma 独立地抽取了另一个随机样本，并求出了 pp 的另一个近似 90%90\% 置信区间。

(b) 求这两个置信区间都不包含 pp 的真值的概率。

(1)
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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将区间宽度表达为两倍临界值乘比例标准误。用90%的表值1.645求样本比例的二次方程，再转换为人数。

答题过程

展开

Put p^=x100\hat p=\frac{x}{100}. Using the 90% critical value z=1.645z=1.645, the width equation is

2(1.645)p^(1−p^)100=0.15792.2(1.645)\sqrt{\frac{\hat p(1-\hat p)}{100}}=0.15792.

Hence

p^(1−p^)= 100(0.157922(1.645))2= 0.2304.\begin{align*} \hat p(1-\hat p)=&\,100\bigg(\frac{0.15792}{2(1.645)}\bigg)^2\\[4mm] =&\,0.2304. \end{align*}

Thus

p^2−p^+0.2304=0⟹(p^−0.36)(p^−0.64)=0.\hat p^2-\hat p+0.2304=0 \quad\Longrightarrow\quad (\hat p-0.36)(\hat p-0.64)=0.

Therefore the two possible numbers are x=36 or 64\boxed{x=36\text{ or }64}.

(b)

解法一

思路

展开

两个区间来自明确独立的样本，各自未包含真值的概率为0.1，相乘即可。

答题过程

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Each 90% interval fails to contain pp with probability 0.10. Since the samples were taken independently,

 P(neither contains p)= (0.10)2=0.01.\begin{align*} &\,\mathrm{P}(\text{neither contains }p)\\[4mm] =&\,(0.10)^2=\boxed{0.01}. \end{align*}