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CIE 9709 2022 March Paper 62 Q3

A Level / CIE / S2

CIE 9709 2022 March Paper 62 Paper · Question 3

题目

Problem

A random sample of 500500 households in a certain town was chosen. Using this sample, a confidence interval for the proportion, pp, of all households in that town that owned two or more cars was found to be 0.355<p<0.4450.355<p<0.445.

Find the confidence level of this confidence interval. Give your answer correct to the nearest integer.

(5)
题目中文翻译

从某镇随机抽取 500500 户家庭。利用这个样本,求得该镇拥有两辆或更多汽车的家庭占全镇家庭的比例 pp 的一个置信区间为 0.355<p<0.4450.355<p<0.445。

求这个置信区间的置信水平,答案精确到最接近的整数。

(5)
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

解法一

思路

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区间中点给样本比例0.4,半宽0.045除以标准误得到临界值,再转回双尾置信水平。

答题过程

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The sample proportion and the margin of error are

p^= 0.355+0.4452=0.4,h= 0.445−0.3552=0.045.\begin{align*} \widehat p=&\,\frac{0.355+0.445}{2}=0.4,\\[4mm] h=&\,\frac{0.445-0.355}{2}=0.045. \end{align*}

Using h=zp^(1−p^)/nh=z\sqrt{\widehat p(1-\widehat p)/n},

z=0.0450.4(0.6)/500=2.05396….z=\frac{0.045}{\sqrt{0.4(0.6)/500}}=2.05396\ldots.

Hence the confidence level is

 100(2Φ(2.05396…)−1)%= 96.002…%.\begin{align*} &\,100\bigl(2\Phi(2.05396\ldots)-1\bigr)\%\\[4mm] =&\,96.002\ldots\%. \end{align*}

To the nearest integer, this is 96%\boxed{96\%}.