题目

Problem

A random sample of $250$ people living in Barapet was chosen. It was found that $78$ of these people owned a BETEC phone.

(a) Calculate an approximate $98\%$ confidence interval for the proportion of people living in Barapet who own a BETEC phone.

(3)

(b) Manjit claims that more than $40\%$ of the people living in Barapet own a BETEC phone.

Use your answer to part (a) to comment on this claim.

(1)

题目中文翻译

从 Barapet 的居民中随机抽取 $250$ 人，发现其中有 $78$ 人拥有一部 BETEC 手机。

(a) 计算 Barapet 居民中拥有 BETEC 手机的人所占比例的近似 $98\%$ 置信区间。

(3)

(b) Manjit 声称，Barapet 有超过 $40\%$ 的居民拥有 BETEC 手机。

根据 (a) 的答案评价这一说法。

(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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用样本比例78/250及比例标准误建立98%正态近似区间。

答题过程

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The sample proportion is $\hat p=\frac{78}{250}=0.312$ . With $z=\Phi^{-1}(0.99)=2.326348\ldots$ , an approximate 98% confidence interval is

\begin{align*} &\,\hat p\pm z\sqrt{\frac{\hat p(1-\hat p)}{250}}\\[4mm] =&\,0.312\pm2.326348\ldots\sqrt{\frac{0.312(0.688)}{250}}\\[4mm] =&\,0.312\pm0.0681673\ldots. \end{align*}

Thus the interval is

\boxed{(0.244,\ 0.380)}\quad\text{(3 s.f.).}

(b)

解法一

思路

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区间上限低于0.4，可认为大于40%的说法不太可能正确，但不能作确定断言。

答题过程

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Manjit’s claim is unlikely to be true: the entire 98% confidence interval lies below 0.40.

CIE 9709 2024 March Paper 62 Q2

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程