题目

Problem

The diameters, $x$ millimetres, of a random sample of $200$ discs made by a certain machine were recorded. The results are summarised below.

$n=200\qquad\sum x=2520\qquad\sum x^2=31\,852$

(a) Calculate a $95\%$ confidence interval for the population mean diameter.

(6)

(b) Jean chose $40$ random samples and used each sample to calculate a $95\%$ confidence interval for the population mean diameter.

How many of these $40$ confidence intervals would be expected to include the true value of the population mean diameter?

(1)

题目中文翻译

从某台机器制造的圆片中随机抽取 $200$ 片，记录其直径 $x$ 毫米。结果汇总如下。

$n=200\qquad\sum x=2520\qquad\sum x^2=31\,852$

(a) 计算总体平均直径的 $95\%$ 置信区间。

(6)

(b) Jean 抽取了 $40$ 个随机样本，并用每个样本计算总体平均直径的一个 $95\%$ 置信区间。

预计这 $40$ 个置信区间中有多少个包含总体平均直径的真值？

(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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先求样本均值与无偏方差，再用大样本正态均值区间公式。

答题过程

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The sample estimates are

\begin{align*} \bar x=&\,\frac{2520}{200}=12.6,\\[4mm] s^2=&\,\frac{31852-2520^2/200}{199}=\frac{100}{199}. \end{align*}

Using the large-sample normal approximation, the 95% confidence interval is

12.6\pm1.96\sqrt{\frac{100/199}{200}} =12.6\pm0.0982459\ldots.

Thus the interval for the population mean diameter is

\boxed{(12.5,\ 12.7)\text{ mm}}\quad\text{(3 s.f.).}

(b)

解法一

思路

展开

每个95%区间的覆盖概率为0.95，期望覆盖个数为总数乘该概率。

答题过程

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The expected number is

40(0.95)=\boxed{38}.

CIE 9709 2022 June Paper 61 Q1

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程