题目

Problem

Each of a random sample of $80$ adults gave an estimate, $h$ metres, of the height of a particular building. The results were summarised as follows.

$n=80\qquad\sum h=2048\qquad\sum h^2=52\,760$

(a) Calculate unbiased estimates of the population mean and variance.

(3)

(b) Using this sample, the upper boundary of an $\alpha\%$ confidence interval for the population mean is $26.0$ .

Find the value of $\alpha$ .

(4)

题目中文翻译

随机抽取 $80$ 名成年人，每人对某栋建筑的高度给出一个估计值，记为 $h$ 米。结果汇总如下。

$n=80\qquad\sum h=2048\qquad\sum h^2=52\,760$

(a) 计算总体均值和方差的无偏估计值。

(3)

(b) 用该样本求得总体均值的一个 $\alpha\%$ 置信区间，其上限为 $26.0$ 。

求 $\alpha$ 的值。

(4)

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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用样本均值及分母79的无偏方差公式。

答题过程

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The unbiased estimate of the population mean is

\bar h=\frac{2048}{80}=\boxed{25.6\text{ m}}.

The unbiased estimate of the population variance is

\begin{align*} s^2=&\,\frac{52760-2048^2/80}{79}\\[4mm] =&\,\frac{1656}{395}\\[4mm] \approx&\,\boxed{4.19\text{ m}^2}\quad\text{(3 s.f.).} \end{align*}

(b)

解法一

思路

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把上限26代入均值区间公式，使用未舍入方差求z，再换算中央置信概率。

答题过程

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The random sample is large, so the Central Limit Theorem gives an approximately normal sampling distribution for the sample mean. Using the unrounded variance from part (a),

25.6+z\sqrt{\frac{1656/395}{80}}=26.0.

Hence

\begin{align*} z=&\,\frac{26.0-25.6}{\sqrt{(1656/395)/80}}\\[4mm] =&\,1.747324\ldots. \end{align*}

Thus

\alpha=100\bigl(2\Phi(z)-1\bigr) =91.94188\ldots,

giving $\boxed{\alpha=91.9}$ (3 s.f.).

CIE 9709 2022 November Paper 62 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