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CIE 9231 2024 Nov Paper 41 Q1

A Level / CIE / FS

CIE 9231 2024 Nov Paper 41 Paper · Question 1

题目

Problem

Ellie is investigating the heights of two types of beech tree, AA and BB, in a certain region. She has chosen a random sample of 60 beech trees of type AA in the region, recorded their heights, xx m, and calculated unbiased estimates for the population mean and population variance as 35.635.6 m and 4.954.95 m2^2 respectively.

Ellie also chooses a random sample of 50 beech trees of type BB in the region and records their heights, yy m. Her results are summarised as follows.

∑y=1654∑y2=54 850\sum y = 1654 \qquad \sum y^2 = 54\,850

Find a 95% confidence interval for the difference between the population mean heights of type AA and type BB beech trees in the region.

[6]
题目中文翻译

Ellie 正在调查某地区两种山毛榉树 AA 和 BB 的高度。她在该地区随机选取了 60 棵 AA 型山毛榉树,记录其高度 xx m,并计算出总体均值和总体方差的无偏估计分别为 35.635.6 m 和 4.954.95 m2^2。

Ellie 还在该地区随机选取了 50 棵 BB 型山毛榉树,并记录其高度 yy m。她的结果总结如下。

∑y=1654∑y2=54 850\sum y = 1654 \qquad \sum y^2 = 54\,850

求该地区 AA 型和 BB 型山毛榉树总体平均高度之差的 95% 置信区间。

(6)
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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先求 BB 组的无偏方差。两组样本量都较大,用均值差的正态近似,独立样本的两个方差项相加。

答题过程

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For type BB, the sample mean is yˉ=1654/50=33.08\bar y=1654/50=33.08 and the unbiased variance estimate is

sB2=54850−16542/5049=2.768979….s_B^2=\frac{54850-1654^2/50}{49}=2.768979\ldots.

Using the two independent samples and the large-sample normal approximation, the estimated standard error of xˉ−yˉ\bar x-\bar y is

4.9560+sB250=0.371321….\sqrt{\frac{4.95}{60}+\frac{s_B^2}{50}}=0.371321\ldots.

The 95%95\% confidence interval for μA−μB\mu_A-\mu_B is

 (35.6−33.08)±1.96(0.371321…)= (1.79221…,3.247789…).\begin{align*} &\,(35.6-33.08)\pm1.96(0.371321\ldots)\\ =&\,(1.79221\ldots,3.247789\ldots). \end{align*}

Thus the interval is (1.79,3.25)(1.79,3.25) m.