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CIE 9709 2023 November Paper 62 Q4

A Level / CIE / S2

CIE 9709 2023 Nov Paper 62 Paper · Question 4

题目

Problem

The height HH, in metres, of mature trees of a certain variety is normally distributed with standard deviation 0.670.67. In order to test whether the population mean of HH is greater than 4.234.23, the heights of a random sample of 200200 trees are measured.

(a) Write down suitable null and alternative hypotheses for the test.

(1)

The sample mean height, h‾\overline{h} metres, of the 200200 trees is found and the test is carried out. The result of the test is to reject the null hypothesis at the 5%5\% significance level.

(b) Find the set of possible values of h‾\overline{h}.

(3)

(c) Ajit said, ‘In (b) we had to assume that H‾\overline{H} is normally distributed, so it was necessary to use the Central Limit Theorem.’

Explain whether you agree with Ajit.

(1)
题目中文翻译

某品种成熟树木的高度 HH（以米为单位）服从标准差为 0.670.67 的正态分布。为了检验 HH 的总体均值是否大于 4.234.23，随机抽取 200200 棵树并测量它们的高度。

(a) 写出适合该检验的原假设和备择假设。

(1)

求得这 200200 棵树的样本平均高度为 h‾\overline{h} 米，并进行检验。检验的结果是在 5%5\% 的显著性水平下拒绝原假设。

(b) 求 h‾\overline{h} 的所有可能取值所构成的集合。

(3)

(c) Ajit 说：“在 (b) 中，我们必须假设 H‾\overline{H} 服从正态分布，因此有必要使用中心极限定理。”

说明你是否同意 Ajit 的看法。

(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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用总体平均树高为参数，检验是否大于4.23。

答题过程

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Let μ\mu be the population mean mature-tree height, in metres.

H0:μ=4.23,H1:μ>4.23.\boxed{H_0:\mu=4.23,\qquad H_1:\mu>4.23}.

(b)

解法一

思路

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拒绝原假设要求右尾标准化统计量超过5%临界值，将不等式还原为样本均值范围，保留精确分位数边界。

答题过程

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Under H0H_0,

H‾∼N(4.23,0.672200).\overline H\sim\mathrm{N}\bigg(4.23,\frac{0.67^2}{200}\bigg).

The upper-tail 5% critical value is Φ−1(0.95)\Phi^{-1}(0.95). Thus rejection requires

hˉ−4.230.67/200>Φ−1(0.95).\frac{\bar h-4.23}{0.67/\sqrt{200}}>\Phi^{-1}(0.95).

Therefore the possible sample means satisfy

hˉ>4.23+Φ−1(0.95)0.67200.\boxed{\bar h>4.23+\Phi^{-1}(0.95)\frac{0.67}{\sqrt{200}}}.

The boundary is approximately 4.3084.308 m, or 4.314.31 m to 3 significant figures.

(c)

解法一

思路

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总体已经给定为正态，因此样本均值精确正态，不需要中心极限定理。

答题过程

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I disagree. The population of tree heights is given to be normally distributed, so the sample mean is exactly normally distributed. The Central Limit Theorem is unnecessary.