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CIE 9231 2025 June Paper 44 Q1

A Level / CIE / FS

CIE 9231 2025 June Paper 44 Paper · Question 1

题目

Problem

A random sample of 12 observations of a normal random variable is taken. The results give unbiased estimates for the population mean and variance as 10.24 and 0.52 respectively.

Test, at the 10% significance level, the null hypothesis that the population mean is 10.6 against the alternative hypothesis that the population mean is less than 10.6.

[4]
题目中文翻译

从一个正态随机变量中抽取容量为 12 的随机样本。结果给出总体均值和总体方差的无偏估计分别为 10.24 和 0.52。

在 10% 显著性水平下,检验原假设“总体均值为 10.6”,备择假设为“总体均值小于 10.6”。

(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

解法一

思路

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样本来自正态总体,方差未知。使用自由度 1111 的左尾单样本 tt 检验。

答题过程

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For H0 ⁣:μ=10.6H_0\colon\mu=10.6 against H1 ⁣:μ<10.6H_1\colon\mu<10.6, the statistic is

t=10.24−10.60.52/12=−1.72938….t=\frac{10.24-10.6}{\sqrt{0.52/12}}=-1.72938\ldots.

The lower 10%10\% critical value with 12−1=1112-1=11 degrees of freedom is −1.363-1.363. Since −1.72938…<−1.363-1.72938\ldots<-1.363, we reject H0H_0. There is sufficient evidence at the 10%10\% level that the population mean is less than 10.610.6.