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CIE 9231 2026 June Paper 43 Q1

A Level / CIE / FS

CIE 9231 2026 June Paper 43 Paper · Question 1

题目

Problem

A random sample of 99 observations of a normally distributed random variable XX is as follows.

 19.421.324.020.718.9 21.222.523.420.8\begin{align*} &\,19.4\quad21.3\quad24.0\quad20.7\quad18.9\\ &\,21.2\quad22.5\quad23.4\quad20.8 \end{align*}

The population mean of XX is μ\mu.

Test, at the 2.5%2.5\% significance level, the null hypothesis μ=20\mu=20 against the alternative hypothesis μ>20\mu>20.

(6)
题目中文翻译

从服从正态分布的随机变量 XX 中随机抽取的 99 个观测值如下。

 19.421.324.020.718.9 21.222.523.420.8\begin{align*} &\,19.4\quad21.3\quad24.0\quad20.7\quad18.9\\ &\,21.2\quad22.5\quad23.4\quad20.8 \end{align*}

XX 的总体均值为 μ\mu。

在 2.5%2.5\% 的显著性水平下,检验原假设 μ=20\mu=20,备择假设为 μ>20\mu>20。

(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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总体正态而方差未知,用单样本 tt 检验。备择假设为均值大于 2020,所以取右尾。

答题过程

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The sample gives ∑x=192.2\sum x=192.2 and ∑x2=4127.64\sum x^2=4127.64. Thus

 xˉ=192.2/9=21.35555…, s2=4127.64−192.22/98=2.88777….\begin{align*} &\,\bar x=192.2/9=21.35555\ldots,\\ &\,s^2=\frac{4127.64-192.2^2/9}{8}=2.88777\ldots. \end{align*}

The test statistic for H0 ⁣:μ=20H_0\colon\mu=20 against H1 ⁣:μ>20H_1\colon\mu>20 is

t=xˉ−20s2/9=2.39307….t=\frac{\bar x-20}{\sqrt{s^2/9}}=2.39307\ldots.

The upper 2.5%2.5\% critical value with 88 degrees of freedom is 2.3062.306. Since the statistic is larger, we reject H0H_0. There is sufficient evidence at the 2.5%2.5\% level that μ>20\mu>20.