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CIE 9231 2025 June Paper 43 Q1

A Level / CIE / FS

CIE 9231 2025 June Paper 43 Paper · Question 1

题目

Problem

A person’s eye colour may be categorised as “brown”, “blue” or “other”. A scientist claims that these eye colours are uniformly distributed and hence are equally likely to occur in the population. A survey of 120 people from this population found that 38 people had brown eyes, 52 people had blue eyes and 30 people had eyes which were neither brown nor blue.

Use the data to carry out a goodness of fit test at the 5% significance level to test the scientist’s claim.

[6]
题目中文翻译

一个人的眼睛颜色可以分为“棕色”、“蓝色”或“其他”。一位科学家声称这些眼睛颜色服从均匀分布,因此在总体中出现的可能性相同。对该总体中 120 人进行的一项调查发现,38 人有棕色眼睛,52 人有蓝色眼睛,30 人的眼睛既不是棕色也不是蓝色。

使用这些数据,在 5% 显著性水平下进行拟合优度检验,检验这位科学家的声称。

(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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均匀分布下三种眼睛颜色的期望人数相同。将观测人数与各自的期望人数比较。

答题过程

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Test H0H_0 that the three eye-colour categories have equal population probabilities against H1H_1 that they do not. Under H0H_0 each expected count is 120/3=40120/3=40. The statistic is

 χ2=(38−40)240+ (52−40)240+ (30−40)240=6.20.\begin{align*} &\,\chi^2=\frac{(38-40)^2}{40}\\ +&\,\frac{(52-40)^2}{40}\\ +&\,\frac{(30-40)^2}{40}=6.20. \end{align*}

There are 3−1=23-1=2 degrees of freedom, with upper 5%5\% critical value 5.9915.991. Since 6.20>5.9916.20>5.991, we reject H0H_0. There is sufficient evidence at the 5%5\% level that the three eye-colour categories are not equally likely in this population.