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CIE 9709 2022 March Paper 62 Q2

A Level / CIE / S2

CIE 9709 2022 March Paper 62 Paper · Question 2

题目

Problem

Harry has a five-sided spinner with sectors coloured blue, green, red, yellow and black. Harry thinks the spinner may be biased. He plans to carry out a hypothesis test with the following hypotheses.

H0 ⁣: P(the spinner lands on blue)=15H1 ⁣: P(the spinner lands on blue)≠15\begin{aligned} \mathrm{H}_0\colon &\,\mathrm{P}(\text{the spinner lands on blue})=\frac15\\ \mathrm{H}_1\colon &\,\mathrm{P}(\text{the spinner lands on blue})\ne\frac15 \end{aligned}

Harry spins the spinner 300300 times. It lands on blue on 4545 spins.

Use a suitable approximation to carry out Harry’s test at the 5%5\% significance level.

(5)
题目中文翻译

Harry 有一个五边形转盘,扇区的颜色分别为蓝色、绿色、红色、黄色和黑色。Harry 认为转盘可能有偏。他计划用下列假设进行假设检验。

H0 ⁣: P(转盘停在蓝色扇区)=15H1 ⁣: P(转盘停在蓝色扇区)≠15\begin{aligned} \mathrm{H}_0\colon &\,\mathrm{P}(\text{转盘停在蓝色扇区})=\frac15\\ \mathrm{H}_1\colon &\,\mathrm{P}(\text{转盘停在蓝色扇区})\ne\frac15 \end{aligned}

Harry 转动转盘 300300 次,有 4545 次停在蓝色扇区。

用适当的近似方法,在 5%5\% 的显著性水平下进行 Harry 的检验。

(5)
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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原假设下蓝色次数服从二项分布,用均值60、方差48的正态近似;双尾5%检验比较修正后的z与±1.96。

答题过程

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Let BB be the number of blue outcomes. We test

H0:p=15,H1:p≠15,H_0:p=\frac15,\qquad H_1:p\ne\frac15,

where pp is the probability of landing on blue. Under H0H_0, B∼B(300,0.2)B\sim\mathrm{B}(300,0.2). Since np=60np=60 and n(1−p)=240n(1-p)=240 are large, use Y∼N(60,48)Y\sim\mathrm{N}(60,48). For the observed lower-tail count of 45, continuity correction gives

z=45.5−6048=−2.09289….z=\frac{45.5-60}{\sqrt{48}}=-2.09289\ldots.

The two-tailed 5% critical values are ±1.96\pm1.96. Since z<−1.96z<-1.96, we reject H0H_0. There is sufficient evidence at the 5% significance level that the spinner is biased with respect to blue.