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CIE 9709 2022 November Paper 63 Q2

A Level / CIE / S2

CIE 9709 2022 Nov Paper 63 Paper · Question 2

题目

Problem

A spinner has five sectors, each printed with a different colour. Susma and Sanjay both wish to test whether the spinner is biased so that it lands on red on fewer spins than it would if it were fair. Susma spins the spinner 4040 times. She finds that it lands on red exactly 44 times.

(a) Use a binomial distribution to carry out the test at the 5%5\% significance level.

(5)

Sanjay also spins the spinner 4040 times. He finds that it lands on red rr times.

(b) Use a binomial distribution to find the largest value of rr that lies in the rejection region for the test at the 5%5\% significance level.

(3)
题目中文翻译

一个转盘有五个扇区，每个扇区印有不同的颜色。Susma 和 Sanjay 都想检验这个转盘是否有偏，使它转到红色的次数比公平时更少。Susma 转动转盘 4040 次，发现它恰好有 44 次停在红色扇区。

(a) 用二项分布在 5%5\% 的显著性水平下进行检验。

(5)

Sanjay 也转动转盘 4040 次，发现它有 rr 次停在红色扇区。

(b) 用二项分布求在 5%5\% 的显著性水平下落入该检验拒绝域的最大 rr 值。

(3)
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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公平转盘落在红色的概率为1/5，检验红色概率是否更低，使用不超过4次的二项左尾。

答题过程

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Let pp be the probability of landing on red.

H0:p=0.2,H1:p<0.2.H_0:p=0.2,\qquad H_1:p<0.2.

Under H0H_0, the number RR of red results in 40 spins has distribution B(40,0.2)\mathrm{B}(40,0.2). For the observed count of 4,

 P(R≤4∣H0)= (0.8)40+40(0.2)(0.8)39 +(402)(0.2)2(0.8)38 +(403)(0.2)3(0.8)37 +(404)(0.2)4(0.8)36= 0.0759145….\begin{align*} &\,\mathrm{P}(R\leq4\mid H_0)\\[4mm] =&\,(0.8)^{40}+40(0.2)(0.8)^{39}\\[4mm] &\,+\binom{40}{2}(0.2)^2(0.8)^{38}\\[4mm] &\,\hspace{2pt}+\binom{40}{3}(0.2)^3(0.8)^{37}\\[4mm] &\,\hspace{4pt}+\binom{40}{4}(0.2)^4(0.8)^{36}\\[4mm] =&\,0.0759145\ldots. \end{align*}

Since 0.0759145…>0.050.0759145\ldots>0.05, we do not reject H0H_0. There is insufficient evidence at the 5% significance level that the spinner lands on red less often than a fair spinner would.

(b)

解法一

思路

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比较相邻累计概率，4次未入拒绝域而3次进入，因此最大值为3。

答题过程

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Under H0H_0, R∼B(40,0.2)R\sim\mathrm{B}(40,0.2). From part (a),

P(R≤4)=0.0759145…>0.05.\mathrm{P}(R\leq4)=0.0759145\ldots>0.05.

But

P(R≤3)= P(R≤4) −(404)(0.2)4(0.8)36= 0.0284621…<0.05.\begin{align*} \mathrm{P}(R\leq3)=&\,\mathrm{P}(R\leq4)\\[4mm] &\,-\binom{40}{4}(0.2)^4(0.8)^{36}\\[4mm] =&\,0.0284621\ldots<0.05. \end{align*}

Since the cumulative probability increases with rr, the largest value in the rejection region is r=3\boxed{r=3}.