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CIE 9709 2023 November Paper 62 Q3

A Level / CIE / S2

CIE 9709 2023 Nov Paper 62 Paper · Question 3

题目

Problem

A researcher read a magazine article which stated that boys aged 11 to 33 prefer green to orange. It claimed that, when offered a green cube and an orange cube to play with, a boy is more likely to choose the green one.

The researcher disagrees with this claim. She believes that boys of this age are equally likely to choose either colour. In order to test her belief, the researcher carried out a hypothesis test at the 5%5\% significance level. She offered a green cube and an orange cube to each of 1010 randomly chosen boys aged 11 to 33, and recorded the number, XX, of boys who chose the green cube.

Out of the 1010 boys, 88 boys chose the green cube.

(a)

(i) Assuming that the researcher’s belief that either colour cube is equally likely to be chosen is valid, a student correctly calculates that P(X=8)=0.0439\mathrm{P}(X=8)=0.0439, correct to 33 significant figures.

He says that, because this value is less than 0.050.05, the null hypothesis should be rejected.

Explain why this statement is incorrect.

(1)

(ii) Carry out the test on the researcher’s claim that either colour cube is equally likely to be chosen.

(5)

(b) Another researcher claims that a Type I error was made in carrying out the test.

Explain why this cannot be true.

(1)

A similar test, at the 5%5\% significance level, was carried out later using 1010 other randomly chosen boys aged 11 to 33.

(c) Find the probability of a Type I error.

(2)
题目中文翻译

一名研究人员读到一篇杂志文章,文章称 11 至 33 岁的男孩比起橙色更喜欢绿色。文章声称,给男孩一个绿色方块和一个橙色方块玩时,他更可能选择绿色方块。

这名研究人员不同意这一说法。她认为这个年龄段的男孩选择两种颜色的可能性相同。为了检验自己的看法,她在 5%5\% 的显著性水平下进行了一次假设检验。她随机选取 1010 名 11 至 33 岁的男孩,给每人一个绿色方块和一个橙色方块,并记录选择绿色方块的男孩人数 XX。

这 1010 名男孩中,有 88 人选择了绿色方块。

(a)

(i) 假设研究人员的看法成立,即两种颜色的方块被选中的概率相同。一名学生正确算得 P(X=8)=0.0439\mathrm{P}(X=8)=0.0439,精确到 33 位有效数字。

他说,由于这个值小于 0.050.05,应拒绝原假设。

解释这一说法为什么不正确。

(1)

(ii) 对研究人员所称两种颜色的方块被选中的概率相同这一说法进行检验。

(5)

(b) 另一名研究人员声称,该检验中发生了第一类错误。

解释为什么这一说法不可能成立。

(1)

后来,又随机选取了另外 1010 名 11 至 33 岁的男孩,在 5%5\% 的显著性水平下进行了一次类似的检验。

(c) 求发生第一类错误的概率。

(2)
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)(i)

解法一

思路

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检验应使用至少与观测结果一样偏向备择的右尾概率,而非单点概率。

答题过程

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The comparison must use the tail probability P(X≥8)\mathrm{P}(X\geq8), rather than the single-value probability P(X=8)\mathrm{P}(X=8).

(a)(ii)

解法一

思路

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以选择绿色概率一半为原假设,以偏好绿色为备择。精确计算至少8人的二项右尾概率并比较5%。

答题过程

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Let pp be the probability that a boy of this age chooses the green cube.

H0:p=0.5,H1:p>0.5.H_0:p=0.5,\qquad H_1:p>0.5.

Under H0H_0, X∼B(10,0.5)X\sim\mathrm{B}(10,0.5). For the observed value 8,

 P(X≥8∣H0)= (108)(0.5)10 +(109)(0.5)10+(0.5)10= 45+10+11024= 0.0546875.\begin{align*} &\,\mathrm{P}(X\geq8\mid H_0)\\[4mm] =&\,\binom{10}{8}(0.5)^{10}\\[4mm] &\,+\binom{10}{9}(0.5)^{10}+(0.5)^{10}\\[4mm] =&\,\frac{45+10+1}{1024}\\[4mm] =&\,0.0546875. \end{align*}

Since 0.0546875>0.050.0546875>0.05, we do not reject H0H_0. There is insufficient evidence at the 5% significance level that boys of this age prefer green, so the researcher’s equal-choice claim is not rejected.

(b)

解法一

思路

展开

第一类错误指拒绝真实原假设,本次没有拒绝原假设。

答题过程

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A Type I error requires rejection of a true null hypothesis. Here H0H_0 was not rejected, so a Type I error cannot have occurred.

(c)

解法一

思路

展开

用相邻右尾概率找出拒绝域,实际第一类错误概率为原假设下落入该域的概率。

答题过程

展开

Under H0H_0, X∼B(10,0.5)X\sim\mathrm{B}(10,0.5). We have

P(X≥8)=0.0546875>0.05,\mathrm{P}(X\geq8)=0.0546875>0.05,

whereas

P(X≥9)= (109)(0.5)10+(0.5)10= 111024<0.05.\begin{align*} \mathrm{P}(X\geq9)=&\,\binom{10}{9}(0.5)^{10}+(0.5)^{10}\\[4mm] =&\,\frac{11}{1024}<0.05. \end{align*}

Thus the rejection region is X≥9X\geq9, and

P(Type I error)= 111024≈ 0.0107(3 s.f.).\begin{align*} \mathrm{P}(\text{Type I error})=&\,\frac{11}{1024}\\[4mm] \approx&\,\boxed{0.0107}\quad\text{(3 s.f.).} \end{align*}