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CIE 9709 2025 June Paper 65 Q6

A Level / CIE / S2

CIE 9709 2025 June Paper 65 Paper · Question 6

题目

Problem

It is claimed that 28%28\% of voters in a certain town support the Forward Now political party. A researcher suspects that the true figure is less than 28%28\%. She interviews a random sample of 3030 voters from the town and she finds that 44 voters in the sample say that they support the Forward Now party. She plans to carry out a hypothesis test at the 10%10\% significance level.

(a) Use a binomial distribution to carry out the test.

(5)

(b) State, with a reason, whether it is possible that a Type I error was made in carrying out the test.

(1)

Later the researcher carries out a similar test at the 10%10\% significance level, using a new random sample of 3030 voters from the town.

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

(2)
题目中文翻译

有人声称某镇有 28%28\% 的选民支持 Forward Now 政党。一名研究人员怀疑真实比例低于 28%28\%。她在镇上随机抽取 3030 名选民进行访问，发现样本中有 44 名选民表示支持 Forward Now 政党。她计划在 10%10\% 的显著性水平下进行假设检验。

(a) 用二项分布进行检验。

(5)

(b) 指出进行该检验时是否可能犯第一类错误，并说明理由。

(1)

后来，研究人员在镇上重新随机抽取了 3030 名选民，在 10%10\% 的显著性水平下进行类似的检验。

(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)

解法一

思路

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怀疑比例降低，对观察到的四名支持者计算左尾二项概率。

答题过程

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Let pp be the population proportion supporting Forward Now. Test H0:p=0.28H_0:p=0.28 against H1:p<0.28H_1:p<0.28.

Under H0H_0, the count KK has distribution B(30,0.28)\mathrm B(30,0.28). For K=4K=4,

P(K≤4)= ∑j=04(30j)(0.28)j(0.72)30−j= (0.72)30+30(0.28)(0.72)29 +(302)(0.28)2(0.72)28 +(303)(0.28)3(0.72)27 +(304)(0.28)4(0.72)26≈ 0.0495411.\begin{align*} \mathrm P(K\le4)=&\,\sum_{j=0}^4\binom{30}{j}(0.28)^j(0.72)^{30-j}\\[4mm] =&\,(0.72)^{30}+30(0.28)(0.72)^{29}\\[4mm] &\,+\binom{30}{2}(0.28)^2(0.72)^{28}\\[4mm] &\,\hspace{2pt}+\binom{30}{3}(0.28)^3(0.72)^{27}\\[4mm] &\,\hspace{4pt}+\binom{30}{4}(0.28)^4(0.72)^{26}\\[4mm] \approx&\,0.0495411. \end{align*}

Since 0.0495411<0.100.0495411<0.10, reject H0H_0. There is sufficient evidence at the 10%10\% level that the population proportion of voters supporting Forward Now is less than 28%28\%.

(b)

解法一

思路

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本次拒绝了原假设；若原假设实际上为真，就可能犯第一类错误。

答题过程

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Yes. The test rejected H0H_0, so a Type I error may have been made if the true population support proportion is still 28%28\%.

(c)

解法一

思路

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核对下一个整数，确定实际拒绝域，再求原假设下落入拒绝域的概率。

答题过程

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Part (a) gives P(K≤4)≈0.0495411<0.10\mathrm P(K\le4)\approx0.0495411<0.10. But under H0H_0,

P(K≤5)= P(K≤4) +(305)(0.28)5(0.72)25≈ 0.116059>0.10.\begin{align*} \mathrm P(K\le5) =&\,\mathrm P(K\le4)\\[4mm] &\,+\binom{30}{5}(0.28)^5(0.72)^{25}\\[4mm] \approx&\,0.116059>0.10. \end{align*}

Thus the critical region is K≤4K\le4, and

 P(Type I error)= P(K≤4∣p=0.28)≈ 0.0495.\begin{align*} &\,\mathrm P(\text{Type I error})\\[4mm] =&\,\mathrm P(K\le4\mid p=0.28)\\[4mm] \approx&\,\boxed{0.0495}. \end{align*}