CalcGospel 國際數學圖譜
返回

CIE 9709 2025 June Paper 63 Q7

A Level / CIE / S2

CIE 9709 2025 June Paper 63 Paper · Question 7

题目

Problem

In the past, one quarter of job applicants at a certain firm had first-class degrees. A change is made in the job description and a director of the firm believes that, on average, the proportion of job applicants with first class degrees has decreased.

In the month following the change, there were 3535 job applicants, and rr of these had first-class degrees. The firm carried out a hypothesis test at the 4%4\% significance level to test the director’s belief.

(a) Use a binomial distribution to find the largest value of rr that would provide sufficient evidence that the director’s belief is correct.

(6)

In another month, the director carries out a similar test at the 4%4\% significance level using the 3535 job applicants from that month.

(b) Explain the meaning of a Type I error in this context, and state the probability of a Type I error.

(2)

(c) Given that the proportion of job applicants with first class degrees this year is actually 0.050.05, find the probability of a Type II error.

(2)
题目中文翻译

过去，某公司的求职者中有四分之一拥有一等学位。职位描述修改后，公司的一名董事认为，拥有一等学位的求职者所占的比例平均有所下降。

修改后的一个月内共有 3535 名求职者，其中 rr 人拥有一等学位。公司在 4%4\% 的显著性水平下进行假设检验，以检验该董事的看法。

(a) 用二项分布求能够提供充分证据支持该董事看法的 rr 的最大值。

(6)

在另一个月，董事用该月的 3535 名求职者的数据，在 4%4\% 的显著性水平下进行类似的检验。

(b) 说明在此情境中第一类错误的含义，并写出发生第一类错误的概率。

(2)

(c) 已知今年拥有一等学位的求职者所占的实际比例为 0.050.05，求发生第二类错误的概率。

(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

展开

比例降低对应左尾二项检验。找最大可拒绝观测值时，必须检查边界三与相邻的四的累计概率。

答题过程

展开

Let pp be the population proportion of applicants with first-class degrees after the change. Test H0:p=0.25H_0:p=0.25 against H1:p<0.25H_1:p<0.25.

Under H0H_0, the count RR among 3535 applicants has distribution B(35,0.25)\mathrm B(35,0.25). We have

P(R≤3)= ∑j=03(35j)(0.25)j(0.75)35−j= (0.75)35+35(0.25)(0.75)34 +(352)(0.25)2(0.75)33 +(353)(0.25)3(0.75)32≈ 0.0136113<0.04.\begin{align*} \mathrm P(R\le3) =&\,\sum_{j=0}^3\binom{35}{j}(0.25)^j(0.75)^{35-j}\\[4mm] =&\,(0.75)^{35}+35(0.25)(0.75)^{34}\\[4mm] &\,\hspace{2pt}+\binom{35}{2}(0.25)^2(0.75)^{33}\\[4mm] &\,\hspace{4pt}+\binom{35}{3}(0.25)^3(0.75)^{32}\\[4mm] \approx&\,0.0136113<0.04. \end{align*}

But

P(R≤4)= P(R≤3) +(354)(0.25)4(0.75)31≈ 0.0410055>0.04.\begin{align*} \mathrm P(R\le4) =&\,\mathrm P(R\le3)\\[4mm] &\,+\binom{35}{4}(0.25)^4(0.75)^{31}\\[4mm] \approx&\,0.0410055>0.04. \end{align*}

The critical region is R≤3R\le3, so the largest observed value providing sufficient evidence is r=3\boxed{r=3}.

(b)

解法一

思路

展开

第一类错误是在真实比例没有降低时却判为降低；概率等于原假设下落入(a)拒绝域的概率。

答题过程

展开

A Type I error is concluding that the population proportion of applicants with first-class degrees has decreased when it has not.

Using the critical region from (a),

 P(Type I error)= P(R≤3∣p=0.25)≈ 0.0136.\begin{align*} &\,\mathrm P(\text{Type I error})\\[4mm] =&\,\mathrm P(R\le3\mid p=0.25)\\[4mm] \approx&\,\boxed{0.0136}. \end{align*}

(c)

解法一

思路

展开

真实比例变为 0.050.05，拒绝域仍是至多三人。第二类错误是未落入拒绝域，即至少四人。

答题过程

展开

When p=0.05p=0.05, R∼B(35,0.05)R\sim\mathrm B(35,0.05). A Type II error occurs when R≥4R\ge4. Therefore

 P(Type II error)= 1−∑j=03(35j)(0.05)j(0.95)35−j= 1−(0.95)35−35(0.05)(0.95)34 −(352)(0.05)2(0.95)33 −(353)(0.05)3(0.95)32≈ 0.0958.\begin{align*} &\,\mathrm P(\text{Type II error})\\[4mm] =&\,1-\sum_{j=0}^3\binom{35}{j}(0.05)^j(0.95)^{35-j}\\[4mm] =&\,1-(0.95)^{35}-35(0.05)(0.95)^{34}\\[4mm] &\,-\binom{35}{2}(0.05)^2(0.95)^{33}\\[4mm] &\,\hspace{2pt}-\binom{35}{3}(0.05)^3(0.95)^{32}\\[4mm] \approx&\,\boxed{0.0958}. \end{align*}