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CIE 9709 2026 June Paper 63 Q5

A Level / CIE / S2

CIE 9709 2026 June Paper 63 Paper · Question 5

题目

Problem

The lengths, X cmX\,\mathrm{cm}, of Wamba snakes have mean μ cm\mu\,\mathrm{cm}. Research has shown that, for the whole population of Wamba snakes, μ=41.5\mu=41.5. A researcher claims that, for the Wamba snakes in a particular region, the mean length is less than 41.541.5. She measures the lengths of a random sample of 150150 Wamba snakes in this region. The results are summarised below.

n=150∑x=6030∑x2=252 014n=150\qquad \sum x=6030\qquad \sum x^2=252\,014

(a) Find unbiased estimates of the population mean and variance of XX for Wamba snakes in this region, showing that the unbiased estimate of variance is 64.48364.483 correct to 33 decimal places.

(3)

(b) Assuming that Var⁡(X)=64.483\operatorname{Var}(X)=64.483 for the whole population of Wamba snakes, test the researcher’s claim at the 2%2\% significance level.

(5)

(c) Use your answer to 5(b) to explain whether it is possible for the researcher to make a Type I error.

(1)

Later the researcher carries out a similar test at the 2%2\% significance level, using the same hypotheses and another random sample of 150150 Wamba snakes from this region.

(d) Given that the population mean length of Wamba snakes for this region is 38.7 cm38.7\,\mathrm{cm}, find the probability of a Type II error.

(5)
题目中文翻译

Wamba 蛇的体长为 X cmX\,\mathrm{cm},均值为 μ cm\mu\,\mathrm{cm}。研究表明,对 Wamba 蛇的整个总体而言,μ=41.5\mu=41.5。一名研究人员声称,某个地区的 Wamba 蛇的平均体长小于 41.541.5。她在这个地区随机抽取了 150150 条 Wamba 蛇,测量其体长。结果汇总如下。

n=150∑x=6030∑x2=252 014n=150\qquad \sum x=6030\qquad \sum x^2=252\,014

(a) 求该地区 Wamba 蛇体长 XX 的总体均值和方差的无偏估计值,并证明方差的无偏估计值保留 33 位小数为 64.48364.483。

(3)

(b) 假设 Wamba 蛇的整个总体满足 Var⁡(X)=64.483\operatorname{Var}(X)=64.483,在 2%2\% 的显著性水平下检验研究人员的说法。

(5)

(c) 用你对 5(b) 的答案说明研究人员是否有可能犯第一类错误。

(1)

后来,研究人员采用相同的假设,从这个地区另随机抽取了 150150 条 Wamba 蛇,在 2%2\% 的显著性水平下进行类似的检验。

(d) 已知这个地区 Wamba 蛇的总体平均体长为 38.7 cm38.7\,\mathrm{cm},求发生第二类错误的概率。

(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

(a)

解法一

思路

展开

均值用样本均值估计,方差使用无偏公式。先保留分式,最后再四舍五入到题目指定的小数位。

答题过程

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The unbiased estimate of the population mean is

x‾=6030150=40.2 cm.\boxed{\overline x=\frac{6030}{150}=40.2\ \text{cm}}.

For the variance,

s2= 1149(252014−60302150)= 252014−242406149= 9608149≈ 64.483 cm2(3 decimal places).\begin{align*} s^2 =&\,\frac1{149}\bigg(252014-\frac{6030^2}{150}\bigg)\\[4mm] =&\,\frac{252014-242406}{149}\\[4mm] =&\,\frac{9608}{149}\\[4mm] \approx&\,\boxed{64.483\ \text{cm}^2}\quad\text{(3 decimal places)}. \end{align*}

(b)

解法一

思路

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虽然总体分布未说明,但样本量 150150 足够大,可对样本均值作正态近似;本问明确要求把给定方差当总体方差使用。

答题过程

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Test

H0:μ=41.5,H1:μ<41.5,H_0:\mu=41.5,\qquad H_1:\mu<41.5,

where μ\mu is the population mean length of Wamba snakes in this region. By the Central Limit Theorem, under H0H_0,

X‾≈N(41.5,64.483150).\overline X\approx\mathrm N\bigg(41.5,\frac{64.483}{150}\bigg).

The observed test statistic is

z=40.2−41.564.483/150≈−1.983.z=\frac{40.2-41.5}{\sqrt{64.483/150}}\approx-1.983.

At the 2%2\% significance level, the lower-tail critical value is Φ−1(0.02)≈−2.054\Phi^{-1}(0.02)\approx-2.054. Since −1.983>−2.054-1.983>-2.054, do not reject H0H_0.

There is insufficient evidence at the 2%2\% level that the population mean length of Wamba snakes in this region is less than 41.5 cm41.5\ \text{cm}.

(c)

解法一

思路

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本问针对(b)这一次已经作出的检验结论。第一类错误必须先拒绝原假设,而(b)没有拒绝。

答题过程

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No. A Type I error means rejecting a true null hypothesis. In (b), H0H_0 was not rejected, so the decision made there cannot be a Type I error.

(d)

解法一

思路

展开

先在原假设下找拒绝域的样本均值边界;真实均值改变后,拒绝域不变,只改变计算第二类错误概率时使用的分布。

答题过程

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Let cc be the critical sample mean. The test uses the same hypotheses, sample size and variance, so

c−41.564.483/150= Φ−1(0.02),c= 41.5 +Φ−1(0.02)64.483150≈ 40.153444.\begin{align*} \frac{c-41.5}{\sqrt{64.483/150}} =&\,\Phi^{-1}(0.02),\\[4mm] c=&\,41.5\\[4mm] &\,+\Phi^{-1}(0.02)\sqrt{\frac{64.483}{150}}\\[4mm] \approx&\,40.153444. \end{align*}

The rejection region is X‾<c\overline X<c. When the true mean is 38.738.7, the sample mean is approximately N(38.7,64.483/150)\mathrm N(38.7,64.483/150). A Type II error occurs when X‾≥c\overline X\ge c, so

P(Type II error)≈ 1−Φ(40.153444−38.764.483/150)= 1−Φ(2.216774…)≈ 0.0133.\begin{align*} \mathrm P(\text{Type II error}) \approx&\,1-\Phi\bigg(\frac{40.153444-38.7}{\sqrt{64.483/150}}\bigg)\\[4mm] =&\,1-\Phi(2.216774\ldots)\\[4mm] \approx&\,\boxed{0.0133}. \end{align*}