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CIE 9709 2024 March Paper 62 Q7

A Level / CIE / S2

CIE 9709 2024 March Paper 62 Paper · Question 7

题目

Problem

The heights, in centimetres, of adult females in Litania have mean μ\mu and standard deviation σ\sigma. It is known that in 20042004 the values of μ\mu and σ\sigma were 163.21163.21 and 6.956.95 respectively. The government claims that the value of μ\mu this year is greater than it was in 20042004. In order to test this claim a researcher plans to carry out a hypothesis test at the 1%1\% significance level. He records the heights of a random sample of 300300 adult females in Litania this year and finds the value of the sample mean.

(a) State the probability of a Type I error.

(1)

You should assume that the value of σ\sigma after 20042004 remains at 6.956.95.

(b) Given that the value of μ\mu this year is actually 164.91164.91, find the probability of a Type II error.

(5)
题目中文翻译

Litania 的成年女性身高(以厘米为单位)的均值为 μ\mu,标准差为 σ\sigma。已知 20042004 年 μ\mu 和 σ\sigma 的值分别为 163.21163.21 和 6.956.95。政府声称今年的 μ\mu 值高于 20042004 年。为了检验这一说法,一名研究人员计划在 1%1\% 的显著性水平下进行假设检验。他从今年 Litania 的成年女性中随机抽取 300300 人,记录她们的身高,并求出样本均值。

(a) 写出发生第一类错误的概率。

(1)

假定 20042004 年后 σ\sigma 的值保持为 6.956.95。

(b) 已知今年的 μ\mu 值实际为 164.91164.91,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

连续均值检验的第一类错误概率等于所设显著性水平。

答题过程

展开 P(Type I error)=0.01.\boxed{\mathrm{P}(\text{Type I error})=0.01}.

(b)

解法一

思路

展开

先在原假设均值163.21下确定右尾拒绝边界,再在真实均值164.91下计算样本均值未超过边界的概率。

答题过程

展开

The hypotheses are

H0:μ=163.21,H1:μ>163.21.H_0:\mu=163.21,\qquad H_1:\mu>163.21.

By the Central Limit Theorem, the sample mean is approximately normal, with standard error

sH‾=6.95300.s_{\overline H}=\frac{6.95}{\sqrt{300}}.

Let cc be the critical sample mean. For an upper-tail 1% test,

c= 163.21+Φ−1(0.99)6.95300= 164.143467….\begin{align*} c=&\,163.21+\Phi^{-1}(0.99)\frac{6.95}{\sqrt{300}}\\[4mm] =&\,164.143467\ldots. \end{align*}

We reject H0H_0 when H‾>c\overline H>c. If the true mean is 164.91 cm, a Type II error occurs when H‾≤c\overline H\leq c, so

 P(Type II error)≈ Φ(164.143467…−164.916.95/300)= 0.0280458…= 0.0280(3 s.f.).\begin{align*} &\,\mathrm{P}(\text{Type II error})\\[4mm] \approx&\,\Phi\bigg(\frac{164.143467\ldots-164.91}{6.95/\sqrt{300}}\bigg)\\[4mm] =&\,0.0280458\ldots\\[4mm] =&\,\boxed{0.0280}\quad\text{(3 s.f.).} \end{align*}