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CIE 9709 2024 November Paper 62 Q7

A Level / CIE / S2

CIE 9709 2024 Nov Paper 62 Paper · Question 7

题目

Problem

The heights of one-year-old trees of a certain variety are known to have mean 2.3 m2.3\,\mathrm{m}. A scientist believes that, on average, trees of this age and variety in her region are slightly taller than in other places. She plans to carry out a hypothesis test, at the 2%2\% significance level, in order to test her belief.

(a) State the probability that she will make a Type I error.

(1)

She takes a random sample of 100100 such trees in her region and measures their heights, h mh\,\mathrm{m}. Her results are summarised below.

n=100∑h=238∑h2=580n=100\qquad\sum h=238\qquad\sum h^2=580

(b) Carry out the test at the 2%2\% significance level.

(7)

(c) The scientist carries out the test correctly, but another scientist claims that she has made a Type II error.

Comment on this claim.

(1)
题目中文翻译

已知某品种的一年生树木的平均高度为 2.3 m2.3\,\mathrm{m}。一名科学家认为,在她所在地区,同龄同品种的树木平均比其他地方的略高。她计划在 2%2\% 的显著性水平下进行假设检验,以检验自己的看法。

(a) 写出她犯第一类错误的概率。

(1)

她在所在地区随机抽取 100100 棵这样的树,测量其高度 h mh\,\mathrm{m}。结果汇总如下。

n=100∑h=238∑h2=580n=100\qquad\sum h=238\qquad\sum h^2=580

(b) 在 2%2\% 的显著性水平下进行检验。

(7)

(c) 这名科学家正确地完成了检验,但另一名科学家声称她犯了第二类错误。

评价这一说法。

(1)
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

第一类错误的概率为检验的显著性水平。

答题过程

展开 P(Type I error)=0.02.\boxed{\mathrm{P}(\text{Type I error})=0.02}.

(b)

解法一

思路

展开

以总体平均身高作右尾检验。用样本数据计算无偏方差和标准误,基于大样本正态近似比较临界值。

答题过程

展开

Let μ\mu be the population mean height, in metres, of one-year-old trees of this variety in the scientist’s region.

H0:μ=2.3,H1:μ>2.3.H_0:\mu=2.3,\qquad H_1:\mu>2.3.

The sample estimates are

hˉ=238100=2.38.\bar h=\frac{238}{100}=2.38.

For the variance,

s2= 580−2382/10099= 113825=0.1369697….\begin{align*} s^2=&\,\frac{580-238^2/100}{99}\\[4mm] =&\,\frac{113}{825}=0.1369697\ldots. \end{align*}

By the Central Limit Theorem, the sample mean is approximately normal. The test statistic is

z=2.38−2.3(113/825)/100=2.16161….z=\frac{2.38-2.3}{\sqrt{(113/825)/100}} =2.16161\ldots.

The upper-tail 2% critical value is 2.0542.054. Since 2.16161…>2.0542.16161\ldots>2.054, we reject H0H_0.

There is sufficient evidence at the 2% significance level that the population mean height in the scientist’s region is greater than 2.3 m.

(c)

解法一

思路

展开

第二类错误须在未拒绝原假设时发生,而本次已经拒绝原假设。

答题过程

展开

The claim is incorrect. A Type II error occurs when a false null hypothesis is not rejected, whereas H0H_0 was rejected in part (b).