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IAL 2022 June S3 Q2

A Level / Edexcel / S3

IAL 2022 June Paper · Question 2

题目

Problem

An experiment is conducted to compare the heat retention of two brands of flasks, brand A and brand B. Both brands of flask have a capacity of 750 ml.

In the experiment 750 ml of boiling water is poured into the flask, which is then sealed. Four hours later the temperature, in °C, of the water in the flask is recorded. A random sample of 100 flasks from brand A gives the following summary statistics, where x is the temperature of the water in the flask after four hours.

∑x=7690∑(x−xˉ)2=669.24\sum x = 7690 \qquad \sum (x-\bar x)^2 = 669.24

(a) Find unbiased estimates for the mean and variance of the temperature of the water, after four hours, for brand A.

(3)

A random sample of 80 flasks from brand B gives the following results, where y is the temperature of the water after four hours.

yˉ=75.9sy=2.2\bar y = 75.9 \qquad s_y=2.2

(b) Test, at the 1% significance level, whether there is a difference in the mean water temperature after four hours between brand A and brand B. You should state your hypotheses, test statistic and critical value clearly.

(7)

(c) Explain why it is reasonable to assume that σ2=s2\sigma^2=s^2 in this situation.

(1)
(Total for Question 2 is 11 marks)
题目中文翻译

进行了一项实验，用于比较两种品牌保温瓶的保温效果，A 品牌和 B 品牌。两种保温瓶的容量都为 750 毫升。

实验中向保温瓶中倒入 750 毫升沸水，然后密封。 四小时后记录瓶中水的温度（摄氏度）。 A 品牌的 100 个保温瓶的随机样本给出以下汇总统计，其中 x 表示四小时后的水温。

∑x=7690∑(x−xˉ)2=669.24\sum x = 7690 \qquad \sum (x-\bar x)^2 = 669.24

(a) 求 A 品牌四小时后水温的均值和方差的无偏估计。

B 品牌的 80 个保温瓶随机样本给出如下结果，其中 y 表示四小时后的水温。

yˉ=75.9sy=2.2\bar y = 75.9 \qquad s_y=2.2

(b) 在 1% 显著性水平下，检验 A 品牌与 B 品牌四小时后水温均值是否存在差异。 你应清楚写出原假设、检验统计量和临界值。

(c) 解释为什么在这种情况下假设 σ2=s2\sigma^2=s^2 是合理的。

解答

(a)

解法一

思路

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样本均值是总体均值的无偏估计。题目给出的离均差平方和以样本均值为中心，因此总体方差的无偏估计要除以 n−1=99n-1=99，而不是除以 100100。

答题过程

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The unbiased estimate of the population mean is

xˉ=∑xn=7690100=76.9.\bar{x}=\frac{\sum x}{n}=\frac{7690}{100}=\boxed{76.9}.

The unbiased estimate of the population variance is

sx2= ∑(x−xˉ)2n−1= 669.2499= 6.76.\begin{align*} s_x^2 =&\,\frac{\sum(x-\bar{x})^2}{n-1}\\ =&\,\frac{669.24}{99}\\ =&\,\boxed{6.76}. \end{align*}

(b)

解法一

思路

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题目问两品牌的总体平均水温是否“存在差异”，所以进行双尾检验。两个样本都很大，可用各自的样本方差估计总体方差，并用标准正态分布近似检验两个总体均值之差。

答题过程

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Let μA\mu_A and μB\mu_B be the population mean water temperatures after four hours for brands A and B respectively.

H0: μA=μB,H1: μA≠μB.\begin{aligned} H_0:&\ \mu_A=\mu_B,\\ H_1:&\ \mu_A\ne\mu_B. \end{aligned}

Under H0H_0, the test statistic is

Z= xˉ−yˉsx2100+sy280= 76.9−75.96.76100+2.2280= 2.793…\begin{align*} Z =&\,\frac{\bar{x}-\bar{y}} {\sqrt{\dfrac{s_x^2}{100}+\dfrac{s_y^2}{80}}}\\ =&\,\frac{76.9-75.9} {\sqrt{\dfrac{6.76}{100}+\dfrac{2.2^2}{80}}}\\ =&\,2.793\ldots \end{align*}

For a two-tailed test at the 1%1\% significance level, the critical values are

z=±2.5758.z=\pm 2.5758.

Since

2.793…>2.5758,2.793\ldots>2.5758,

the test statistic lies in the critical region. Therefore, H0H_0 is rejected. There is sufficient evidence at the 1%1\% significance level to suggest that the mean water temperature after four hours is different for the two brands.

(c)

解法一

思路

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这里需要说明为什么可以用样本方差代替未知的总体方差。理由必须同时涉及两组样本：两组样本容量分别为 100100 和 8080，都足够大。

答题过程

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Both samples are reasonably large, so each sample variance should provide a reliable estimate of its corresponding population variance. It is therefore reasonable to use s2s^2 as an approximation to σ2\sigma^2.