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IAL 2023 June S3 Q6

A Level / Edexcel / S3

IAL 2023 June Paper · Question 6

题目

Problem

Roxane, a scientist, carries out an investigation into the fat content of different brands of crisps.

Roxane took random samples of different brands of crisps and recorded, in grams, the fat content (x) of a 30 gram serving. The table below shows some results for just two of these brands.

Brand∑x\sum x∑x2\sum x^2xˉ\bar xsSample size
A3501753.97445.00.2470
B331.51694.65α\alphaβ\beta65

(a) Calculate the value of α\alpha and the value of β\beta

(3)

Roxane claims that these results show that the crisps from brand A have a lower fat content than the crisps from brand B, as the mean fat content for brand A is, statistically, significantly less than the mean fat content for brand B.

(b) Stating your hypotheses clearly, carry out a suitable test, at the 5% level of significance, to assess Roxane’s claim. You should state your test statistic and critical value.

(7)

(c) For the test in part (b), state whether or not it is necessary to assume that the fat content of crisps is normally distributed. Give a reason for your answer.

(2)

(d) State an assumption you have made in carrying out the test in part (b).

(1)
(Total for Question 6 is 13 marks)
题目中文翻译

科学家 Roxane 正在调查不同品牌薯片的脂肪含量。

她从不同品牌的薯片中抽取随机样本，并记录每 30 克一份的脂肪含量 xx（单位：克）。 下表给出了其中两个品牌的部分结果。

品牌∑x\sum x∑x2\sum x^2xˉ\bar xs样本量
A3501753.97445.00.2470
B331.51694.65α\alphaβ\beta65

(a) 计算 α\alpha 和 β\beta 的值。

Roxane 认为，这些结果表明 A 品牌薯片的脂肪含量比 B 品牌更低，因为 A 品牌的平均脂肪含量在统计上显著小于 B 品牌。

(b) 清楚写出原假设，在 5% 显著性水平下进行适当检验，以评估 Roxane 的说法。 你应写出检验统计量和临界值。

(c) 对 (b) 小题的检验，是否有必要假设薯片的脂肪含量服从正态分布？请给出理由。

(d) 写出你在进行 (b) 小题检验时所作出的一个假设。

解答

(a)

解法一

思路

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α\alpha 是 B 品牌的样本均值，直接用 ∑x/n\sum x/n 计算；β\beta 是样本标准差，先用分母为 n−1n-1 的无偏样本方差公式求 β2\beta^2，再取正平方根。

答题过程

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For brand B,

α= ∑xn= 331.565= 5.1.\begin{align*} \alpha =&\,\frac{\sum x}{n}\\ =&\,\frac{331.5}{65}\\ =&\,\boxed{5.1}. \end{align*}

Also,

β2= ∑x2−nx‾ 2n−1= 1694.65−65(5.1)264= 0.0625.\begin{align*} \beta^2 =&\,\frac{\sum x^2-n\overline{x}^{\,2}}{n-1}\\ =&\,\frac{1694.65-65(5.1)^2}{64}\\ =&\,0.0625. \end{align*}

Since β\beta is a standard deviation, β>0\beta>0. Therefore,

β=0.0625=0.25.\beta=\sqrt{0.0625}=\boxed{0.25}.

(b)

解法一

思路

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Roxane 声称 A 品牌的总体平均脂肪含量较低，所以进行左尾两总体均值差检验。两个样本量都较大，以样本方差估计总体方差，计算标准化检验统计量并与 5%5\% 单尾临界值比较。

答题过程

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Let μA\mu_A and μB\mu_B be the population mean fat contents for brands A and B respectively. The hypotheses are

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

The estimated variance of X‾A−X‾B\overline X_A-\overline X_B is

SE⁡2= 0.24270 +0.25265= 0.00178439….\begin{align*} \operatorname{SE}^2 =&\,\frac{0.24^2}{70}\\ &\,\hspace{2pt}+\frac{0.25^2}{65}\\ =&\,0.00178439\ldots. \end{align*}

Hence

SE⁡=0.00178439…=0.0422421….\operatorname{SE} =\sqrt{0.00178439\ldots} =0.0422421\ldots.

Therefore, the test statistic is

z= 5.0−5.10.0422421…= −2.3673…= −2.37.\begin{align*} z =&\,\frac{5.0-5.1}{0.0422421\ldots}\\ =&\,-2.3673\ldots\\ =&\,-2.37. \end{align*}

For a one-tailed test at the 5%5\% significance level, the critical value is −1.6449-1.6449, giving the critical region

z⩽−1.6449.z\leqslant-1.6449.

Since −2.37<−1.6449-2.37<-1.6449, the test statistic lies in the critical region, so H0H_0 is rejected. There is sufficient evidence at the 5%5\% significance level to support Roxane’s claim that crisps from brand A have a lower mean fat content than crisps from brand B.

(c)

解法一

思路

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两个样本量分别为 70 和 65，都足够大，因此中央极限定理可保证两个样本均值近似服从正态分布，无须额外假设原始脂肪含量总体为正态分布。

答题过程

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No. Both sample sizes are large, so the central limit theorem applies and the distributions of the sample means are approximately normal.

(d)

解法一

思路

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检验中以两组样本方差代替了未知的总体方差，因此需要假设它们能充分近似各自的总体方差。

答题过程

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It is assumed that the sample variance is approximately equal to the corresponding population variance for both brands.