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CIE 9231 2024 Nov Paper 43 Q6

A Level / CIE / FS

CIE 9231 2024 Nov Paper 43 Paper · Question 6

题目

Problem

Ansal is investigating the wingspans of Monarch butterflies in two different regions, XX and YY. He takes a random sample of 8 Monarch butterflies from region XX and records their wingspans, xx cm. His results are as follows.

8.27.07.38.87.88.59.27.48.2 \quad 7.0 \quad 7.3 \quad 8.8 \quad 7.8 \quad 8.5 \quad 9.2 \quad 7.4

Ansal also takes a random sample of 9 Monarch butterflies from region YY and records their wingspans, yy cm. His results are summarised as follows.

∑y=71.10∑y2=567.13\sum y = 71.10 \qquad \sum y^2 = 567.13

Ansal suspects that the mean wingspan of Monarch butterflies from region XX is greater than the mean wingspan of Monarch butterflies from region YY. It is known that the wingspans of Monarch butterflies in regions XX and YY are normally distributed with equal population variances.

Test, at the 10% significance level, whether Ansal’s suspicion is supported by the data.

[8]
题目中文翻译

Ansal 正在调查两个不同地区 XX 和 YY 的帝王蝶翼展。他从地区 XX 随机抽取 8 只帝王蝶,并记录它们的翼展 xx cm。结果如下。

8.27.07.38.87.88.59.27.48.2 \quad 7.0 \quad 7.3 \quad 8.8 \quad 7.8 \quad 8.5 \quad 9.2 \quad 7.4

Ansal 还从地区 YY 随机抽取 9 只帝王蝶,并记录它们的翼展 yy cm。结果总结如下。

∑y=71.10∑y2=567.13\sum y = 71.10 \qquad \sum y^2 = 567.13

Ansal 怀疑地区 XX 的帝王蝶平均翼展大于地区 YY 的帝王蝶平均翼展。已知地区 XX 和 YY 的帝王蝶翼展服从正态分布,且总体方差相等。

在 10% 显著性水平下检验数据是否支持 Ansal 的怀疑。

(8)
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

解法一

思路

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两个独立样本来自等方差的正态总体。合并方差后,对地区 XX 的平均翼展较大做右尾检验。

答题过程

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Let μX\mu_X and μY\mu_Y be the population mean wingspans in the two regions. Test H0 ⁣:μX=μYH_0\colon\mu_X=\mu_Y against H1 ⁣:μX>μYH_1\colon\mu_X>\mu_Y. The sample means are 64.2/8=8.02564.2/8=8.025 and 71.10/9=7.971.10/9=7.9. The pooled variance estimate is

 sp2=(519.46−64.22/8)+(567.13−71.102/9)8+9−2= 0.646333….\begin{align*} &\,s_p^2=\frac{(519.46-64.2^2/8)+(567.13-71.10^2/9)}{8+9-2}\\ =&\,0.646333\ldots. \end{align*}

The test statistic is

t=8.025−7.9sp2(1/8+1/9)=0.319980….t=\frac{8.025-7.9}{\sqrt{s_p^2(1/8+1/9)}}=0.319980\ldots.

With 1515 degrees of freedom, the upper 10%10\% critical value is 1.3411.341. Since 0.319980…<1.3410.319980\ldots<1.341, we fail to reject H0H_0. There is insufficient evidence at the 10%10\% level that the population mean wingspan is greater in region XX.