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CIE 9231 2022 June Paper 42 Q5

A Level / CIE / FS

CIE 9231 2022 June Paper 42 Paper · Question 5

题目

Problem

Raman is researching the heights of male giraffes in a particular region. Raman assumes that the heights of male giraffes in this region are normally distributed. He takes a random sample of 8 male giraffes from the region and measures the height, in metres, of each giraffe. These heights are as follows.

 5.2, 5.8, 4.9, 6.1, 5.5, 5.9, 5.4, 5.6.\begin{align*} &\,5.2,\ 5.8,\ 4.9,\ 6.1,\\ &\,5.5,\ 5.9,\ 5.4,\ 5.6. \end{align*}

(a) Find a 90% confidence interval for the population mean height of male giraffes in this region.

[5]

Raman claims that the population mean height of male giraffes in the region is less than 5.9 metres.

(b) Test at the 2.5% significance level whether this sample provides sufficient evidence to support Raman’s claim.

[4]
题目中文翻译

Raman 正在研究某地区雄性长颈鹿的身高。Raman 假设该地区雄性长颈鹿的身高服从正态分布。他从该地区随机抽取 8 只雄性长颈鹿,并测量每只长颈鹿的身高,单位为米。这些身高如下。

 5.2, 5.8, 4.9, 6.1, 5.5, 5.9, 5.4, 5.6.\begin{align*} &\,5.2,\ 5.8,\ 4.9,\ 6.1,\\ &\,5.5,\ 5.9,\ 5.4,\ 5.6. \end{align*}

(a) 求该地区雄性长颈鹿总体平均身高的 90% 置信区间。

(5)

Raman 声称该地区雄性长颈鹿的总体平均身高小于 5.9 米。

(b) 在 2.5% 显著性水平下,检验这个样本是否提供了足够证据支持 Raman 的声称。

(4)
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

先求样本均值与无偏方差。总体正态、样本量为 88,用自由度 77 的 tt 区间。

答题过程

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The sums are ∑x=44.4\sum x=44.4 and ∑x2=247.48\sum x^2=247.48. Thus xˉ=5.55\bar x=5.55 and

s2=247.48−44.42/87=0.151428….s^2=\frac{247.48-44.4^2/8}{7}=0.151428\ldots.

The two-sided 90%90\% critical value with 77 degrees of freedom is 1.8951.895. The interval is

 5.55±1.895s2/8= (5.28928…,5.81071…),\begin{align*} &\,5.55\pm1.895\sqrt{s^2/8}\\ =&\,(5.28928\ldots,5.81071\ldots), \end{align*}

or (5.29,5.81)(5.29,5.81) m.

(b)

解法一

思路

展开

继续用 (a) 的标准误。平均身高小于 5.95.9 是左尾备择。

答题过程

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Test H0 ⁣:μ=5.9H_0\colon\mu=5.9 against H1 ⁣:μ<5.9H_1\colon\mu<5.9. The statistic is

t=5.55−5.9s2/8=−2.54395….t=\frac{5.55-5.9}{\sqrt{s^2/8}}=-2.54395\ldots.

For 77 degrees of freedom the lower 2.5%2.5\% critical value is −2.365-2.365. Since the statistic is lower, we reject H0H_0. There is sufficient evidence at the 2.5%2.5\% level that the population mean height of male giraffes in the region is below 5.95.9 m.