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CIE 9231 2024 June Paper 43 Q6

A Level / CIE / FS

CIE 9231 2024 June Paper 43 Paper · Question 6

题目

Problem

Seva is investigating the lengths of the tails of adult wallabies in two regions of Australia, X and Y. He chooses a random sample of 50 adult wallabies from region X and records the lengths, xx cm, of their tails. He also chooses a random sample of 40 adult wallabies from region Y and records the lengths, yy cm, of their tails. His results are summarised as follows.

x=1080x2=23480y=940y2=22220\sum x = 1080 \qquad \sum x^2 = 23\,480 \qquad \sum y = 940 \qquad \sum y^2 = 22\,220

It cannot be assumed that the population variances of the two distributions are the same.

(a) Find a 90% confidence interval for the difference between the population mean lengths of the tails of adult wallabies in regions X and Y.

[6]

The population mean lengths of the tails of adult wallabies in regions X and Y are μX\mu_X cm and μY\mu_Y cm respectively.

(b) Test, at the 10% significance level, the null hypothesis μYμX=1.1\mu_Y - \mu_X = 1.1 against the alternative hypothesis μYμX>1.1\mu_Y - \mu_X > 1.1. State your conclusion in the context of the question.

[4]
题目中文翻译

Seva 正在研究澳大利亚两个地区 X 和 Y 中成年袋鼠尾巴的长度。他从地区 X 随机抽取 50 只成年袋鼠,并记录它们尾巴的长度 xx cm。他还从地区 Y 随机抽取 40 只成年袋鼠,并记录它们尾巴的长度 yy cm。结果概括如下。

x=1080x2=23480y=940y2=22220\sum x = 1080 \qquad \sum x^2 = 23\,480 \qquad \sum y = 940 \qquad \sum y^2 = 22\,220

不能假设两个分布的总体方差相同。

(a) 求地区 X 与地区 Y 成年袋鼠尾巴总体平均长度之差的 90% 置信区间。

地区 X 和地区 Y 成年袋鼠尾巴的总体平均长度分别为 μX\mu_X cm 和 μY\mu_Y cm。

(b) 在 10% 显著性水平下,检验原假设 μYμX=1.1\mu_Y - \mu_X = 1.1,备择假设为 μYμX>1.1\mu_Y - \mu_X > 1.1。结合题目背景说明你的结论。

解答