Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2022 June Paper 63 Q3

A Level / CIE / S2

CIE 9709 2022 June Paper 63 Paper · Question 3

题目

Problem

Batteries of type AA are known to have a mean life of 150150 hours. It is required to test whether a new type of battery, type BB, has a shorter mean life than type AA batteries.

(a) Give a reason for using a sample rather than the whole population in carrying out this test.

(1)

A random sample of 120120 type BB batteries are tested and it is found that their mean life is 147147 hours, and an unbiased estimate of the population variance is 225 hours2225\,\mathrm{hours}^2.

(b) Test, at the 2%2\% significance level, whether type BB batteries have a shorter mean life than type AA batteries.

(5)

(c) Calculate a 94%94\% confidence interval for the population mean life of type BB batteries.

(3)
题目中文翻译

已知 AA 型电池的平均寿命为 150150 小时。需要检验新型电池,即 BB 型电池的平均寿命是否比 AA 型电池的短。

(a) 给出一个理由,说明为什么进行这项检验时使用样本,而非整个总体。

(1)

随机抽取 120120 节 BB 型电池进行测试,得到平均寿命为 147147 小时,总体方差的一个无偏估计值为 225 小时2225\,\text{小时}^2。

(b) 在 2%2\% 的显著性水平下检验 BB 型电池的平均寿命是否比 AA 型电池的短。

(5)

(c) 计算 BB 型电池总体平均寿命的一个 94%94\% 置信区间。

(3)
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

寿命测试会耗尽电池,测试全部总体会把所有电池用完。

答题过程

展开

Testing a battery’s lifetime uses it up, so testing the whole population would leave all the batteries unusable.

(b)

解法一

思路

展开

以B型总体平均寿命作左尾检验,使用无偏方差225和样本量120的标准误。

答题过程

展开

Let μB\mu_B be the population mean life of type BB batteries, in hours.

H0:μB=150,H1:μB<150.H_0:\mu_B=150,\qquad H_1:\mu_B<150.

Using the large-sample normal approximation,

z=147−150225/120=−2.19089….z=\frac{147-150}{\sqrt{225/120}} =-2.19089\ldots.

The lower-tail 2% critical value is −2.054-2.054. Since −2.19089…<−2.054-2.19089\ldots<-2.054, we reject H0H_0.

There is sufficient evidence at the 2% significance level that type BB batteries have a shorter population mean life than type AA batteries.

(c)

解法一

思路

展开

94%区间两端各3%,使用累计概率0.97的正态临界值。

答题过程

展开

With z=Φ−1(0.97)=1.880794…z=\Phi^{-1}(0.97)=1.880794\ldots, the 94% confidence interval is

147±z225120=147±2.57538….147\pm z\sqrt{\frac{225}{120}} =147\pm2.57538\ldots.

Thus the interval is

(144, 150) hours(3 s.f.).\boxed{(144,\ 150)\text{ hours}}\quad\text{(3 s.f.).}