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IAL 2024 Jan S3 Q6

A Level / Edexcel / S3

IAL 2024 Jan Paper · Question 6

题目

Problem

A random sample of 8 three-month-old golden retriever dogs is taken. The heights of the golden retrievers are recorded.

Using this sample, a 95% confidence interval for the mean height, in cm, of three-month-old golden retrievers is found to be (45.72, 53.88)

(a) Find a 99% confidence interval for the mean height. You may assume that the heights are normally distributed with known population standard deviation.

(5)

Some summary statistics for the weights, x kg, of this sample are given below.

∑x\sum x91.2
∑x2\sum x^21145.16
n8

(b) Calculate unbiased estimates of the mean and the variance of the weights of three-month-old golden retrievers.

(3)

A further random sample of 24 three-month-old golden retrievers is taken. The unbiased estimates of the mean and the variance of the weights, in kg, from this sample are found to be 10.8 and 17.64 respectively.

(c) Estimate the standard error of the mean weight for the combined sample of 32 three-month-old golden retrievers.

(7)
(Total for Question 6 is 15 marks)
题目中文翻译

抽取了 8 只三个月大的金毛寻回犬作为随机样本。 记录了这些金毛寻回犬的身高。

利用这个样本，求得三个月大金毛寻回犬平均身高（单位 cm）的 95% 置信区间为 (45.72, 53.88)。

(a) 求平均身高的 99% 置信区间。 你可以假设身高服从正态分布且总体标准差已知。

下列给出了该样本体重 x kg 的一些统计量。

∑x\sum x91.2
∑x2\sum x^21145.16
n8

(b) 求三个月大金毛寻回犬体重的均值和方差的无偏估计。

又抽取了另外 24 只三个月大的金毛寻回犬。 该样本中体重（单位 kg）的均值和方差的无偏估计分别为 10.8 和 17.64。

(c) 估计 32 只三个月大金毛寻回犬合并样本中体重均值的标准误。

解答

(a)

解法一

思路

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同一个样本、同一个已知总体标准差下，置信区间的中心不变，误差界与标准正态临界值成正比。因此先由原区间求出样本均值和 95%95\% 误差界，再按临界值之比把误差界调整到 99%99\%。

答题过程

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The sample mean is the midpoint of the given confidence interval:

x‾=45.72+53.882=49.8.\overline{x}=\frac{45.72+53.88}{2}=49.8.

The margin of error for the 95%95\% confidence interval is

53.88−45.722=4.08.\frac{53.88-45.72}{2}=4.08.

For the same sample and known population standard deviation, the margin of error is proportional to the standard normal critical value. Hence the margin of error for a 99%99\% confidence interval is

4.08×2.57581.96=5.3618….4.08\times\frac{2.5758}{1.96}=5.3618\ldots.

Therefore, the 99%99\% confidence interval is

49.8±5.3618…=(44.438…, 55.161…).\begin{align*} 49.8&\pm5.3618\ldots\\ &=(44.438\ldots,\ 55.161\ldots). \end{align*}

Thus a 99%99\% confidence interval for the mean height is

(44.4, 55.2) cm.\boxed{(44.4,\ 55.2)\text{ cm}}.

(b)

解法一

思路

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均值的无偏估计是样本均值；方差的无偏估计则使用分母 n−1n-1，并由给出的 ∑x\sum x 与 ∑x2\sum x^2 直接计算。

答题过程

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The unbiased estimate of the population mean is

x‾=∑xn=91.28=11.4 kg.\overline{x}=\frac{\sum x}{n}=\frac{91.2}{8}=\boxed{11.4\text{ kg}}.

The unbiased estimate of the population variance is

s2=∑x2−nx‾2n−1=1145.16−8(11.4)27=15.0685….\begin{align*} s^2 &=\frac{\sum x^2-n\overline{x}^{2}}{n-1}\\ &=\frac{1145.16-8(11.4)^2}{7}\\ &=15.0685\ldots. \end{align*}

Therefore, the unbiased estimate of the variance is

15.1 kg2.\boxed{15.1\text{ kg}^2}.

(c)

解法一

思路

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先把第二组样本的无偏方差还原为 ∑x2\sum x^2，再合并两组的 ∑x\sum x、∑x2\sum x^2 和样本量。由此求得合并样本的无偏方差，最后用 s2/n\sqrt{s^2/n} 估计样本均值的标准误。

答题过程

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For the second sample,

∑x=24(10.8)=259.2.\sum x=24(10.8)=259.2.

Since its unbiased sample variance is 17.6417.64,

∑x2=(24−1)(17.64)+24(10.8)2=3205.08.\begin{align*} \sum x^2 &=(24-1)(17.64)+24(10.8)^2\\ &=3205.08. \end{align*}

For the combined sample,

n=8+24=32,∑x=91.2+259.2=350.4,∑x2=1145.16+3205.08=4350.24.\begin{aligned} n&=8+24=32,\\ \sum x&=91.2+259.2=350.4,\\ \sum x^2&=1145.16+3205.08=4350.24. \end{aligned}

Hence the combined sample mean is

x‾=350.432=10.95,\overline{x}=\frac{350.4}{32}=10.95,

and the combined unbiased sample variance is

s2=4350.24−32(10.95)231=16.56.\begin{align*} s^2 &=\frac{4350.24-32(10.95)^2}{31}\\ &=16.56. \end{align*}

Therefore, the estimated standard error of the mean weight is

SE⁡(X‾)=s2n=16.5632=0.71937….\begin{align*} \operatorname{SE}(\overline X) &=\sqrt{\frac{s^2}{n}}\\ &=\sqrt{\frac{16.56}{32}}\\ &=0.71937\ldots. \end{align*}

Thus the estimated standard error is

0.719 kg.\boxed{0.719\text{ kg}}.