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Allez😃Sea · 2020年04月19日

问一道题:NO.PZ2015120604000118

问题如下:

Deli Bur is a high school. A recent survey of 25 student of the high school indicates that the mean time they spend going to school is 40 minutes . This sample's standard deviation is 8 minutes. The distribution of the population is supposed to be normal. The 99% confidence interval for the mean time that all Deli Bur students  spend going to the school is:

选项:

A.

30.72 to 34.55

B.

38.52 to 54.48

C.

35.52 to 44.48

解释:

C is correct.

Because the simple size is less than 30, so the confidence interval fo rthe  population whose variance is unknow is :

x - ± t α/2 (s/ n ) .

To calculate critical value: t 0.005  and df = 24 is 2.797.

So, the confidence interval is 40 ± 2.797(8 / 5) = 40 ± 4.48 = 35.52 to 44.48

老师,题目里给的是“The distribution of the population is supposed to be normal. ”那是不是就是说正态分布?不是应该直接用z分布吗?为什么要用t来求?

1 个答案
已采纳答案

星星_品职助教 · 2020年04月20日

同学你好,

你列的这句话是在说整体(population)服从正态分布。不过我们首要看的是“总体方差是否已知”,已知用z,未知用t。除非题干明确说了已经有known population variance,否则都是用t。

只有当小样本的时候(N<30,类似本题)才会考虑总体分布,因为非正态总体,小样本是不可估计的。这也是这道题强调总体服从正态的原因。

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NO.PZ2015120604000118 问题如下 li Bur is a high school. A recent survey of 25 stunt of the high school incates ththe metime they spengoing to school is 40 minutes . This sample's stanrviation is 8 minutes. The stribution of the population is supposeto normal. The 99% confinintervfor the metime thall li Bur stunts spengoing to the school is: A.30.72 to 34.55 B.38.52 to 54.48 C.35.52 to 44.48 C is correct.Because the simple size is less th30, so the confinintervfor the population whose varianis unknow is : x - ± t α/2 (s/ n ) . To calculate criticvalue: t 0.005 an = 24 is 2.797.So, the confinintervis 40 ± 2.797(8 / 5) = 40 ± 4.48 = 35.52 to 44.48本题由于样本数量小于30,且总体方差未知。可知均值X bar=40,对应的X bar的标准差(即标准误)=8/√25=1.6。 而criticvalue需要查表求得,如果是正态分布,这个值就是2.58。但这道题对应的是t分布,需要查的是t表(总体方差未知用t)。本题α=1%,则α/2=0.005。t分布需要考虑自由度,=n-1=24。通过对应单尾概率0.005和自由度24查表可得t criticvalue=2.797.代入公式40±2.797×1.6即可得到答案 如何看出这道题总体方差未知呢?

2023-04-12 19:27 1 · 回答

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