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广超 · 2020年10月29日

问一道题:NO.PZ2015120604000118 [ CFA I ]

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,答案中这个公式,老师写在讲义里了,可是基础班的课程里没讲这个公式是怎么来的。基础班Estimation的三个视频里没有关于这个公式的内容。Z a/2 是怎么来的?代表什么?
1 个答案
已采纳答案

星星_品职助教 · 2020年10月30日

同学你好,

置信区间的公式在正态分布知识点里讲过原理。以99%的置信区间为例,通过查表,可知落在在均值μ周围±2.58个标准差范围内的概率为99%。即99%置信区间对应[μ-2.58σ,μ+2.58σ]。其余68%,90%,95%置信区间同理,其余概率基本上都不需要考虑。

这里面2.58对应的就是Z α/2。在置信区间公式里,这个Z α/2叫做reliability factor,也可以叫做critical value,后者更常用。

α是significance level,表示置信区间外的总面积(拒绝域)为1%。由于对称,那么单尾拒绝域的面积就是α/2=1%/2=0.5%. 在正态分布(用Z表示)下,查表即可得这个概率面积对应的分位点为2.58.

-------------

对应到这道题,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critical value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,df=n-1=24。通过对应单尾概率0.005和自由度24查表可得t critical value=2.797.

代入公式40±2.797×1.6即可得到答案C选项。

 

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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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