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为了求职冲呀 · 2020年03月10日

问一道题：NO.PZ2015120604000118

  1. CFA I
  2. Quantitative

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，请问查表在那个知识点有详细讲到吗？这个知识点里何老师是快速过了一遍，还是没有听懂。在这道题里面df24，为什么数字是p=0.005的2.797呢？

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1 个答案

星星_品职助教 · 2020年03月10日

同学你好，

这道题就是考察置信区间的公式，只是里面的临界值由于是t分布，所以需要查t表。

查t表需要知道自由度和对应概率。检验均值的t自由度是n-1，所以就是25-1=24

由于是99%的置信区间，所以如果查单尾的t表，就查单尾面积为0.5%的，如果是双尾的t表，就查双尾面积为1%的。

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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

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 34.55B.38.52 to 54.48C.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即可得到答案百分之99对应的不是2.58吗，为啥答案是2.7几

2023-04-10 19:58 1 · 回答

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

2022-12-05 20:52 2 · 回答

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2022-09-22 16:13 2 · 回答

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2022-09-13 11:00 1 · 回答