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CFApipa · 2022年11月19日

why not using 1/2 times α for this formula? shouldn't it be 1/2 times 1.96?

NO.PZ2015120604000117

问题如下:

A random sample is 100 CFA candidate's exam scoring. The mean of the scoring is 64. The standard deviation of the population scoring is 15. The distribution is supposed to be normal. The 95% confidence interval for the population mean should be:

选项:

A.

61.06 to 66.94.

B.

61.06 to 69.94.

C.

65.06 to 66.94.

解释:

A is correct.

Because the variance of the population is know and n ≥ 30,so The confidence interval is:

x¯ ± z α/2 (σ/ n )

z α/2 = z 0.025 =1.96

64 ± 1.96(15/10) = 64 ± 2.94 = 61.06 to 66.94.

因为样本方差已知且n>30,我们直接使用置信区间计算:

64 ± 1.96(15/10) = 64 ± 2.94 = 61.06 至 66.94.



1 个答案
已采纳答案

星星_品职助教 · 2022年11月19日

同学你好,

这道题的1.96即为α/2所对应的关键值。

具体而言,本题α=5%,由于是求置信区间,所以可知单侧尾部面积为α/2=2.5%。单侧尾部面积2.5%对应的关键值就是1.96。

与之对比,如果单侧尾部面积为5%(即直接是α),则此时关键值就是1.645了。

-----

上述内容可总结为下面的规律。

数量科目中,正态分布下,关键值为:

90%置信区间:±1.645-----对应单尾面积5%;

95%置信区间:±1.96-------对应单尾面积2.5%;

99%置信区间:±2.58-------对应单尾面积0.5%;

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NO.PZ2015120604000117问题如下 A ranm sample is 100 CFA cante's exscoring. The meof the scoring is 64. The stanrviation of the population scoring is 15. The stribution is supposeto normal. The 95% confinintervfor the population meshoulbe: A.61.06 to 66.94.B.61.06 to 69.94.C.65.06 to 66.94. A is correct.Because the varianof the population is know ann ≥ 30,so The confinintervis: x¯ ± z α/2 (σ/ n ) z α/2 = z 0.025 =1.96 64 ± 1.96(15/10) = 64 ± 2.94 = 61.06 to 66.94.因为样本方差已知且n 30,我们直接使用置信区间计算64 ± 1.96(15/10) = 64 ± 2.94 = 61.06 至 66.94. 根据题目分数只可能是正数,拒绝域应该是在右边,95%的单尾应该对应的是1.65呀?

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