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@_@ · 2021年06月06日

老师请问样本标准差和样本均值标准分别是怎么定义的,区别是什么?

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 sopposed 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.

老师请问样本标准差和样本均值标准分别是怎么定义的,区别是什么?

2 个答案
已采纳答案

星星_品职助教 · 2021年06月06日

同学你好,

抽一次样,里面有n个数据(sample size),这n个数据的均值就是这个样本的均值,这n个数据的标准差就是样本标准差(sample standard deviation)。

抽很多次样,每个样本里都有一个样本均值,把这些样本均值再集中在一起求标准差,就是样本均值的标准差,即标准误(standard error)

@_@ · 2021年06月06日

终于明白了,谢谢老师

星星_品职助教 · 2021年06月06日

to:@_@ 好的加油~

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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.

2022-11-19 10:58 1 · 回答

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2022-10-20 09:10 1 · 回答

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