开发者:上海品职教育科技有限公司 隐私政策详情

应用版本:4.2.11(IOS)｜3.2.5(安卓)APP下载

返回

书茜 · 2022年02月04日

样本的标准差和样本均值的标准误是同一个公式吧？

  1. CFA I
  2. Quantitative

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.

如果只抽一次，一次数量大于等于30，那么，这一次数据的标准差等于总体标准差除以根号n么？如果抽100次，每一次数量都大于等于30，每一次都会有一个均值，这些均值又形成一个新的分布，这个新的分布的方差叫做标准误，也等于总体的标准差除以根号n吗？

添加评论
1 个答案
已采纳答案

星星_品职助教 · 2022年02月04日

同学你好，

1）只抽一次，只能得到这组数据自身的标准差s。但可以知道的是这组数据的均值服从正态分布，样本均值 的均值为总体均值μ，标准差是总体标准差σ除以根号n。这是中心极限定理的内容。

如果此时σ不知道，就用这组数据自身的标准差s来做近似替代。此时标准误=s/根号n。

2）中心极限定理是对“抽100次”的情况做的总结。有了这个定理后，此后就不需要反复抽样。即使只抽一次我们也知道样本均值的标准差（标准误）就是总体标准差σ除以根号n。



  • 1

    回答

  • 0

    关注

  • 461

    浏览
相关问题

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

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？

2022-10-20 09:10 1 · 回答

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. 老师您好，PPT上的公司用的是sigma，所以我的理解是直接用population S好，为什么用SE呢？

2022-04-03 04:05 1 · 回答

NO.PZ2015120604000117 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？

2021-06-06 14:57 2 · 回答