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awen · 2017年12月31日

问一道题:NO.PZ2015120604000117 [ CFA I ]

答案选择A。但是提示A错误,请进行改正。


问题如下图:

    

选项:

A.

B.

C.

解释:



2 个答案
已采纳答案

源_品职助教 · 2017年12月31日

好滴,收到。

awen · 2018年01月02日

尽快改正哦,还没有改好哦

源_品职助教 · 2018年01月03日

你在看看,应该改过来了。

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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

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 61.06 to 69.94. 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. 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?

2022-02-04 12:33 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