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

Ariel · 2023年01月18日

怎么区分置信间，好像还有一个正态抽样前的置信区间是（u-1.65标准差 u+1.65标准差）

  1. CFA I
  2. Quantitative

NO.PZ2015120604000130

问题如下：

We want to use z-statistic to construct confidence interval for a normally distribution. Assume the sample size is 100, sample mean is 15% and the standard deviation of sample is 25%. The significance level is supposed to be 10% , the confidence interval is :

选项：

A.

- 10.88% to 19.13%.

B.

10.88% to 19.13%.

C.

10.88% to 20.57%.

解释：

B is correct.

In this case, we can know that the Confidence Interval=[Point Estimate +/- (reliability factor)*Standard error]=


什么时候会考这个区间，一般问这个区间的问题是怎么问的

这个区间的标准差指的是总体标准差吗

添加评论
1 个答案
已采纳答案

星星_品职助教 · 2023年01月18日

同学你好，

如果是针对总体做置信区间，则中心是μ，对应的是总体标准差σ。就是提问中的“u-1.65标准差 u+1.65标准差”。

如果是针对样本均值做置信区间，则中心是样本均值 X bar，对应的也是X bar的标准差，即标准误。标准误的公式为s / 跟号n。也就是答案解析中的形式。

考试的定量题目一般都是要求根据公式来做置信区间的相关计算。例如求置信区间，求出上区间端点值，计算区间宽度等等。

  • 1

    回答

  • 0

    关注

  • 374

    浏览
相关问题

NO.PZ2015120604000130 问题如下 We want to use z-statistito construconfinintervfor a normally stribution. Assume the sample size is 100, sample meis 15% anthe stanrviation of sample is 25%. The significanlevel is supposeto 10% , the confinintervis : A.- 10.88% to 19.13%. B.10.88% to 19.13%. C.10.88% to 20.57%. B is correct.In this case, we cknow ththe ConfinInterval=[Point Estimate +/- (reliability factor)*Stanrerror]= SignificanLevel都是要除以2吗？

2022-11-01 18:08 1 · 回答

NO.PZ2015120604000130 问题如下 We want to use z-statistito construconfinintervfor a normally stribution. Assume the sample size is 100, sample meis 15% anthe stanrviation of sample is 25%. The significanlevel is supposeto 10% , the confinintervis : A.- 10.88% to 19.13%. B.10.88% to 19.13%. C.10.88% to 20.57%. B is correct.In this case, we cknow ththe ConfinInterval=[Point Estimate +/- (reliability factor)*Stanrerror]=x¯±zα/2σn=15%±1.65×25%100=10.88%,19.13% 请题干以及该题涉及的原理所对应知识点

2022-10-13 23:10 1 · 回答

NO.PZ2015120604000130 问题如下 We want to use z-statistito construconfinintervfor a normally stribution. Assume the sample size is 100, sample meis 15% anthe stanrviation of sample is 25%. The significanlevel is supposeto 10% , the confinintervis : A.- 10.88% to 19.13%. B.10.88% to 19.13%. C.10.88% to 20.57%. B is correct.In this case, we cknow ththe ConfinInterval=[Point Estimate +/- (reliability factor)*Stanrerror]=x¯±zα/2σn=15%±1.65×25%100=10.88%,19.13% 这里的Stanrerror 不应该是25%/ √100 吗？解析里的式子是什么意思

2022-10-09 10:47 1 · 回答

NO.PZ2015120604000130问题如下We want to use z-statistito construconfinintervfor a normally stribution. Assume the sample size is 100, sample meis 15% anthe stanrviation of sample is 25%. The significanlevel is supposeto 10% , the confinintervis :A.- 10.88% to 19.13%.B.10.88% to 19.13%.C.10.88% to 20.57%.B is correct.In this case, we cknow ththe ConfinInterval=[Point Estimate +/- (reliability factor)*Stanrerror]=x¯±zα/2σn=15%±1.65×25%100=10.88%,19.13%老师，既然题目说是正态分布，而且是选择题。想问下能否利用正态分布的特性，代入解题？正太分布中轴是均值，题干给出的置信区间应该也是轴对称。直接带入，分位点-均值19.13-15=4.13均值-分位点到均值的距离15-4.13=10.87然后选出最符合的的

2022-09-27 15:46 1 · 回答