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

卓娅 · 2023年06月06日

类似题型的计算步骤

  1. CFA I
  2. Quantitative

NO.PZ2017092702000114

问题如下：

For a sample size of 65 with a mean of 31 taken from a normally distributed population with a variance of 529, a 99% confidence interval for the population mean will have a lower limit closest to:

选项：

A.

23.64.

B.

25.41.

C.

30.09.

解释：

A is correct.

To solve, use the structure of Confidence interval = Point estimate ± Reliability factor × Standard error, which, for a normally distributed population with known variance, is represented by the following formula:X‾±zα/2σn\overline X\pm z_{\alpha/2}\frac\sigma{\sqrt n}

For a 99% confidence interval, use z0.005 = 2.58. Also, σ = 529\sqrt{529} = 23.

Therefore, the lower limit =31−2.58×2365=23.639831-2.58\times\frac{23}{\sqrt{65}}=23.6398

我们需要用到【置信区间结构】的计算公式解决本题：

the structure of Confidence interval = Point estimate ± Reliability factor × Standard error

即：X‾±zα/2σn\overline X\pm z_{\alpha/2}\frac\sigma{\sqrt n}

当置信区间=99%的时候，Z0.005=2.58，且 σ = 529\sqrt{529} = 23.

所以，下限为： =31−2.58∗2365=23.6398=31-2.58*\frac{23}{\sqrt{65}}=23.6398

老师，这种题型的计算思考步骤是什么呢，总是容易搞混，看题找不到方向？谢谢老师

添加评论
1 个答案
已采纳答案

袁园_品职助教 · 2023年06月08日

嗨，爱思考的PZer你好：


题干问总体均值置信区间的下限，核心考点是中心极限定理和标准误。

  1. 你看到求总体均值的置信区间就要反应到是中心极限定理。因为中心极限定理给了我们从样本均值推论到总体均值的一个原则。
  2. 如何理解中心极限定理？例如总体有1万个人，如果只抽1个样本，里面有100个人（sample size）的话，那么被抽出来的是哪100个人其实是不确定的。如果抽200次，那么前后就会有200个样本均值，每个样本均值取值都是由具体那一次被抽出来的200人决定的。所以样本均值本身也是随机变量的概念。于是就有自己的分布，均值（也就是“样本均值X bar”的均值，和自己的标准差（样本均值X bar的标准差，也就叫做标准误））。
  3. 中心极限定理说的是从一个总体中抽样，如果样本容量足够大（n＞30），则样本均值作为一个随机变量就服从正态分布，样本均值的均值（注意此处理解）就为总体均值，样本均值的方差为总体方差/样本容量，即样本均值的标准差（也就是标准误）为总体标准差/根号下n。有了中心极限定理之后，哪怕只抽一次样，也可以确定抽的这一次样所得到的样本均值也服从以上的正态分布。而如果放宽了假设，当总体方差未知时，就可以用只抽一次样被抽出来的那个样本的标准差（注意样本的标准差和样本均值的标准差是两个概念，后者才是标准误）来代替总体的标准差。也就是样本均值的标准差可以被代替为 样本标准差S / 根号n。

这是一个很重要的知识点，何老师在基础班里讲的很详细，如果这里不理解的话，一定要回头再听一遍基础班的中心极限定理部分。你看到给了样本均值求总体均值的范围就是要意识到这是在考中心极限定理哦

----------------------------------------------
加油吧，让我们一起遇见更好的自己！

  • 1

    回答

  • 0

    关注

  • 396

    浏览
相关问题

NO.PZ2017092702000114 问题如下 For a sample size of 65 with a meof 31 taken from a normally stributepopulation with a varianof 529, a 99% confinintervfor the population mewill have a lower limit closest to: A.23.64. B.25.41. C.30.09. A is correct.To solve, use the structure of Confininterv= Point estimate ± Reliability factor × Stanrerror, which, for a normally stributepopulation with known variance, is representethe following formula:X‾±zα/2σn\overline X\pm z_{\alpha/2}\frac\sigma{\sqrt n}X±zα/2​n​σ​ For a 99% confininterval, use z0.005 = 2.58. Also, σ = 529\sqrt{529}529​ = 23. Therefore, the lower limit =31−2.58×2365=23.639831-2.58\times\frac{23}{\sqrt{65}}=23.639831−2.58×65​23​=23.6398 我们需要用到【置信区间结构】的计算公式解决本题the structure of Confininterv= Point estimate ± Reliability factor × Stanrerror即X‾±zα/2σn\overline X\pm z_{\alpha/2}\frac\sigma{\sqrt n}X±zα/2​n​σ​当置信区间=99%的时候，Z0.005=2.58，且 σ = 529\sqrt{529}529​ = 23. 所以，下限为 =31−2.58∗2365=23.6398=31-2.58*\frac{23}{\sqrt{65}}=23.6398=31−2.58∗65​23​=23.6398 请问为什么 No.PZ2017092702000113 (选择题)，这道题的置信区间计算需要使用“分母根号下n-1”，即需要用自由度求解，但是这道题不需要呢？请问老师区别在哪里？谢谢老师

2023-04-12 16:32 1 · 回答

NO.PZ2017092702000114 问题如下 For a sample size of 65 with a meof 31 taken from a normally stributepopulation with a varianof 529, a 99% confinintervfor the population mewill have a lower limit closest to: A.23.64. B.25.41. C.30.09. A is correct.To solve, use the structure of Confininterv= Point estimate ± Reliability factor × Stanrerror, which, for a normally stributepopulation with known variance, is representethe following formula:X‾±zα/2σn\overline X\pm z_{\alpha/2}\frac\sigma{\sqrt n}X±zα/2​n​σ​ For a 99% confininterval, use z0.005 = 2.58. Also, σ = 529\sqrt{529}529​ = 23. Therefore, the lower limit =31−2.58×2365=23.639831-2.58\times\frac{23}{\sqrt{65}}=23.639831−2.58×65​23​=23.6398 我们需要用到【置信区间结构】的计算公式解决本题the structure of Confininterv= Point estimate ± Reliability factor × Stanrerror即X‾±zα/2σn\overline X\pm z_{\alpha/2}\frac\sigma{\sqrt n}X±zα/2​n​σ​当置信区间=99%的时候，Z0.005=2.58，且 σ = 529\sqrt{529}529​ = 23. 所以，下限为 =31−2.58∗2365=23.6398=31-2.58*\frac{23}{\sqrt{65}}=23.6398=31−2.58∗65​23​=23.6398 请问老师，我有两个问题，第一个是如果题干没有告诉我应该用到什么分布的话，我怎么判断是应该用z分布还是t分布呢？第二个问题是，我现在很混乱z分布和t分布的计算有什么区别呢》唯一的区别就是t分布的计算是自由度-1是吗，z分布就直接用sample zise的数量？

2022-12-13 06:49 1 · 回答

NO.PZ2017092702000114问题如下 For a sample size of 65 with a meof 31 taken from a normally stributepopulation with a varianof 529, a 99% confinintervfor the population mewill have a lower limit closest to:A.23.64. B.25.41. C.30.09. A is correct.To solve, use the structure of Confininterv= Point estimate ± Reliability factor × Stanrerror, which, for a normally stributepopulation with known variance, is representethe following formula:X‾±zα/2σn\overline X\pm z_{\alpha/2}\frac\sigma{\sqrt n}X±zα/2​n​σ​ For a 99% confininterval, use z0.005 = 2.58. Also, σ = 529\sqrt{529}529​ = 23. Therefore, the lower limit =31−2.58×2365=23.639831-2.58\times\frac{23}{\sqrt{65}}=23.639831−2.58×65​23​=23.6398 我们需要用到【置信区间结构】的计算公式解决本题the structure of Confininterv= Point estimate ± Reliability factor × Stanrerror即X‾±zα/2σn\overline X\pm z_{\alpha/2}\frac\sigma{\sqrt n}X±zα/2​n​σ​当置信区间=99%的时候，Z0.005=2.58，且 σ = 529\sqrt{529}529​ = 23. 所以，下限为 =31−2.58∗2365=23.6398=31-2.58*\frac{23}{\sqrt{65}}=23.6398=31−2.58∗65​23​=23.6398 ​为什么样本标准差不是除以根号64呢？

2022-11-06 16:45 1 · 回答

NO.PZ2017092702000114问题如下 For a sample size of 65 with a meof 31 taken from a normally stributepopulation with a varianof 529, a 99% confinintervfor the population mewill have a lower limit closest to:A.23.64. B.25.41. C.30.09. A is correct.To solve, use the structure of Confininterv= Point estimate ± Reliability factor × Stanrerror, which, for a normally stributepopulation with known variance, is representethe following formula:X‾±zα/2σn\overline X\pm z_{\alpha/2}\frac\sigma{\sqrt n}X±zα/2​n​σ​ For a 99% confininterval, use z0.005 = 2.58. Also, σ = 529\sqrt{529}529​ = 23. Therefore, the lower limit =31−2.58×2365=23.639831-2.58\times\frac{23}{\sqrt{65}}=23.639831−2.58×65​23​=23.6398 我们需要用到【置信区间结构】的计算公式解决本题the structure of Confininterv= Point estimate ± Reliability factor × Stanrerror即X‾±zα/2σn\overline X\pm z_{\alpha/2}\frac\sigma{\sqrt n}X±zα/2​n​σ​当置信区间=99%的时候，Z0.005=2.58，且 σ = 529\sqrt{529}529​ = 23. 所以，下限为 =31−2.58∗2365=23.6398=31-2.58*\frac{23}{\sqrt{65}}=23.6398=31−2.58∗65​23​=23.6398 ​为什么这个n用的65，但是上一题t分布里面n用的n37，查表用的 36呢？

2022-07-07 15:19 1 · 回答