求lower limit时是不是应该置信区间的范围除以2？-有问必答-品职教育专注CFA ESG FRM CPA 考研等财经培训课程

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/2nσ 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}}=23.639831−2.58×6523=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/2nσ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∗6523=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

2023-06-06 23:49 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/2nσ 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}}=23.639831−2.58×6523=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/2nσ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∗6523=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/2nσ 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}}=23.639831−2.58×6523=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/2nσ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∗6523=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/2nσ 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}}=23.639831−2.58×6523=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/2nσ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∗6523=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/2nσ 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}}=23.639831−2.58×6523=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/2nσ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∗6523=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

求lower limit时是不是应该置信区间的范围除以2？

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