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 · 2021年11月28日

置信区间

  1. CFA I
  2. Quantitative

NO.PZ2017092702000113

问题如下：

For a sample size of 17, with a mean of 116.23 and a variance of 245.55, the width of a 90% confidence interval using the appropriate t-distribution is closest to:

选项：

A.

13.23.

B.

13.27.

C.

13.68.

解释：

B is correct.

The confidence interval is calculated using the following equation:X‾±tα/2sn\overline X\pm t_{\alpha/2}\frac s{\sqrt n}

Sample standard deviation (s) = 245.55\sqrt{245.55} = 15.670.

For a sample size of 17, degrees of freedom equal 16, so t0.05 = 1.746.

The confidence interval is calculated as：


Therefore, the interval spans 109.5943 to 122.8656, meaning its width is equal to approximately 13.271. (This interval can be alternatively calculated as 6.6357 × 2).

样本标准差的计算如下：

245.55\sqrt{245.55} = 15.670.

当样本=17，自由度=16，那么 t0.05 = 1.746.

自信区间计算如下：


因此，自信区间为：109.5943 - 122.8656，这意味着其宽度大约等于 13.271。 （这个区间也可以计算为 6.6357 × 2）。

老师，最后的公式用的是置信区间，但是置信区间不是+/-1.746S么，怎么成了标准差/n根号

添加评论
4 个答案
已采纳答案

星星_品职助教 · 2021年11月29日

同学你好，

通过题干描述可知这是在求一个sample mean的置信区间（For a sample size of 17, with a mean of.....）。所以对应的标准差也是sample mean的标准差，即标准误。

根据中心极限定理，sample mean的标准误为总体标准差σ除以根号n。

当总体标准差σ未知时，可用sample standard deviation s进行替代。

所以最终对应的标准差就是 s除以根号n。



卓娅 · 2021年12月01日

谢谢老师

星星_品职助教 · 2022年09月01日

@琳 已经在单独的提问中回复，同样的问题提问一次即可。

琳 · 2022年09月01日

请问老师 置信区间，置信水平，置信度分别代表什么，能用例子或者图示解答吗？谢谢

星星_品职助教 · 2021年12月01日

@卓娅 好的，学习加油~

  • 4

    回答

  • 0

    关注

  • 539

    浏览
相关问题

NO.PZ2017092702000113 问题如下 For a sample size of 37, with a meof 116.23 ana varianof 245.55, the wih of a 90% confinintervusing the appropriate t-stribution is closest to: A.8.5480. B.8.6970 C.8.8456. B is correct. The confinintervis calculateusing the following equation:X‾±tα/2sn\overline X\pm t_{\alpha/2}\frs{\sqrt n}X±tα/2​n​s​ Sample stanrviation (s) = 245.55\sqrt{245.55}245.55​ = 15.670. For a sample size of 37, grees of freem equ36, so t0.05 = 1.688. The confinintervis calculateasTherefore, the intervspans 120.5785 to 111.8815, meaning its wih is equto approximately 8.6970. (This intervcalternatively calculate4.3485 × 2). 样本标准差的计算如下245.55\sqrt{245.55}245.55​ = 15.670. 当样本=37，自由度=36，那么 t0.05 = 1.688. 置信区间计算如下因此，置信区间为111.8815 - 120.5785，这意味着其宽度大约等于 8.6970。 （这个区间也可以计算为 4.3485 × 2）。 为什么解答过程中提到了自由度36，是用来做什么的呢

2023-11-11 14:52 1 · 回答

NO.PZ2017092702000113问题如下 For a sample size of 37, with a meof 116.23 ana varianof 245.55, the wih of a 90% confinintervusing the appropriate t-stribution is closest to: A.8.5480.B.8.6970C.8.8456. B is correct. The confinintervis calculateusing the following equation:X‾±tα/2sn\overline X\pm t_{\alpha/2}\frs{\sqrt n}X±tα/2​n​s​ Sample stanrviation (s) = 245.55\sqrt{245.55}245.55​ = 15.670. For a sample size of 37, grees of freem equ36, so t0.05 = 1.688. The confinintervis calculateasTherefore, the intervspans 120.5785 to 111.8815, meaning its wih is equto approximately 8.6970. (This intervcalternatively calculate4.3485 × 2). 样本标准差的计算如下245.55\sqrt{245.55}245.55​ = 15.670. 当样本=37，自由度=36，那么 t0.05 = 1.688. 置信区间计算如下因此，置信区间为111.8815 - 120.5785，这意味着其宽度大约等于 8.6970。 （这个区间也可以计算为 4.3485 × 2）。 百分之90，为什么不能是1.65呢

2023-09-12 11:09 1 · 回答

NO.PZ2017092702000113问题如下 For a sample size of 37, with a meof 116.23 ana varianof 245.55, the wih of a 90% confinintervusing the appropriate t-stribution is closest to: A.8.5480.B.8.6970C.8.8456. B is correct. The confinintervis calculateusing the following equation:X‾±tα/2sn\overline X\pm t_{\alpha/2}\frs{\sqrt n}X±tα/2​n​s​ Sample stanrviation (s) = 245.55\sqrt{245.55}245.55​ = 15.670. For a sample size of 37, grees of freem equ36, so t0.05 = 1.688. The confinintervis calculateasTherefore, the intervspans 120.5785 to 111.8815, meaning its wih is equto approximately 8.6970. (This intervcalternatively calculate4.3485 × 2). 样本标准差的计算如下245.55\sqrt{245.55}245.55​ = 15.670. 当样本=37，自由度=36，那么 t0.05 = 1.688. 置信区间计算如下因此，置信区间为111.8815 - 120.5785，这意味着其宽度大约等于 8.6970。 （这个区间也可以计算为 4.3485 × 2）。 是图里阴影部分吗？为啥

2023-06-22 09:52 2 · 回答

NO.PZ2017092702000113 问题如下 For a sample size of 37, with a meof 116.23 ana varianof 245.55, the wih of a 90% confinintervusing the appropriate t-stribution is closest to: A.8.5480. B.8.6970 C.8.8456. B is correct. The confinintervis calculateusing the following equation:X‾±tα/2sn\overline X\pm t_{\alpha/2}\frs{\sqrt n}X±tα/2​n​s​ Sample stanrviation (s) = 245.55\sqrt{245.55}245.55​ = 15.670. For a sample size of 37, grees of freem equ36, so t0.05 = 1.688. The confinintervis calculateasTherefore, the intervspans 120.5785 to 111.8815, meaning its wih is equto approximately 8.6970. (This intervcalternatively calculate4.3485 × 2). 样本标准差的计算如下245.55\sqrt{245.55}245.55​ = 15.670. 当样本=37，自由度=36，那么 t0.05 = 1.688. 置信区间计算如下因此，置信区间为111.8815 - 120.5785，这意味着其宽度大约等于 8.6970。 （这个区间也可以计算为 4.3485 × 2）。 这个题目没有给出T分布的表格，怎么算出来呢

2023-06-04 11:40 1 · 回答

NO.PZ2017092702000113 问题如下 For a sample size of 37, with a meof 116.23 ana varianof 245.55, the wih of a 90% confinintervusing the appropriate t-stribution is closest to: A.8.5480. B.8.6970 C.8.8456. B is correct. The confinintervis calculateusing the following equation:X‾±tα/2sn\overline X\pm t_{\alpha/2}\frs{\sqrt n}X±tα/2​n​s​ Sample stanrviation (s) = 245.55\sqrt{245.55}245.55​ = 15.670. For a sample size of 37, grees of freem equ36, so t0.05 = 1.688. The confinintervis calculateasTherefore, the intervspans 120.5785 to 111.8815, meaning its wih is equto approximately 8.6970. (This intervcalternatively calculate4.3485 × 2). 样本标准差的计算如下245.55\sqrt{245.55}245.55​ = 15.670. 当样本=37，自由度=36，那么 t0.05 = 1.688. 置信区间计算如下因此，置信区间为111.8815 - 120.5785，这意味着其宽度大约等于 8.6970。 （这个区间也可以计算为 4.3485 × 2）。 为什么p等于0.05

2022-12-13 07:38 1 · 回答