问一道题：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：

13.23.

13.27.

13.68.

解释：

B is correct.

The confidence interval is calculated using the following equation: $\overline X\pm t_{\alpha/2}\frac s{\sqrt n}$ Sample standard deviation (s) = $\sqrt{245.55}$ = 15.670. For a sample size of 17, degrees of freedom equal 16, so t_0.05= 1.746. The confidence interval is calculated as

$116.23\text{ }\pm\text{ }1.746\frac{15.67}{\sqrt{17}}\text{ }=\text{ }116.23\text{ }\pm\text{ }6.6357$ 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).

老师，这道题的SAMPLE SIZE是17，应该是大于30，才服从T分布吧？

添加评论

1 个答案

星星_品职助教 · 2020年03月27日

同学你好，

是否服从t分布和sample size无关。N大于30只是t分布趋近于正态分布的条件。对于本题而言，题干中已经说明了要求用t分布，直接用t分布即可。

添加评论

1
回答
0
关注
1068
浏览

我要回答关注问题

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/2ns 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/2ns 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/2ns 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/2ns 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

2022-12-13 07:38 1 · 回答

问一道题：NO.PZ2017092702000113

1 个答案

1

0

1068

相关问题