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Dang.D · 2022年10月01日

怎么判断他是双尾的呢?

NO.PZ2017092702000113

问题如下:

For a sample size of 37, 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.

8.5480.

B.

8.6970

C.

8.8456.

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 37, degrees of freedom equal 36, so t0.05 = 1.688.

The confidence interval is calculated as:


Therefore, the interval spans 120.5785 to 111.8815, meaning its width is equal to approximately 8.6970. (This interval can be alternatively calculated as 4.3485 × 2).

样本标准差的计算如下:

245.55\sqrt{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)。

题里没给出,是不默认双尾

1 个答案

星星_品职助教 · 2022年10月01日

同学你好,

数量科目中,所有的置信区间(confidence interval)都对应双尾情况。

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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)。 原版题中没给出表的话怎么计算出1.688,如果需要查表的话为什么不一起放表在题里面呢

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