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ditto · 2023年11月11日

这道题和自由度有什么关系呢

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)。

为什么解答过程中提到了自由度36,是用来做什么的呢

1 个答案
已采纳答案

星星_品职助教 · 2023年11月11日

同学你好,

t分布的置信区间公式如下,其中的 tα/2 为t分布的关键值(reliability factor)。该值需要查表。查表时就会用到自由度(df=n-1即37-1=36)。

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