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:
Sample standard deviation (s) = = 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:
= 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