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Helen 🎈 · 2023年06月22日

老师,我不明白为啥求的是两边的?

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)。


是图里阴影部分吗?为啥

2 个答案

星星_品职助教 · 2023年06月26日

无论是什么题型,什么分布,最终都要看单尾面积是多少,从而得出对应的关键值。

在正态分布下:

90%置信区间相当于对应单尾面积为5%,关键值为±1.645;

95%置信区间相当于对应单尾面积为2.5%,关键值为±1.96;

99%置信区间相当于对应单尾面积为0.5%,对应关键值为±2.58;

遇到假设检验的题型,也可以按照上述原理去判断单尾面积和关键值。

t分布的关键值只能查表,但原理相同。

星星_品职助教 · 2023年06月22日

同学你好,

本题求的是90%置信区间的宽度,也就是画图中90%的中间的那部分,和两端阴影部分的尾部无关。

根据置信区间的公式计算出置信区间后(参考答案解析),用上端点值减下端点值,即可得出该置信区间的距离/宽度。

Helen 🎈 · 2023年06月23日

老师能不能详细解释下,为什么Z分布的90%的a/2直接对应的1.65。而t分布的要用1-90%再除以2,再查表。 题做多了我会算,但我有点稀里糊涂,你能不能画图解释下?其实对于这个显著性水平,置信度我还不能完全理解。

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