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@_@ · 2021年06月06日

老师，总体方差未知，T分布条件足够的情况下用T分布，如果条件不足且样本容量大于30，才用z分布，请问b选项为什么用z分布？

  1. CFA I
  2. Quantitative

NO.PZ2015120604000147

问题如下：

Here is a table discribing sample statistics from two bonds' rate of return which are both normally distributed over the past decades. If an investor is considering whether the mean of bond B is greater than 26%,

which of the following conclusion is least appropriate (significant level=5%) ?

选项：

A.

It is approptiate to use the Z-test.

B.

The mean of bond B is not significant greater than 26%.

C.

It is a one-tailed test.

解释：

B is correct.

It is appropraite to ues the one-tailed Z-test because the sample size of bond B is bigger than 30.

The null hypothesis is: H0: μ ≤ 26%

The test statistic:Z=X-μSxn =(0.32-0.26)0.2249=1.91

 α= 5% indicates the critical value is equal to ±1.65.

Because 1.91>1.65, therefore, we should reject the null hypothesis.

老师，总体方差未知，T分布条件足够的情况下用T分布，如果条件不足且样本容量大于30，才用z分布，请问b选项为什么用z分布？

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已采纳答案

星星_品职助教 · 2021年06月06日

同学你好，

理解无误，B选项最合适的方法是用t分布。这里z分布只是一个近似的替代。

严谨的方法是应该用t分布计算t统计量，然后查表t关键值做出判断。

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从解题的角度来说，

总体方差未知时，对于只是判断“是否拒绝原假设”的问题，用z替代t去判断比较简单。这是因为n＞30的时候z分布和t分布相差不大，而z的关键值背过不需要花时间查表。

这种题目最合适的分布还是t，但用z不会影响最终的判断结果。

但对于精细计算的题目，例如计算t分布下的置信区间，这个时候必须用t分布，用z分布替代往往是算不出答案的。


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