本题未var未知，是因为样本量大于30，所以用Z分布也可以，是么？

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：

It is approptiate to use the Z-test.

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

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: $H_{0} : μ$ ≤ $26%$

The test statistic: $Z = \frac{X - μ}{\frac{S_{x}}{\sqrt{n}}}$ $= \frac{(0.32 - 0.26)}{\frac{0.22}{\sqrt{49}}} = 1.91$

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

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

本题未var未知，是因为样本量大于30，所以用Z分布也可以，是么？

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星星_品职助教 · 2021年03月14日

同学你好，

对的。

总体方差未知背景下，最准确的检验应该是t-test。但因为本题的n很大（n=49＞30），所以也是可以用z检验替代t检验的。

但如果选项中同时存在t检验和z检验，那还是要选最适合的t检验的

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本题未var未知，是因为样本量大于30，所以用Z分布也可以，是么？

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