问一道题：NO.PZ2015120604000147 [ CFA I ]

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.

为什么图中题目的原假设就是u=，但这一题原假设是跟consider的意思反过来了？？怎么区分？

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星星_品职助教 · 2020年06月29日

同学你好，

这道题的原假设其实无论从哪个方向建立都不会影响最后的结果。由于计算出来的t-statistics（1.91）＞t critical value（1.65），这代表着随意抽一个数结果就大于了均值26%，这已经可以说明假设检验的结果是均值要显著大于26%（significant greater ）了。

如果考虑如何建立原假设，根据题干中的描述“If an investor is considering whether the mean of bond B is greater than 26%”，需要把想证明的放在备择假设里，也就是Ha：μ＞26%，所以H0即为 μ≤26%。但如何建立原假设并不是这道题的考点。

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问一道题：NO.PZ2015120604000147 [ CFA I ]

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