问一道题：NO.PZ2015120604000145 [ 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 A is equal to 22%,

which of the following conclusion is least appropriate (significant level=1%) ?

选项：

The null hypothesis can be rejected.

It is appropriate to use a two-tailed t-test.

The test statistic value is 1.333.

解释：

A is correct.

The null hypothesis: $H_{0} : μ = 22 %.$

Because the sample size is 25, which is less than 30, so it is appropriate to use the two-tailed t-test.

$t_{24} = \frac{x - μ_{0}}{\frac{s}{\sqrt{n}}} = \frac{0.26 - 0.22}{\frac{0.15}{\sqrt{25}}} = 1.333$

±t₂₄ at α= 0.01= ±2.797;

Because -2.797 <1.333<+2.797, therefore, H₀ cannot be rejected.

为什么不是2.58？题目不是说符合正态分布？

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1 个答案

星星_品职助教 · 2021年03月14日

同学你好，

根据题干，这道题符合正态分布的是“two bonds rate of return”，这是“总体”服从正态分布的概念。

而假设检验看的是检验统计量的分布，在这道题里就是bond A return的样本均值的分布。并不是bond A return自身（也就是总体）的分布。

由于本题总体方差未知，所以检验统计量服从的是t分布。

如果sample size n＞30，可以选择用z-test来近似替代t-test，但本题bond A也就是sample 1的sample size=25。所以不能用z-test。

t分布的关键值是需要根据自由度（df=25-1=24）和significance level（1%）查表求得的。即本题中的±2.797。

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问一道题：NO.PZ2015120604000145 [ CFA I ]

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