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扑扑扑 · 2020年08月02日

问一道题：NO.PZ2015120604000144

  1. CFA I
  2. Quantitative

问题如下：

Quintina conducts a hypothesis to test whether the variance of the annually return is 0.0144. Then she selects a  random  25 year's data as a sample, Quintina figures out the variance of the sample is 0.013924.Assume the level of significance is 5%, which of the following conclusion is least accurate?

选项：

A.

The population variance is not significant different from 0.0144.

B.

The population and sample variances are significantly different.

C.

This hypothesis is appropriate to use the chi-square test.

解释：

B is correct.

The null hypothesis is H0: σ2=0.0144. It is appropriate to use the two-tailed chi-square test.

The test statistic is:   χ 2 = (n−1) s 2 σ 0 2 ={(25-1)*0.013924}/0.0144= 23.2067

The critical chi-square critical values are 12.4 and 39.36.

Because the test statistic falls between these two values,so Quintina fails to reject the null hypothesis.

So we conclude that the population variance is not significantly different from 0.0144.

请问这题怎么算啊？检验统计量是带入这个公式吗？分子24*0.013924？分母是啥？另外能示范下怎么查表吗？老师上课没说

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已采纳答案

星星_品职助教 · 2020年08月04日

同学你好，

N=25，sample variance= 0.013924，要检验方差是不是等于一个确定的数（0.0144），所以选择卡方检验，C选项的描述正确。

代入卡方检验统计量公式（就是你截图那个），卡方统计量=（24*0.013924）/0.0144=23.2067，用这个值跟查表求得的卡方临界值相比就可以。

卡方查表以原版书后附表为例，找到自由度24对应的那一行。由于这是双尾检验（备择假设为σ0≠0.0144），所以拒绝域有两个，面积各为2.5%，卡方查表是看的是右尾概率面积，所以左尾的2.5%拒绝域对应的右尾面积为97.5%，对应查临界值为12.401；右尾拒绝域的右侧面积就是2.5%，对应临界值为39.364。

由于计算出来卡方检验统计量23.2067没有落在这两个拒绝域里，所以无法拒绝原假设，A选项描述正确，B选项描述错误。本题要求选 least accurate的一项，即B选项。

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