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胖婷肥周 · 2024年01月14日

完全没有看懂题目，为什么会选择FUND3和FUND4呢？

  1. CFA I
  2. Quantitative

NO.PZ2017092702000162

问题如下：

The following table shows the sample correlations between the monthly returns for four different mutual funds and the S&P 500. The correlations are based on 36 monthly observations. The funds are as follows:


Test the null hypothesis that each of these correlations, individually, is equal to zero against the alternative hypothesis that it is not equal to zero. Use a 5 percent significance level.





选项：

解释：

The critical t-value for n − 2 = 34 df, using a 5 percent significance level and a two-tailed test, is 2.032. First, take the smallest correlation in the table, the correlation between Fund 3 and Fund 4, and see if it is significantly different from zero. Accoding to the formula of correlaion t-test, its calculated t-value is t=1.903. This correlation is not significantly different from zero. If we take the next lowest correlation, between Fund 2 and Fund 3, this correlation of 0.4156 has a calculated t-value of 2.664. So this correlation is significantly different from zero at the 5 percent level of significance. All of the other correlations in the table (besides the 0.3102) are greater than 0.4156, so they too are significantly different from zero.

完全没有看懂题目，为什么会选择FUND3和FUND4呢？


看不出来最后是问什么的，是要把所有的5个基金，两两组合，分别看是否在拒绝域吗？要做10次吗？

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1 个答案

品职助教_七七 · 2024年01月15日

嗨，努力学习的PZer你好：


是要把所有的5个基金，两两组合，分别看是否在拒绝域吗？要做10次吗？------对，题目要求检验这5只基金彼此两两组合成的10个相关系数是否显著，理论上应该做10次，实际不用做那么多次。

为什么会选择FUND3和FUND4呢？--------如果|test statistic|＞|critical value|，就可以得到相关系数不等于0的结论。所以，应该从表格给出的相关系数中最小的开始计算test statistic。本题表格中的最小相关系数为Fund 3和Fund 4的0.3102。算出test statistic=1.9026。由于本题的关键值查表后为2.032，所以还不能拒绝等于0的原假设。

此后接着计算次小的相关系数，也就是Fund 2和Fund 3的r=0.4156。得到t-statistic=2.6643，由于这个数字大于了t critical value=2.032，所以此时就可以拒绝等于0的原假设。由于这个值就已经可以拒绝原假设了，所以剩下8个更大的都可以拒绝原假设，不用继续计算了。实际上只需要检验这两个，就可以得到所有10个相关系数所有的结果。



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