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杨甜甜 · 2020年02月17日

问一道题：NO.PZ201512020300000303

  1. CFA II
  2. Quantitative

* 问题详情，请 查看题干

问题如下：

3.The most appropriate null hypothesis and the most appropriate conclusion regarding Hansen’s belief about the magnitude of the initial return relative to that of the pre-offer price adjustment (reflected by the coefficient bj) are:

选项：

Null Hypothesis
Conclusion about bj(0.05 Level of Significance)
A.
H0: bj=0.5
Reject H0
B.
H0: bj≥0.5
Fail to reject H0
C.
H0: bj≥0.5
Reject H0

解释：

C is correct.

To test Hansen’s belief about the direction and magnitude of the initial return, the test should be a one-tailed test. The alternative hypothesis is H1: bj<0.5b_j<0.5, and the null hypothesis is H0: bj≥0.5b_j\geq0.5. The correct test statistic is: t = (0.435-0.50)/0.0202 = -3.22, and the critical value of the t-statistic for a one-tailed test at the 0.05 level is -1.645. The test statistic is significant, and the null hypothesis can be rejected at the 0.05 level of significance.

老师，你好，H0不是我们想要拒绝的吗？题里面说increase less than0.5，想要拒绝不应该是<0.5吗？

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1 个答案

星星_品职助教 · 2020年02月18日

同学你好，

这种题目用排除法更好一些。

把想拒绝的放在原假设确实是设立H0的思想，但二级一般不这么考了。这种题型一般是只给三个原假设，分别是大于小于和等于0，然后问用哪个原假设。

本题更多想考察的还是是否可以拒绝原假设

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NO.PZ201512020300000303 问题如下 3.The most appropriate null hypothesis anthe most appropriate conclusion regarng Hansen’s belief about the magnitu of the initireturn relative to thof the pre-offer priaustment (reflectethe coefficient bj) are: Null HypothesisConclusion about bj(0.05 Level of Significance) A.H0: bj=0.5RejeH0 B.H0: bj≥0.5Fail to rejeH0 C.H0: bj≥0.5RejeH0 C is correct.To test Hansen’s belief about the rection anmagnitu of the initireturn, the test shoula one-tailetest. The alternative hypothesis is H1: 0.5b_j 0.5bj​ 0.5, anthe null hypothesis is H0: bj≥0.5b_j\geq0.5bj​≥0.5. The corretest statistic is: t = (0.435-0.50)/0.0202 = -3.22, anthe criticvalue of the t-statistic for a one-tailetest the 0.05 level is -1.645. The test statistic is significant, anthe null hypothesis crejectethe 0.05 level of significance. 我计算的Sbj=1/(1725^0.5)=0.024077

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