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lele1234 · 2024年07月09日

Sbj=0.0202怎么算的

  1. CFA II
  2. Quantitative

* 问题详情，请 查看题干

NO.PZ201512020300000303

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.

我计算的Sbj=1/(1725^0.5)=0.024077

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1 个答案

品职助教_七七 · 2024年07月09日

嗨，爱思考的PZer你好：


标准误为题目表格直接给出，不需计算。

1/√n是AR模型中序列相关t检验的标准误公式，和本题无关。

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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. 请问 这一题计算出来的检验统计量-3.22，为什么不和表里Pre-offer priaustment的检验结果21.53作对比？

2024-01-31 16:01 1 · 回答

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. 总是把h0和ha的设定有点搞混，不清楚这道题为什么要设定h0为bj大于等于0.5，结论是这样得出来的

2023-04-27 00:12 2 · 回答

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. criticvalue为什么是负的1.645，而不是1.645？如果是双尾检验，95%的置信区间，criticvalue是否等于1.96？

2023-01-23 21:09 2 · 回答

NO.PZ201512020300000303 老师好 这里T st为啥不是 减去 0.005 虽然题目里说 0.5 percent 也有提到用in cimal. \" corretest statistic is: t = (0.435-0.50)/0.0202 = -3.22\"

2021-11-11 05:26 2 · 回答