问题如下图：
选项：
A. 
B. 
C. 
为什么0.5 percent, 直接带入公式，而不是转化为0.005 in decimal再带入？ 解释：
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
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？
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
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？
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\"
