为什么是0.5?不是1?
问题如下图：
选项：
A. 
B. 
C. 
解释：
NO.PZ2015120204000015 问题如下 Baseon past research, Hansen selects the following inpennt variables to preIPO initireturns: Unrwriter rank = 1–10, where 10 is highest rankPre-offer priaustment (Expressea cimal) = (Offer pri– Initifiling price)/Initifiling priceOffer size ($ millions) = Shares sol× Offer priceFraction retaine(Expressea cimal) = Fraction of totcompany shares retaineinsirsHe also believes thfor ea1 percent increase in pre-offer priaustment, the initireturn will increase less th0.5 percent, holng other variables constant. Hansen wishes to test this hypothesis the 0.05 level of significance.Hansen collects a sample of 1,725 recent IPOs for his regression mol.\Hansen’s Regression Results pennt Variable: IPO InitiReturn (Expressein cimForm, i.e., 1% = 0.01)SelecteValues for the t-stribution ( = ∞)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.C 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. H0 =0.5，H0.5 的情况下，t critic不是与Ha的方向相反么，答案应该是rejeH0。这样的逻辑是什么地方出错了？
NO.PZ2015120204000015 问题如下 Baseon past research, Hansen selects the following inpennt variables to preIPO initireturns: Unrwriter rank = 1–10, where 10 is highest rankPre-offer priaustment (Expressea cimal) = (Offer pri– Initifiling price)/Initifiling priceOffer size ($ millions) = Shares sol× Offer priceFraction retaine(Expressea cimal) = Fraction of totcompany shares retaineinsirsHe also believes thfor ea1 percent increase in pre-offer priaustment, the initireturn will increase less th0.5 percent, holng other variables constant. Hansen wishes to test this hypothesis the 0.05 level of significance.Hansen collects a sample of 1,725 recent IPOs for his regression mol.\Hansen’s Regression Results pennt Variable: IPO InitiReturn (Expressein cimForm, i.e., 1% = 0.01)SelecteValues for the t-stribution ( = ∞)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.C 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. 题干表格里面的t-statistic代表的是什么呢？
NO.PZ2015120204000015 问题如下 Baseon past research, Hansen selects the following inpennt variables to preIPO initireturns: Unrwriter rank = 1–10, where 10 is highest rankPre-offer priaustment (Expressea cimal) = (Offer pri– Initifiling price)/Initifiling priceOffer size ($ millions) = Shares sol× Offer priceFraction retaine(Expressea cimal) = Fraction of totcompany shares retaineinsirsHe also believes thfor ea1 percent increase in pre-offer priaustment, the initireturn will increase less th0.5 percent, holng other variables constant. Hansen wishes to test this hypothesis the 0.05 level of significance.Hansen collects a sample of 1,725 recent IPOs for his regression mol.\Hansen’s Regression Results pennt Variable: IPO InitiReturn (Expressein cimForm, i.e., 1% = 0.01)SelecteValues for the t-stribution ( = ∞)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.C 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. 我就是看了这个提示，为了统一单位，以为应该把0.5%改成0.005计算的。
NO.PZ2015120204000015问题如下 Baseon past research, Hansen selects the following inpennt variables to preIPO initireturns: Unrwriter rank = 1–10, where 10 is highest rankPre-offer priaustment (Expressea cimal) = (Offer pri– Initifiling price)/Initifiling priceOffer size ($ millions) = Shares sol× Offer priceFraction retaine(Expressea cimal) = Fraction of totcompany shares retaineinsirsHe also believes thfor ea1 percent increase in pre-offer priaustment, the initireturn will increase less th0.5 percent, holng other variables constant. Hansen wishes to test this hypothesis the 0.05 level of significance.Hansen collects a sample of 1,725 recent IPOs for his regression mol.\Hansen’s Regression Results pennt Variable: IPO InitiReturn (Expressein cimForm, i.e., 1% = 0.01)SelecteValues for the t-stribution ( = ∞)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=0.5RejeH0B.H0: bj≥0.5Fail to rejeH0C.H0: bj≥0.5RejeH0C is correct.C 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.请问这个题右尾左尾怎么判断
NO.PZ2015120204000015 问题如下 Baseon past research, Hansen selects the following inpennt variables to preIPO initireturns: Unrwriter rank = 1–10, where 10 is highest rankPre-offer priaustment (Expressea cimal) = (Offer pri– Initifiling price)/Initifiling priceOffer size ($ millions) = Shares sol× Offer priceFraction retaine(Expressea cimal) = Fraction of totcompany shares retaineinsirsHe also believes thfor ea1 percent increase in pre-offer priaustment, the initireturn will increase less th0.5 percent, holng other variables constant. Hansen wishes to test this hypothesis the 0.05 level of significance.Hansen collects a sample of 1,725 recent IPOs for his regression mol.\Hansen’s Regression Results pennt Variable: IPO InitiReturn (Expressein cimForm, i.e., 1% = 0.01)SelecteValues for the t-stribution ( = ∞)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.C 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. The corretest statistic is: t = (0.435-0.50)/0.0202 = -3.22为什么是减去0.5，另外这个题目的t分布的值是怎么查找出来的，为什么是负数
