为何判断p是0.025而不是0.05？
问题如下图：
选项：
A. 
B. 
C. 
解释：
NO.PZ2015120204000022 1.72 to 4.36. 1.93 to 4.15. B is correct. The confinintervis computea1±s(a1)×(95%,∞)a_1\pm s(a_1)\times(95\%,\infty)a1±s(a1)×(95%,∞). From Exhibit 1, = 3.04 ant(a1) = 4.52, resulting in a stanrerror of = s(a1) = 3.04/4.52 = 0.673. The criticvalue for t from Exhibit 3 is 1.96 for p = 0.025. The confinintervfor is 3.04 ± 0.673 × 1.96 = 3.04 ± 1.31908 or from 1.72092 to 4.35908.老师请问stanrerror为什么是3.04\4.52 ? 谢谢
NO.PZ2015120204000022问题如下lExcess stock market returnt=a0+a1fault sprea−1 +a2Term sprea−1 +a3Pres party mmyt−1 +e{l}Excess\text{ }stock\text{ }market\text{ }return_t\\=a_0+a_1fault\text{ }sprea{t-1}\text{ }+a_2Term\text{ }sprea{t-1}\text{ }+a_3Pres\text{ }party\text{ }mmy_{t-1}\text{ }+elExcess stock market returnt=a0+a1fault sprea−1 +a2Term sprea−1 +a3Pres party mmyt−1 +efault spreis equto the yielon Bbon minus the yielon Abon. Term spreis equto the yielon a 10-yeconstant-maturity US Treasury inx minus the yielon a 1-yeconstant-maturity US Treasury inx. Pres party mmy is equto 1 if the US Presint is a member of the mocratic Party an0 if a member of the RepublicParty.The regression is estimatewith 431 observations.Exhibit 1.Multiple Regression OutputExhibit 2. Table of the Stunt’s t-stribution (One-TaileProbabilities for = ∞)The 95 percent confinintervfor the regression coefficient for the fault spreis closest to:A.0.13 to 5.95.B.1.72 to 4.36.C.1.93 to 4.15.B is correct.The confinintervis computea1±s(a1)×(95%,∞)a_1\pm s(a_1)\times(95\%,\infty)a1±s(a1)×(95%,∞). From Exhibit 1, = 3.04 ant(a1) = 4.52, resulting in a stanrerror of = s(a1) = 3.04/4.52 = 0.673. The criticvalue for t from Exhibit 3 is 1.96 for p = 0.025. The confinintervfor is 3.04 ± 0.673 × 1.96 = 3.04 ± 1.31908 or from 1.72092 to 4.35908.题目显示为one-tail，则95%confinlevel应该对应的p是0.05、t是1.645，否则选0.025为双尾与表格不符，是否应为C
NO.PZ2015120204000022 1.72 to 4.36. 1.93 to 4.15. B is correct. The confinintervis computea1±s(a1)×(95%,∞)a_1\pm s(a_1)\times(95\%,\infty)a1±s(a1)×(95%,∞). From Exhibit 1, = 3.04 ant(a1) = 4.52, resulting in a stanrerror of = s(a1) = 3.04/4.52 = 0.673. The criticvalue for t from Exhibit 3 is 1.96 for p = 0.025. The confinintervfor is 3.04 ± 0.673 × 1.96 = 3.04 ± 1.31908 or from 1.72092 to 4.35908. 老师，请问有什么好一点的办法判断什么时候用单尾，什么时候用双尾吗？做题的时候纠结了一下该用5%（单尾）还是2.5%（双尾）对应的criticvalue
NO.PZ2015120204000022 具体对应的有可以背的数据么
NO.PZ2015120204000022 1.72 to 4.36. 1.93 to 4.15. B is correct. The confinintervis computea1±s(a1)×(95%,∞)a_1\pm s(a_1)\times(95\%,\infty)a1±s(a1)×(95%,∞). From Exhibit 1, = 3.04 ant(a1) = 4.52, resulting in a stanrerror of = s(a1) = 3.04/4.52 = 0.673. The criticvalue for t from Exhibit 3 is 1.96 for p = 0.025. The confinintervfor is 3.04 ± 0.673 × 1.96 = 3.04 ± 1.31908 or from 1.72092 to 4.35908. 表一中，P值得作用是什么呢？不太明白为什么要用3.04/t-sta值，来计算。
