问题如下图：所以这道题按照单尾来计算吗？我找的是0.05的？不应该双尾才找0.025吗？这道题怎么选的0.025看其他问题的解析还是不明白？
选项：
A. 
B. 
C. 
解释：
星星_品职助教 · 2019年12月01日
同学你好，
这道题考察的是置信区间的计算。置信“区间”肯定是分布中间的一个“区间”，所以分布的左右是有两个尾巴的，也就是95%置信区间对应的左右双尾的概率一共是5%。单尾对应的概率即为2.5%。
所以这道题题干中说明是单尾的t表（one-tailed），所以就相当于查2.5%对应的critical value。也就是对应1.96。但一旦要是这道题给的是two-tailed t值，就需要用5%对应的1.645了。
这个概念很重要，如果不明白可以继续追问哈，加油。
Junny · 2020年03月06日
老师你好，我还是没弄清楚，置信区间为什么可视同双尾的情况？还是说，我可以这么理解：因为原假设为b1=0，所以需要双尾检验？
星星_品职助教 · 2020年03月06日
这道题考察的是置信区间，95%的置信区间就是分布正中间的面积有95%，左侧剩2.5%，右侧也剩2.5%。所以要查置信区间对应的t临界值就要看双尾的t表的5%，或者单尾的t表的2.5%；假设检验是另一个考点，如果原假设是b1=0，那么拒绝原假设就应该是b1≠0，在分布左侧远小于0和分布右侧远大于0都是不等于0的情况，所以这种情况下拒绝域也是双尾的。查t表方法和之前说的一样。
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。