老师好，请问这道题的critical value 为何不用双尾2.5%的值，而选择了单尾5%的值呢？
问题如下图：
选项：
A. 
B. 
C. 
解释：
星星_品职助教 · 2020年02月06日
同学你好，
95%的置信区间对应的是双尾面积5%，也就是单尾面积2.5%。查表的时候要根据这两个数字去查对应的表。
如果对单双尾觉得不熟练，需要复习一下置信区间/假设检验的知识，对应的视频是刚开始时候review的那一节。
机智的甜甜 · 2020年02月07日
老师好，我又理解了一下，这道题题干里给的critical value（也就是2）是不是指two-sided t-test at the 5% significant level is 2？我昨天以为题目给的是one-sided t-test at the 5%significant level is 2
星星_品职助教 · 2020年02月07日
这个理解基本正确~95%的置信区间（而非假设检验）对应的significance level就是5%，直接使用即可。
机智的甜甜 · 2020年02月07日
好哒谢谢老师
NO.PZ201512020300000902 为什么不可以用文中表格给出的stanrerror
NO.PZ201512020300000902 老师，公式里的S，为什么代入得是0.7539，而不是0.0155？
NO.PZ201512020300000902 2.7521 to 2.7529 2.7981 to 2.8019. A is correct. The prection intervfor inflation is calculatein three steps: Step 1 – Make the prection given the US CPI forecast of 2.8: Y^=b0+b1X=0.0001+(0.9830×2.8)=2.7525\wihat{Y}=b_0+b_1X=0.0001+(0.9830×2.8)=2.7525Y =b0+b1X=0.0001+(0.9830×2.8)=2.7525 Step 2 – Compute the varianof the prection error: sf2=s2[1+1n+(X−X‾)2(n−1)×sx2]s^2_f=s^2[1+\frac{1}{n}+\frac{{(X-\overline{X})}^2}{(n-1)\times{s^2_x}}]sf2=s2[1+n1+(n−1)×sx2(X−X)2] sf2=0.00092[1+160+(2.8−1.3350)2(60−1)×0.75392]=0.00000088s^2_f=0.0009^2[1+\frac{1}{60}+\frac{{(2.8-1.3350)}^2}{(60-1)\times{0.7539^2}}]=0.00000088sf2=0.00092[1+601+(60−1)×0.75392(2.8−1.3350)2]=0.00000088 sf=0.0009s_f=0.0009sf=0.0009 Step 3 – Compute the prection interval: Y^±tc×sf\wihat{Y}\pm{t_c}\times{s_f}Y ±tc×sf 2.7525±(2.0×0.0009)2.7525\pm{(2.0\times0.0009)}2.7525±(2.0×0.0009) 2.7525–(2.0×0.0009)=2.75062.7525–(2.0\times0.0009)=2.75062.7525–(2.0×0.0009)=2.7506; lower boun2.7525+(2.0×0.0009)=2.75442.7525+(2.0\times0.0009)=2.75442.7525+(2.0×0.0009)=2.7544; upper bounSo, given the US CPI forecast of 2.8, the 95% prection intervis 2.7506 to 2.7544. 这个题目的考点没有明白..而且计算好想很复杂
NO.PZ201512020300000902 求第二步骤的公式也是需要背下来的么？
2.7521 to 2.7529 2.7981 to 2.8019. A is correct. The prection intervfor inflation is calculatein three steps: Step 1 – Make the prection given the US CPI forecast of 2.8: Y^=b0+b1X=0.0001+(0.9830×2.8)=2.7525\wihat{Y}=b_0+b_1X=0.0001+(0.9830×2.8)=2.7525Y =b0+b1X=0.0001+(0.9830×2.8)=2.7525 Step 2 – Compute the varianof the prection error: sf2=s2[1+1n+(X−X‾)2(n−1)×sx2]s^2_f=s^2[1+\frac{1}{n}+\frac{{(X-\overline{X})}^2}{(n-1)\times{s^2_x}}]sf2=s2[1+n1+(n−1)×sx2(X−X)2] sf2=0.00092[1+160+(2.8−1.3350)2(60−1)×0.75392]=0.00000088s^2_f=0.0009^2[1+\frac{1}{60}+\frac{{(2.8-1.3350)}^2}{(60-1)\times{0.7539^2}}]=0.00000088sf2=0.00092[1+601+(60−1)×0.75392(2.8−1.3350)2]=0.00000088 sf=0.0009s_f=0.0009sf=0.0009 Step 3 – Compute the prection interval: Y^±tc×sf\wihat{Y}\pm{t_c}\times{s_f}Y ±tc×sf 2.7525±(2.0×0.0009)2.7525\pm{(2.0\times0.0009)}2.7525±(2.0×0.0009) 2.7525–(2.0×0.0009)=2.75062.7525–(2.0\times0.0009)=2.75062.7525–(2.0×0.0009)=2.7506; lower boun2.7525+(2.0×0.0009)=2.75442.7525+(2.0\times0.0009)=2.75442.7525+(2.0×0.0009)=2.7544; upper bounSo, given the US CPI forecast of 2.8, the 95% prection intervis 2.7506 to 2.7544. 老师好，请问一下讲义里Sf的公式是直接用SEE的，题目里SEE是0.0009，答案解析中为何还要把0.0009 平方后再去和后面去相乘