问题如下图：
估计值beta为什么是ci的midpoint？
NO.PZ202001080100001302问题如下Whis the p-value of the t-statistic of β\betaβ?The range of the confinintervis 2c∗se(β^)2c * se(\wihat\beta)2c∗se(β), where c is the criticvalue useto construthe 90% confininterv(5% in eatail, so 1.645). The estimate of β^\wihat\betaβ is the mioint of the confinintervso 1.105. Because μ^−1.645se(β^)=0.32\wihat\mu - 1.645se(\wihat\bet= 0.32μ−1.645se(β)=0.32 anμ^+1.645se(β^)=1.89\wihat\mu + 1.645se(\wihat\bet= 1.89μ+1.645se(β)=1.89, these csolvefor se(β^)=.477se(\wihat\bet= .477se(β)=.477. The t-stis then β^/se(β^)=2.32\wihat\beta / se(\wihat\bet= 2.32β/se(β)=2.32. Finally, the two-sip-value is2(1 - Φ( |t|)) = 2.06%, where Φ( . ) is the stanrnormC function. This p-value confirms the rejection in the previous step because it is less th10%.这个整个计算过程不理解，这是要算个啥
NO.PZ202001080100001302老师，这个P-value是值怎么求的呢？
NO.PZ202001080100001302 没看懂第二个问题的答案，辛苦老师用中文一下，另外，这个计算是考试要求的么
