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NO.PZ202001080100001302

问题如下：

b. What is the p-value of the t-statistic of $\beta$ ?

选项：

解释：

The range of the confidence interval is $2c * se(\widehat\beta)$ , where c is the critical value used to construct the 90% confidence interval (5% in each tail, so 1.645). The estimate of $\widehat\beta$ is the midpoint of the confidence interval so 1.105.

Because $\widehat\mu - 1.645se(\widehat\beta) = 0.32$ and $\widehat\mu + 1.645se(\widehat\beta) = 1.89$ , these can be solved for $se(\widehat\beta) = .477$ . The t-stat is then $\widehat\beta / se(\widehat\beta) = 2.32$ .

Finally, the two-sided p-value is2(1 - Φ( |t|)) = 2.06%, where Φ( . ) is the standard normal CDF function. This p-value confirms the rejection in the previous step because it is less than 10%.

没看懂第二个问题的答案，辛苦老师用中文解释一下，另外，这个计算是考试要求的么

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袁园_品职助教 · 2021年04月11日

同学你好！

因为置信区间是 b的期望±分位数*标准误，所以置信区间的中点就是b的期望，再算出b的标准差

求 t 统计量，根据 t 求 p-value

掌握计算原理比较重要，如果考试要计算会给你表格让你查表的

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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

2022-05-03 07:28 1 · 回答

NO.PZ202001080100001302老师，这个P-value是值怎么求的呢？

2022-03-30 21:26 2 · 回答

估计值beta为什么是ci的mioint？

2020-02-15 23:59 1 · 回答

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