Yan · 2021年03月21日
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NO.PZ201709270100000506
问题如下：
6. Based on the regression output in Exhibit 3 and sales data in Exhibit 4, the forecasted value of quarterly sales for March 2016 for PoweredUP is closest to:
选项：
A.$4.193 billion.
B.$4.205 billion.
C.$4.231 billion.
解释：
C is correct. The quarterly sales for March 2016 is calculated as follows:
这里EXIBIT3的AUTOCORRELATION指的是什么？能具体说一下么？怎么这LAG4是不显著的，而上面EXIBIT2中的LAG4是显著的？
星星_品职助教 · 2021年03月21日
同学你好，
autocorrelation指的是残差项之间的相关性。具体而言，是滞后的残差项εt-1（一直到εt-4）和εt之间彼此的相关性。
Exhibit 3中，四项残差项对应的t-statistic的绝对值都小于t critical value的绝对值（根据Exhibit 3的notes可知t critical value=2.03），所以这四项都不能拒绝ρ=0的原假设，即这四项和εt的相关性都是0。也就是no autocorrelation。
而Exhibit 2中，第四项εt-4和εt的t-statistic的绝对值等于4.3111，大于了此时的t critical value2.02，所以是拒绝ρ=0的原假设，即ρ≠0。这就有了相关性，所以违背了no autocorrelation的假设。
Exhibit 2和3结果不同的原因是Exhibit 3对应的方程进行了修正，加入了一项，反应了出了季节性因素，所以再检测autocorrelation就没问题了。
NO.PZ201709270100000506 问题如下 6. Baseon the regression output in Exhibit 3 ansales ta in Exhibit 4, the forecastevalue of quarterly sales for Mar2016 for PowereP is closest to: A.$4.193 billion. B.$4.205 billion. C.$4.231 billion. C is correct. The quarterly sales for Mar2016 is calculatefollows:beginarraylln Salest-ln Salest-1=b0+b1(ln Salest-1-ln Salest-2)+b2(ln Salest−4-ln Salest-5)ln Salest−ln3.868=0.0092−0.1279(ln3.868−ln3.780)+0.7239(ln3.836−ln3.418)ln Salest=1.44251Salest=e1.44251=4.231begin{array}{l}\ln{\text{ Sales}}_\text{t}{\text{-ln Sales}}_\text{t-1}{\text{=b}}_\text{0}{\text{+b}}_\text{1}{\text{(ln Sales}}_\text{t-1}{\text{-ln Sales}}_\text{t-2}{\text{)+b}}_\text{2}\text{(ln Sales}_{t-4}^{}{\text{-ln Sales}}_\text{t-5}\text{)}\\\ln{\text{ Sales}}_\text{t}-\ln3.868=0.0092-0.1279(\ln3.868-\ln3.780)+0.7239(\ln3.836-\ln3.418)\\\ln{\text{ Sales}}_\text{t}=1.44251\\{\text{Sales}}_\text{t}=e^{1.44251}=4.231beginarraylln Salest-ln Salest-1=b0+b1(ln Salest-1-ln Salest-2)+b2(ln Salest−4-ln Salest-5)ln Salest−ln3.868=0.0092−0.1279(ln3.868−ln3.780)+0.7239(ln3.836−ln3.418)ln Salest=1.44251Salest=e1.44251=4.231 0.0092−0.1279(ln3.868−ln3.780)+0.7239(ln3.836−ln3.418)得出来不是0.089776吗 然后怎么算ln（Sales t）呢？
NO.PZ201709270100000506问题如下6. Baseon the regression output in Exhibit 3 ansales ta in Exhibit 4, the forecastevalue of quarterly sales for Mar2016 for PowereP is closest to: A.$4.193 billion.B.$4.205 billion.C.$4.231 billion.C is correct. The quarterly sales for Mar2016 is calculatefollows:beginarraylln Salest-ln Salest-1=b0+b1(ln Salest-1-ln Salest-2)+b2(ln Salest−4-ln Salest-5)ln Salest−ln3.868=0.0092−0.1279(ln3.868−ln3.780)+0.7239(ln3.836−ln3.418)ln Salest=1.44251Salest=e1.44251=4.231begin{array}{l}\ln{\text{ Sales}}_\text{t}{\text{-ln Sales}}_\text{t-1}{\text{=b}}_\text{0}{\text{+b}}_\text{1}{\text{(ln Sales}}_\text{t-1}{\text{-ln Sales}}_\text{t-2}{\text{)+b}}_\text{2}\text{(ln Sales}_{t-4}^{}{\text{-ln Sales}}_\text{t-5}\text{)}\\\ln{\text{ Sales}}_\text{t}-\ln3.868=0.0092-0.1279(\ln3.868-\ln3.780)+0.7239(\ln3.836-\ln3.418)\\\ln{\text{ Sales}}_\text{t}=1.44251\\{\text{Sales}}_\text{t}=e^{1.44251}=4.231beginarraylln Salest-ln Salest-1=b0+b1(ln Salest-1-ln Salest-2)+b2(ln Salest−4-ln Salest-5)ln Salest−ln3.868=0.0092−0.1279(ln3.868−ln3.780)+0.7239(ln3.836−ln3.418)ln Salest=1.44251Salest=e1.44251=4.231 为什么t-statics不合格的系数也会被放进公式，b0和b1的t检验就过不了啊
NO.PZ201709270100000506 问题如下 6. Baseon the regression output in Exhibit 3 ansales ta in Exhibit 4, the forecastevalue of quarterly sales for Mar2016 for PowereP is closest to: A.$4.193 billion. B.$4.205 billion. C.$4.231 billion. C is correct. The quarterly sales for Mar2016 is calculatefollows:beginarraylln Salest-ln Salest-1=b0+b1(ln Salest-1-ln Salest-2)+b2(ln Salest−4-ln Salest-5)ln Salest−ln3.868=0.0092−0.1279(ln3.868−ln3.780)+0.7239(ln3.836−ln3.418)ln Salest=1.44251Salest=e1.44251=4.231begin{array}{l}\ln{\text{ Sales}}_\text{t}{\text{-ln Sales}}_\text{t-1}{\text{=b}}_\text{0}{\text{+b}}_\text{1}{\text{(ln Sales}}_\text{t-1}{\text{-ln Sales}}_\text{t-2}{\text{)+b}}_\text{2}\text{(ln Sales}_{t-4}^{}{\text{-ln Sales}}_\text{t-5}\text{)}\\\ln{\text{ Sales}}_\text{t}-\ln3.868=0.0092-0.1279(\ln3.868-\ln3.780)+0.7239(\ln3.836-\ln3.418)\\\ln{\text{ Sales}}_\text{t}=1.44251\\{\text{Sales}}_\text{t}=e^{1.44251}=4.231beginarraylln Salest-ln Salest-1=b0+b1(ln Salest-1-ln Salest-2)+b2(ln Salest−4-ln Salest-5)ln Salest−ln3.868=0.0092−0.1279(ln3.868−ln3.780)+0.7239(ln3.836−ln3.418)ln Salest=1.44251Salest=e1.44251=4.231 ln Salest-1-ln Salest-2的t检验是-1.1252，小于criticvalue2.03，所以接受原假设，即系数等于零。
NO.PZ201709270100000506 问题如下 6. Baseon the regression output in Exhibit 3 ansales ta in Exhibit 4, the forecastevalue of quarterly sales for Mar2016 for PowereP is closest to: A.$4.193 billion. B.$4.205 billion. C.$4.231 billion. C is correct. The quarterly sales for Mar2016 is calculatefollows:beginarraylln Salest-ln Salest-1=b0+b1(ln Salest-1-ln Salest-2)+b2(ln Salest−4-ln Salest-5)ln Salest−ln3.868=0.0092−0.1279(ln3.868−ln3.780)+0.7239(ln3.836−ln3.418)ln Salest=1.44251Salest=e1.44251=4.231begin{array}{l}\ln{\text{ Sales}}_\text{t}{\text{-ln Sales}}_\text{t-1}{\text{=b}}_\text{0}{\text{+b}}_\text{1}{\text{(ln Sales}}_\text{t-1}{\text{-ln Sales}}_\text{t-2}{\text{)+b}}_\text{2}\text{(ln Sales}_{t-4}^{}{\text{-ln Sales}}_\text{t-5}\text{)}\\\ln{\text{ Sales}}_\text{t}-\ln3.868=0.0092-0.1279(\ln3.868-\ln3.780)+0.7239(\ln3.836-\ln3.418)\\\ln{\text{ Sales}}_\text{t}=1.44251\\{\text{Sales}}_\text{t}=e^{1.44251}=4.231beginarraylln Salest-ln Salest-1=b0+b1(ln Salest-1-ln Salest-2)+b2(ln Salest−4-ln Salest-5)ln Salest−ln3.868=0.0092−0.1279(ln3.868−ln3.780)+0.7239(ln3.836−ln3.418)ln Salest=1.44251Salest=e1.44251=4.231 ln3.868−ln3.780 用计算器怎么按 算式列对了，一直得不到正确答案 ， 是不是和计算器设置有关？
NO.PZ201709270100000506 问题如下 6. Baseon the regression output in Exhibit 3 ansales ta in Exhibit 4, the forecastevalue of quarterly sales for Mar2016 for PowereP is closest to: A.$4.193 billion. B.$4.205 billion. C.$4.231 billion. C is correct. The quarterly sales for Mar2016 is calculatefollows:beginarraylln Salest-ln Salest-1=b0+b1(ln Salest-1-ln Salest-2)+b2(ln Salest−4-ln Salest-5)ln Salest−ln3.868=0.0092−0.1279(ln3.868−ln3.780)+0.7239(ln3.836−ln3.418)ln Salest=1.44251Salest=e1.44251=4.231begin{array}{l}\ln{\text{ Sales}}_\text{t}{\text{-ln Sales}}_\text{t-1}{\text{=b}}_\text{0}{\text{+b}}_\text{1}{\text{(ln Sales}}_\text{t-1}{\text{-ln Sales}}_\text{t-2}{\text{)+b}}_\text{2}\text{(ln Sales}_{t-4}^{}{\text{-ln Sales}}_\text{t-5}\text{)}\\\ln{\text{ Sales}}_\text{t}-\ln3.868=0.0092-0.1279(\ln3.868-\ln3.780)+0.7239(\ln3.836-\ln3.418)\\\ln{\text{ Sales}}_\text{t}=1.44251\\{\text{Sales}}_\text{t}=e^{1.44251}=4.231beginarraylln Salest-ln Salest-1=b0+b1(ln Salest-1-ln Salest-2)+b2(ln Salest−4-ln Salest-5)ln Salest−ln3.868=0.0092−0.1279(ln3.868−ln3.780)+0.7239(ln3.836−ln3.418)ln Salest=1.44251Salest=e1.44251=4.231 在经典题36页明确标明了 用这个mol : “ ln Salest-ln Salest-1=b0+b1(ln Salest-1-ln Salest-2)+b2(ln Salest−4-ln Salest-5)”但是课后题这边是“ ln Salest-ln Salest-1=b0+b1(ln Salest-1-ln Salest-2)” ，所以我的问题是，哪一边的出错了？ 考试时，会按照哪种来？
