(c)(5 pts)Based on the estimates calculated in parts (a) and (b), it turns out that∑4i=1ˆe2i= 0.1220. Use this information and the data provided to calculate\var(b2).The formula for\var(b2)is:\var(b2) =ˆσ2∑Ni=1(Ti-T)2You have already computed∑Ni=1(Ti-T)2= 41. To computeˆσ2, use the informationyou have just been given (∑4i=1ˆe2i= 0.1220) and the formula forˆσ2:ˆσ2=∑4i=1ˆe2iN-2=0.12202= 0.061So:\var(b2) =ˆσ2∑Ni=1(Ti-T)2=0.06141= 0.001488(d)(5 pts)Perform a two-tail test of significance ofβ2at the 5% level. What do youconclude?You are testingH0:β2= 0againstH1:β26= 0. Your test statistic is:t=b2-0se(b2).Because this is a two-sided test at the 5% level, you will rejectH0ift > t0.975,N-2ort <-t0.975,N-2. Recall thatN-2 = 2. Reading from the t-table provided, noticethatt0.975,2= 4.303. So, you will rejectH0ift >4.303ort <-4.303. To computeyour test statistic, you need to find the standard error ofb2, which is just equal toq\var(b2) =√0.001488 = 0.0386. So, the value of your test statistic is:t=b2se(b2)=0.25370.0386= 6.5762This is greater than 4.303, so it is in the rejection region. Thus, you rejectH0. Youconclude thatβ2is significant at the 5% level.