I'm also confused.  When B says to compute the test statistic for the Null Hypothesis, do we use the formula for t0 given on the bottom of page 104?  Or do we lookup the t-value using the degrees of freedom and the alpha / 2?  What is the t-value we want to be using for the confidence interval in C?  Is it -2.6 or is it 3.18?

Could you explain how to calculate or read the p-value. I am unable to get this answer from the tables, however I have been able to calculate it in Excel. Since we won't have excel during the exam, some explanation would be great. Thanks

[NEAS: On the exam, you may be asked to define the p-value, not to calculate it (which you can’t do by hand). The p-value is the probability of getting the observed value (or a more extreme value) if the null hypothesis is true and the assumptions of classical regression analysis hold.]

to help you out on part B, though it is 3 weeks later, I just did it, yes you use the formula on the bottom of page 104 with B0=0.  You should have B=-.9 then use SE(B)=Root(.00012/.001)=Root(SE^2/(SUM(x-Xbar)^2))

How do we calculate t_.025?

Here are my calculations. Let me know if you agree. I'm not sure if I used the TDIST() function correctly to solve for p.

A. Var(B) = SE^2/sum(xi-xbar)^2 = 0.00012/0.001 = 0.12
B. t = (B-0)/SE(B) = -0.9/0.12^0.5 = -2.598
C. t_0.025 = tinv(.05,3) = 3.182. beta = B +- t_0.025*SE(B) = -0.9 +- 3.182*0.12^0.5 = -0.9 +- 1.102
D. p = TDIST(2.598,3,2) = 0.0805
E. Var(A) = SE^2/(n*sum(xi-xbar)^2 = .00012/.005 = 0.024
F. t = (A - 0.219)/SE(A) = (0.813-0.219)/0.024^0.5 = 3.384
G. alpha = 0.813 +- 3.182*0.024^0.5 = 0.813 +- 0.493
H. p = TDIST(3.384,3,2) = 0.043

oops i missed a factor in E.


E. Var(A) = sum(xi^2)*SE^2/(n*sum(xi-xbar)^2 = 2.179*0.00012/.005 = 0.052
F. t = (A - 0.219)/SE(A) = (0.813-0.219)/0.052^0.5 = 2.597
G. alpha = 0.813 +- 3.182*0.052^0.5 = 0.813 +- 0.728
H. p = TDIST(2.597,3,2) = 0.0806

I got the same answers. Interesting how the p-values for alpha and beta are so close.

This makes sense when Alpha = Beta(0). You're essentially assuming alpha is just a function of beta, and since every point on the regression line has the same standard error, then by default, alpha must have that same standard error.

I'm not sure how to show my work for part D, because I had to use a p-value calculator for t-test. Excel gave me a value of .08051 for this part, however I believe this is the two-tailed test that they're giving, correct? We would want the one-tailed test for this. Please confirm.

For those who are wondering, you just calculate the p-value from the t-value you got in part B. This would be with df=3. There are online calculators for this. The table won't get you a value close enough.

Yes! I'm not sure how to show my work for part D, because I had to use a p-value calculator for t-test. Excel gave me a value of .08051 for this part, however I believe this is the two-tailed test that they're giving, correct? We would want the one-tailed test for this.

[NEAS: Yes, this is a two tailed test. To "show work," just say that Excel gives this p-value.]