When solving for SE(A) at the top of page 112 in Fox's textbook, there is a sqrt(329,731) in the numerator. Where did this number come from? I can't find the value in the original Module 8 material, and I can't find the original data to test different values out.

Also, when solving for test statistic with B sub 0, how did you calculate the B in the numerator of the formula t sub 0 = (B-B sub 0)/SE(B). The formula for m sub i isn't exactly clear. Is this even the correct test statistic that we are supposed to be calculating?

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.]

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.

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

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

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

How do we calculate t_.025?

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))

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.]