Can somebody please respond to 55Dollar5DollarFootLongs post from 3 years ago. I am getting the exact same answers as him and I do not have a clue what I am doing wrong. I used the exact same equations and process as he did and got the exact same answers so I think I have the same issue. Please let me know what I am doing wrong.

[NEAS: That post has arithmetic errors. Recommendation: Use Excel or R to get the results in the homework assignments. For Excel, use the regression add in. For R, use the lm built-in function. In both R and Excel you can compute all the intermediate values as well, to follow the derivation using the formulas in the textbook.]

Jacob: When using the equation on page 106 and page 107, what is R²? Is R² the value from the Excel regression add-in using these two explanatory variables?



Rachel:

No, the R² in the equation on page 106 and page 107 is the correlation of the two explanatory variables, or the R² from a regression on one on the other. If one have more than two explanatory variables, it is the R² from a regression of the explanatory variable under consideration on all the other explanatory variables.



Illustration:

In this homework assignment, the R² from the regression on the response variable on the two explanatory variables is 0.8712. The correlation of the two explanatory variables is 0.6364. Use 0.6364 in the equation, not 0.8712.

In this homework assignment, the R² from the regression on the response variable on the two explanatory variables is 0.8712. The correlation of the two explanatory variables is 0.6364. Use 0.6364 in the equation, not 0.8712.

I am confused. On the HW it says to show the formulas and the computations. However, you say no need to do work by hand. So can I just give a list of answers to A-F from Excel, or do I need to show the manual work?

[NEAS: Just state the formula or computation that Excel does; you don't have to reproduce the mathematics.]

Why are we using a formula from P.107 for Module 10 when Module 10 covered pages 92-96?

Is anyone doing these by hand? In module 9, NEAS said

[NEAS: Computations can be done with Excel; no need to do any work by hand.]

I'm just using the regression tool for all of this. I interpreted the "no need to do any work by hand" meant that you can just plug the table into Excel and have it compute everything for you.

[NEAS: Homework can be done using Excel's regression add-in or any other computer package. You need to understand the equations to solve final exam problems.]

After reading all the posts, I think I really need some help.

I am getting different answers than everybody that has posted on here. Please tell me what I am doing wrong.

For part A

First I solve for a Ybar by taking the average of all ten Yi;s. Ybar = -0.1795
Then I solve for a TSS by subtracting the Ybar from all Yi's...squaring the ten results and then summing all ten results. TSS = 59.86964

[NEAS: The TSS is correct.]

To get the RegSS you need the fitted Y's. To get the fitted Y's you need the A, B1, & B2 (which I solve for in parts B through D). Assuming my parts B-D are correct here is how I calculated my RegSS: I used the equation Fitted Y = A + B1 * X1i + B2 * X2i. I plugged in all combinations of X1 & X2, subtracted the Ybar from all the fitted Y's, squared the results and added up all ten results. RegSS = 51.68978

[NEAS: RegSS is 52.16; RSS is 7.709.]

To get the RSS, I subtracted the RegSS from the TSS. RSS = 8.179862

To get the Rsquared I Used the equation RegSS/TSS Rsquared = 0.8633872
To get the R aka the Correlation, I take the square root for R squared. Correlation = 0.929178

[NEAS: R squared is 0.8712. the correlation needed for the standard errors is between the two independant variables, not the square root of the R sqaured.]

For Parts B-D I needed the following Variables
sum of (X1i * Yi) = 11.57
sum of (X2i * Yi) = -44.05
Sum of (X1i * X2i) = 355
Sum of X1i squared = 385
Sum of X2i squared = 385
Xbar = 5.5
Ybar = 5.5

For B1 & B2 I used the equations on the top of page 88

B1 = (11.57)(385) - (-44.05)(355)
(385)(385) - (355)(355)

B1 = 0.905054

B2 = (-44.05)(385) - (11.57)(355)
(385)(385) - (355)(355)

B2 = -0.94895

For A I also used the equation on the top of page 88

A = Ybar - B1 X1bar - B2 X2bar
-0.1795 - (0.905054)(5.5) - (-0.94895)(5.5)

A = 0.061905

What am I doing wrong?

Found my error, my A was wrong, I just recalculated it to find it should be .3393535.  Thanks a lot for your help, it was driving me crazy that I had a mistake somkewhere, at least I had it correct theoretically.  Thanks

I'm sorry, I lied, we have the same A. our approach was different (i prefer your way, though I can't match your RegSS; I used the formula on the bottom of page 88 to get my A.), but we got the same correlation of .6363.

As far as the SE(Bj), we're differing on the SE. My RSS is 7.709233, which yields an SE of 1.049437 (RSS / n-k-1), and that gives me my 0.1498 SE(Bj) result.

RSS = sum of Ei^2, where Ei = yi - yihat. And yihat = alpha + B1*xi1 + B2*xi2 (see bottom of page 86).

I hope that helps, amigo.

If you got a different A than me, can you tell me how you got it? i have my Regss and TSS shown in the values, which one differed? I have the same y bar and x bars as you.  and for SE(b1) can you show me the formula you used, I used yoru values but did not get your solution, I have all my work shown above, if you can see some kind of mistake, which I know i made since my answer differes from excels regression pack.

Hey there. For some reason, I got a different A than you. I got 0.339333.

is your ybar = -0.1795?
x1bar = x2bar = 5.5?
I matched your B1 and B2 results.

For E), my RSS is 7.709233, which gives me an SE of 1.049437. Using my SE and correlation (r) of .6364 gives me

SE(Bi) = .149781, for i =1,2