1) Run the regression and verify that the R-squared is one.  Why does
this make sense? Explain your work.

Quarter	Sales ($)
1	1,000
2	1,100
3	1,200
4	1,300
5	1,400
6	1,500
7	1,600
8	1,700
9	1,800
10	1,900
11	2,000
12	2,100
13	2,200
14	2,300
15	2,400

Hi again k9queen, 

Running the regression on this data, calling the variables SALES and
QUARTER, yields the following TSP output:

Dependent variable: SALES
 Current sample:  1 to 15
 Number of observations:  15

        Mean of dep. var. = 1.70000       LM het. test = 5.45081 [.020]
   Std. dev. of dep. var. = .447214      Durbin-Watson = .038710 [<1.00]
 Sum of squared residuals = 2.74355   Jarque-Bera test = .916143 [.633]
    Variance of residuals = .195968    Ramsey's RESET2 = 21.9355 [.000]
 Std. error of regression = .442682     Schwarz B.I.C. = 9.89712
                R-squared = 1.000000    Log likelihood = -8.54309
       Adjusted R-squared = 1.000000

            Estimated    Standard
 Variable  Coefficient     Error       t-statistic   P-value
 QUARTER   .187097       .012571       14.8828       [.000]


This confirms that the R-squared is, indeed, 1.  This can also be
confirmed by computing the R-squared value by hand:

R^2 = RSS / TSS

      ?( Y[hat]_i - Y[bar] )^2
    = ------------------------
       ?(  Y_i    - Y[bar])^2

               ? (e_i)^2
    = 1 -  --------------------
            ? ( Y_i - Y[bar])^2

(sorry this looks ugly, but Y_i means Y sub i, and Y[bar] means Y with
a bar above it.  (e_i)^2 is e sub i squared.  If you write this out on
paper it will make a bit more sense.

e_i is the least squares residual, which can be described as a
within-sample prediction error since it is the difference between the
observed and predicted values of Y, as predicted by the least squares
regression.

In this case the R-squared is equal to 1 because the QUARTER variables
is a perfect predictor of SALES.  This is fairly intuitive because
both of them grow in a linear fashion through the whole data sample. 
If SALES did not grow linearly, but instead grew at a more random rate
then R-squared would fall since the QUARTER variable would no longer
be a perfect predictor.

Your regression line is linear, so as long as growth in the SALES
variable remains linear it can cut through every data point on a plot
of these points.  If it cuts through every point then there is no
prediction error and so no e_i.  This would make R-squared 1 - 0, or
just 1, which we have found it to be in this case.

I hope this helped you out.  Let me know if you have difficulty.

Hibiscus

---

Request for Answer Clarification by k9queen-ga on 19 Nov 2003 07:52 PST

Thanks for in complete explanation- 
I do understand the Y hat and Y bar part!
These types are much easier for me to 
follow - when you can see the constant patterns
on paper.

---

Clarification of Answer by hibiscus-ga on 19 Nov 2003 21:23 PST

Glad this helped you out, k9queen.  It's difficult to try to write
coherent equations using only ASCII text, but I'm glad you figured it
out.

Hibiscus