Extra Problems, Set #2 Regression Analysis
Key
Consider the following data
Sales
Adv
Yi
Xi
3
4
6
5
7
6
5
9
10
9
1
2
3
4
5
6
7
8
9
10
1. Input this data on a spreadsheet. Using the regression option, generate regression
predictions. Write your results as an equation, as we did in class. In particular, write
(a) write the regression equation, with estimated coefficients,
(b) below the regression equation, list in parentheses the standard errors of the coefficient
estimates
(c) To the right of the estimated equation write
i) R2 and adjusted R2
ii) F test statistic, along with its p value
iii) MSE of the estimate.
^
Yi = 2.7333 + .667Xi
(0.83)
(0.13)
R2
F1,8
MSE
= .758, Adj. R2 = .727
= 25 (p = .001)
= 1.211
2. Multivariate Regression. Now add to your above regression in a price variable, with
values: 8, 7.5, 7.25, 7.25, 6, 6.75, 6, 5, 4.4, 5.2. Estimate the new regression equation.
Print regression results. Write out the estimated demand function, as in part (c) above.
^
Yi = 17.683 + .0049Xi – 1.785Pi.
R2
= .868, Adj. R2
= .131
(6.19)
(.29)
(.73)
F2,7
=23.11 (p = .0008)
MSE = 0.954
3. Evaluating regression results. With the data you generated in (2) above do the
following.
a. Interpret the R2.
R2 = .868 implies that 86.8% of the variation in data from average sales is explained by
the regression.
b. What is the difference between R2 and adjusted R2?
Adjusted R2 is the percentage in the variation of data from the mean of average sales
explained by the regression after controlling for the potential loss in explanatory power
caused by adding additional independent variables.
c. Goodness of Fit Test. Looking at the F test statistic, can you conclude that the
regression estimate explains more of the variation in sales than the simple
average of sales? Explain.
F2,7
=23.11 (p = .0008). The F test statistic indicates (roughly) that the regression
equation explains 23.11 times more of the variation in the data than it leaves
unexplained. The p = .0008 indicates that 99.92% percent of the time a regression that
explained no more of the movement in the data than the simple average of sales would
explain as much as this estimate.
d. For each coefficient, write the 95% confidence intervals about the estimates.
From this data, which parameters are significant explainers of sales?
Standard Lower Upper
Coefficients Estimate
Error 95%
95%
Intercept 17.68362* 6.187328 3.052921 32.31431
Y
0.004963 0.291866 -0.68519 0.695116
X
-1.78546* 0.734826 -3.52305 -0.04788
Price (X) is a significant explainer of sales
(Note, the above is taken from regression output. Your confidence bounds will differ a bit
from the above if you add and subtract 2 standard deviations from the parameter
estimate. The above is more accurate.)
e. Observe that we can do the same exercises with this information that we did
previously. In particular, suppose that advertising expenditures are 10 and that the
current price is $6. Calculate the point price elasticity of demand.
Y = 17.683 + .0049(10) – 1.785(6) = 7.022. Thus,  = -1.785(6)/7.022 =-1.525
4. Transform the data you created in question 7 to estimate the equation
ln Y
=
a
+
b ln A + c ln P
a. Print regression results, following the style you used in equation 1(c).
ln Yi
=
3.31 + 0.228 ln Xi -1.01 ln Pi
(1.15) (0.142) (0.525)
R2 = .859 Adj. R2 = .818
F2,7 = 21.28 (p=.001)
MSE = 0.16
b. Calculate the price elasticity of demand for this equation.
 = -1.01
b. At a 95% confidence level, can you conclude that the firm is at the point of
unitary elasticity?
You cannot conclude that the firm is not at the point of unitary elasticity. The 95%
confidence band about price is –2.26 to .22. That range includes –1.
e. What is the advertising elasticity of demand? How confident can you be that
advertising increases sales? Justify your answer.
A = 0.228 The 95% confidence interval extends from –0.11 to .564. This interval
includes 0. Therefore we cannot conclude that advertising increases sales.