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MATH 1140 FINAL EXAM 2
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1MATH 1140 FINAL EXAM 2
Solutions to Exercises in Chapter 8
8.1 (a) The estimated equation is
trt = 110.46 − 10.198 pt + 3.361 at − 0.0268 a2t
(3.74) (1.582) (0.422) (0.0159)
(b) The estimated version of equation 8.4.2 and corresponding sum of squared errors are
trt = 111.71 − 5.057 pt SSER = 20907
(8.85) (4.012)
(c) The SHAZAM, EViews and SAS instructions can be found in the files xr8-1.xls, xr8-
1.szm, xr8-1.wf1 and xr8-1.sas, respectively.
(d) In Section 8.4.3 the model and restrictions are
trt = 1 + 2 pt + 3at + a 2 + et
4t
3 + 804 = 1 1 + 22 + 403 + 16004 = 175
Substituting these restrictions into the model yields
trt = 175 − 22 − 40(1 − 804 ) − 16004 + 2 pt + (1 − 804 )at + a 2 + et
4t
trt − 175 + 40 − at = 2 (−2 + pt ) + 4 (3200 − 1600 − 80at + at 2 ) + et
(trt − at − 135) = 2 ( pt − 2) + 4 (at 2 − 80at + 1600) + et
Estimating this model yields SSER = 2715.1. The F-value for the test in question is
(SSER − SSEU ) J = (2715.1 − 2592.3) 2
F= = 1.753
SSEU (T − K) 2592.3 74
(e) Using computer software, we find that, for p = 2 and a = 20, the predicted total revenue is
tr = 146.59. The standard error of the forecast error is se( f ) = 6.064, and a 95%
prediction interval is (134.5, 158.7).
(f) The following table gives the R2s from auxiliary regressions between the explanatory
variables as well as the sample correlations between the explanatory variables. The
results suggest that there are no substantial correlations between pt and the advertising
variables, but the correlation between at and a 2t is relatively high. Whether or not it is a
problem depends on whether we have been able to estimate the effect of advertising on
total revenue sufficiently precisely. In this regard the standard error of the coefficient of
, 2
MATH 1140 FINAL EXAM 2
at is relatively small, but that for the coefficient of a2 is relatively
t large, suggesting
collinearity could be preventing us from getting a precise estimate of this coefficient.
Auxiliary Sample Correlation with
Variable Regression R2 pt at at2
pt 0.1807 1 0.4246 0.4158
at 0.9364 0.4246 1 0.9674
at2 0.9358 0.4158 0.9674 1
(g) The response of total revenue to price and advertising expenditure is an important
relationship to quantify because it has implications for price setting and for the
appropriate level of advertising for the hamburger chain. The estimated equation in part
(a) suggests the responses of total revenue to price and advertising are, respectively,
(tr) (tr)
= −10.198 = 3.361 − 0.0536a
p a
Thus, demand is price elastic, and there are diminishing returns to advertising
expenditure. On the basis of one-tailed tests, all estimated coefficients are significantly
less than (or greater than) zero at a 5% level of significance.
To see whether advertising contributes significantly to the equation, we need to test
whether the coefficients of at and a2 aret both zero. The p-value for this F-test is
0.00000, indicating that the hypothesis is rejected at any conventional significance level.
For testing whether 40 could be the optimal level of advertising expenditure, we obtain a
p-value of 0.801, and thus this hypothesis is not rejected. An F test of the joint
hypothesis that the optimal values for p and a, respectively, are 2 and 40, yields a p-value
of 0.180. Hence, at a 5% significance level, we cannot reject these values as optimal. A
95% prediction interval for total revenue when p = 2 and a = 20 is (134.6, 158.6). Given
our earlier test conclusion, that for p = 2 and a = 40, E(tr) = 175 is a reasonable value, it
would make sense not to set a as low as 20.
Whether or not collinearity between a and a2 is a problem depends on whether
(tr) a is estimated accurately. For a = 20 and a = 40, this response and the
corresponding standard errors are:
(tr)
When a = 20, = 2.291 se(b3 + 40b4) = 0.259
a
(tr)
When a = 40, = 1.221 se(b3 + 80b4) = 0.874
a
For small values of a, the response is estimated reasonably accurately, but, for larger a, it
becomes quite unreliable.
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