Practice Problems from the Text - some answers
Let’s do problems straight from the text. Specifically, from Chapter 3, work on 3.7, 3.8, 3.10,
3.11 and 3.15. Note that 3.10 and 3.15 rely on actual data analysis and using Stata. The data for each exercise is now on the website (under “Data for Book Exercises”). We will do some of this in class, but it is up to you to do these exercises ahead of 2/12. Stata is available on the machines on the main deck of the library, so you should always have access (ask the librarians if you are unsure where).
3.7) (a) The d.f. here are 14. Therefore, the 5% critical t value is 2.145. So, the 95% confidence interval is:
3.24 ± 2.145(1.634) = (-0.2649, 6.7449)
(b) The preceding interval does include the hypothesized value of B
2
= 0. Therefore, do not reject the null hypothesis.
(c) t = 3.24 / 1.634 = 1.9829. Since car sales are expected to be positively related to real disposable income, the null and alternative hypotheses should be: H
0
: B
2

0 and H
1
: B
2
> 0.
Therefore, an one-tailed t test is appropriate in this case. The 5% one-tailed t value for 14 d.f. is
1.761. Since the computed t value of 1.9829 exceeds the critical value, reject the null hypothesis
(notice that one- and two tail tests sometimes give different results).
3.8) (a) The slope coefficient of 1.0598 means that during the 1956 –1976 period a percentage point increase in the market rate of return was associated with about 1.06 percent points increase in the mean return on the IBM stock. In the same period, if the market rate of return were zero, the average rate of return on the stock would have been about 0.73 percent, which may not make economic sense.
(b) About 47 percent of the variation in the mean return on the IBM stock was explained by the
(variation) in the market return.
(c) H
0
: B
2

1, H
1
: B
2
> 1. Hence: t = (1.0598-1)/0.0728 = 0.8214.
For 238 d.f, this t value is not statistically significant at the 5% level on the basis of the onetailed t test. Thus, during the study period, the beta coefficient of IBM was not statistically different from unity, suggesting that the IBM stock was not volatile or aggressive.

3.10) a) Here’s the output you should obtain (with the missing info from the problem boxed):
. reg GNP M1
Source SS df MS Number of obs = 42
F( 1, 40) = 741.09
Model 540032898 1 540032898 Prob > F = 0.0000
Residual 29147986.7 40 728699.668 R-squared = 0.9488
Adj R-squared = 0.9475
Total 569180885 41 13882460.6 Root MSE = 853.64
GNP Coef. Std. Err. t P>|t| [95% Conf. Interval]
M1 8.750264 .3214292 27.22 0.000 8.100631 9.399897
_cons -995.5183 260.2128 -3.83 0.000 -1521.428 -469.6086
b) H
0
: B
2

0, H
1
: B
2
> 0. The null hypothesis can be rejected (the critical t is smaller than the computed t for any reasonably level of significance). c) No particular economic meaning can be attached to it (why?) d) Estimated GNP
2007
= -995.5183 + 8.7503(750)

5,567 billion.
3.11) (a) Negative. Makes intuitive sense that rise in unemployment is associated with tougher election prospects for incumbents.
(b) Yes. We are testing the hypothesis that B
2

0, with the alternative that B
2
< 0 (which is the political business cycle theory). Here, n = 14 (14 presidential elections starting in 1928 and ending in 1980) and therefore d.f. = 12. The computed t value of -2.67 is statistically significant at the five percent level (one-tailed test gives critical t of 1.78). So we reject the hypothesis in support of the political business cycle theory.
(c) They appear to. But in the 1984 elections the personal popularity of Ronald Reagan was an important factor. These along with other potentially important factors are not in the analysis – we thus might have a specification error and so would need to do multiple regression analysis using more data.
(d) Since t = b i
/se(b i
) under the null hypothesis that the true i B is zero, se(b i
) = b i
/t. In the present example these standard errors are 1.5572 and 0.6367, respectively.

3.15) (a) Here’s the output you should obtain
. reg math_male math_female
Source SS df MS Number of obs = 36
F( 1, 34) = 641.73
Model 1605.9162 1 1605.9162 Prob > F = 0.0000
Residual 85.0838005 34 2.50246472 R-squared = 0.9497
Adj R-squared = 0.9482
Total 1691 35 48.3142857 Root MSE = 1.5819
math_male Coef. Std. Err. t P>|t| [95% Conf. Interval]
math_female .670482 .0264673 25.33 0.000 .616694 .7242701
_cons 198.737 12.87537 15.44 0.000 172.5711 224.9028
(b) Reject the null hypothesis, since the computed t value of 25.3325 far exceeds the critical value even at the 0.001 level of significance.
(c) Estimated math_male
2008
= 198.737 + (0.670482*490) = 527.282

527
(d) Don’t’ worry about it, I will not test on this kind of calculation. But, for the more ambitious among you, see equation 3.52: P[b
1
+ b
2
X
0
– t a/2
*se(Y estimate
)

B
1
+ B
2
X
0

b
1
+ b
2
X
0
+ t a/2
*se(Y estimate
)] = (1-α)