Economics 4818 –Spring 2008

Homework 3 - Answers

Graded problems:

4.2 (i) (1 point)

4.6 (i) (2 points) – 1 point for using t-stats, 1 point for using CI

4.6 (ii) (2 points)

4.8 (iii) (2 points)

Financial Model 1: (3 points) – A,B,C 1 point each

4.2 (i) H0: = 0. H1: > 0.

(ii) The proportionate effect on is .00024(50)= .012. To obtain the percentage effect, we multiply this by 100: 1.2%. Therefore, a 50 point ceteris paribus increase in ros is predicted to increase salary by only 1.2%. Practically speaking, this is a very small effect for such a large change in ros.

(iii) The 10% critical value for a one-tailed test, using df= ¥, is obtained from Table G.2 as 1.282. The t statistic on ros is .00024/.00054 .44, which is well below the critical value. Therefore, we fail to reject H0 at the 10% significance level.

(iv) Based on this sample, the estimated ros coefficient appears to be different from zero only because of sampling variation. On the other hand, including ros may not be causing any harm; it depends on how correlated it is with the other independent variables (although these are very significant even with ros in the equation).

4.6 (i) Using t-statistics: With df= n– 2= 86, we obtain the 5% critical value from Table G.2 with df= 90. Because each test is two-tailed, the critical value is 1.987. The t statistic for H0:= 0 is about -.89, which is much less than 1.987 in absolute value. Therefore, we fail to reject = 0. The t statistic for H0: = 1 is (.976– 1)/.049 -.49, which is even less significant. (Remember, we reject H0 in favor of H1 in this case only if |t|> 1.987.)

Using confidence intervals: The 95% confidence interval (CI) for is -14.47± 1.987(16.27), or (-46.799,17.858). Since 0 is within the 95% CI for, we cannot reject H0:= 0 against the two-sided alternative at the 5% level. The 95% CI for is (.976) ± 1.987(.049), or (.879,1.073). Since 1 is within the 95% CI for, we cannot reject H0: = 1 against the two-sided alternative at the 5% level of significance.

(ii) We use the SSR form of the F statistic. We are testing q= 2 restrictions and the df in the unrestricted model is 86. We are given SSRr= 209,448.99 and SSRur= 165,644.51. Therefore,

which is a strong rejection of H0: from Table G.3c, the 1% critical value with 2 and 90 df is 4.85.

(iii) We use the R-squared form of the F statistic. We are testing q= 3 restrictions and there are 88– 5= 83 df in the unrestricted model. The F statistic is [(.829– .820)/(1– .829)](83/3) 1.46. The 10% critical value (again using 90 denominator df in Table G.3a) is 2.15, so we fail to reject H0 at even the 10% level. In fact, the p-value is about .23.

(iv) If heteroskedasticity were present, Assumption MLR.5 would be violated, and the F statistic would not have an F distribution under the null hypothesis. Therefore, comparing the F statistic against the usual critical values, or obtaining the p-value from the F distribution, would not be especially meaningful.

4.8 (i) We use Property VAR.3 from Appendix B: Var(- 3)= Var ()+ 9 Var ()– 6 Cov (,).

(ii) t = (- 3 - 1)/se(- 3), so we need the standard error of - 3.

(iii) Because = – 3b2, we can write = + 3b2. Plugging this into the population model gives

y = + ( + 3b2)x1 + x2 + x3 + u

= + x1 + (3x1 + x2) + x3 + u.

This last equation is what we would estimate by regressing y on x1, 3x1+ x2, and x3. The coefficient and standard error on x1 are what we want.

C4.6 (i) In the model

log(wage) = + educ + exper + tenure + u

the null hypothesis of interest is H0: = .

(ii) Let = – . Then we can estimate the equation

log(wage) = + educ + exper + (exper + tenure) + u

to obtain the 95% CI for . This turns out to be about .0020± 1.96(.0047), or about -.0072 to .0112. Because zero is in this CI, is not statistically different from zero at the 5% level, and we fail to reject H0: = at the 5% level.

Financial Models:

1. Recall the capital asset pricing model (CAPM): (r – rf ) = α + β ( rM - rf ) + u, where r is the stock or portfolio return, rf is the risk-free return, and rM is the market return (proxied for by the return in the S&P500). Applying the CAPM to the returns of MMM and CSCO, you get

Dependent Variable: MMM-RF
Method: Least Squares
Included observations: 60
Variable / Coefficient / Std. Error / t-Statistic / Prob.
C / 0.002641 / 0.006628 / 0.398414 / 0.6918
S_P500-RF / 0.830848 / 0.227720 / 3.648551 / 0.0006
Dependent Variable: CSCO-RF
Method: Least Squares
Included observations: 60
Variable / Coefficient / Std. Error / t-Statistic / Prob.
C / 0.006491 / 0.009460 / 0.686184 / 0.4953
S_P500-RF / 1.602391 / 0.325036 / 4.929893 / 0.0000

A. For MMM: the estimated market beta is 0.830848

For CSCO: the estimated market beta is 1.602391

B. You can either perform a t-test for statistical significance, that is test H0: beta=0 against a two-sided alternative, or use the reported p-values (denoted in EViews by Prob.). From the p-value for MMM’s beta (p-value = 0.0006<0.05), we determine that it is statistically significant (or statistically different from 0) at 5% level of significance. From the p-value for CSCO’s beta (p-value = 0.0000<0.05), we determine that it is statistically significant (or statistically different from 0) at 5% level of significance.

C. MMM is less volatile (or cyclical) than the market since is its beta is less than 1. In other words, if the economy is up, the MMM’s stock price is expected to be up by less than the economy, and if the economy is down, MMM’s stock price is expected to be down by less than the economy. CSCO is more volatile than the market since is its beta is more than 1. So, if the market is up, CSCO’s stock price is expected to be up by more than the market, and if the market is down, CSCO’s stock price is expected to be down by more than the market.

2. Applying the CAPM for Portfolio 1 and Portfolio 2, we get

Dependent Variable: PORTFOLIO_1-RF
Method: Least Squares
Included observations: 60
Variable / Coefficient / Std. Error / t-Statistic / Prob.
C / 0.008238 / 0.003406 / 2.419133 / 0.0187
S_P500-RF / 0.735550 / 0.117010 / 6.286222 / 0.0000
Dependent Variable: PORTFOLIO_2-RF
Method: Least Squares
Included observations: 60
Variable / Coefficient / Std. Error / t-Statistic / Prob.
C / 0.002432 / 0.005300 / 0.458901 / 0.6480
S_P500-RF / 1.722561 / 0.182100 / 9.459422 / 0.0000

A, B. Portfolio 1's estimated monthly alpha is positive (0.008238) and statistically significant at the 5 % level (p-value = 0.0187<0.05), indicating the manager did beat the market. The annualized excess return for Portfolio 1 was (0.008238 + 1)12 – 1 = 10.35%.

Portfolio 2's estimated monthly alpha is positive (0.002432) but not statistically different from zero (p-value = 0.6480>0.05), indicating the manager did not beat the market.

C. Portfolio 2 has a high market beta, indicating that it had a higher return than the market because it was comprised of high volatility stocks over a period when the market was up.