No. Homework 4 EC 469
Problem No. 1
Importing the data set necessitated that I rename the six variables as follows: VM average verbal S.A.T. score for males; VF - average verbal S.A.T. score for females; VT average verbal S.A.T. score (both males and females); MM - average math S.A.T. score for males; MF - average math S.A.T. score for females; MT average math S.A.T. score (both males and females);
Answer Key Fall 2008
A/. With the above defined variables, Equation 1 which seeks to predict the male verbal score (VM) on the basis of the male math score (MM) takes on the following form: Equation 1: VMt = 0 + 1MMt+ t (i) The Eviews output from estimating this equation with data on average male verbal and math scores between the years 1967 and 1990 is given below: Dependent Variable: VM Method: Least Squares Sample: 1967 1990 Included observations: 24 Variable C MM Coefficient Std. Error -380.4789 63.32969 1.641791 0.126648 t-Statistic -6.007906 12.96341 Prob. 0.0000 0.0000 440.4167 11.91972 5.762127 5.860299 168.0500 0.000000
R-squared 0.884241 Mean dependent var Adjusted R-squared 0.878979 S.D. dependent var S.E. of regression 4.146637 Akaike info criterion Sum squared resid 378.2811 Schwarz criterion Log likelihood -67.14553 F-statistic Durbin-Watson stat 0.794443 Prob(F-statistic) The estimated counterpart of equation (1) is:
VM t = 0 + 1 MM t + et OR
VM t = 380.48 + 1.6418MM t
The above regression shows that the slope of the estimated regression line is 1.6418. The interpretation of this coefficient is that if the average math score goes up by one point, the average verbal score for males goes up by about 1.64 points. It seems that the two variables are positively related. The intercept has no particular meaning unless you can interpret negative verbal score as students being speechless? A more plausible interpretation of the intercept (if any) would be one where the S.A.T administrators actually impose a penalty of 380 points on anyone who shows complete lack of math aptitude (i.e. math score is zero). (ii) Step 1: H0: 1 = 0 and Ha: 1 0 In words: To test that the average math score has absolutely no power in explaining the variation in average verbal score for males we test whether the estimated counterpart of 1 is equal to zero. I have formulated the alternative in a way that reflects two competing hypotheses. On the one hand, a higher math score may also entail a higher verbal score indicating the quality of a well rounded education. On the other hand I personally know individuals who are exceptionally good at math but not very eloquent or well adept at good writing and spelling. So there you have it I have two competing hypotheses for which reason I am choosing a two-tailed t-test.
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Step 2: The test statistic is t =
1 s 1
=
1.6418 = 12.96 t 22 0.126648
Step 3: Degrees of freedom are 22. The critical values for a two-sided test at the 10%, 5%, 1% levels of significance are tc = 1.717, tc = 2.074, and tc = 2.819 respectively. Step 4: Reject H0 if t1 > tc. Since t1 = 12.96, I reject the null hypothesis that 1 is zero at the 10%, the 5%, and the 1% level of significance. (iii) To obtain the verbal score prediction for 1991 we use our estimated equation
VM t = 380.48 + 1.6418MM t and plug the value of 433 for MMt. Based on this
calculation, I predict that the average verbal score for males in 1991 in going to be approximately 330.42.
(iv) We have 22 degrees of freedom, 1 = 1.6418 , SE( 1 ) = 0.126648, a 95% two sided critical t value for 22 d.f. is tc = 2.074. Then we compute:
Confidence interval
= 1 tc * SE( 1 ) = 1.6418 0.262668 = {1.379132; 1.904468} In words, I expect that 95% of the time the true coefficient will fall between 1.379132 and 1.904468.
B/. Using the above defined variables, Equation 2 which seeks to predict the female verbal score (VF) on the basis of the female math score (MF) takes on the following form: Equation 2: VFt = 0 + 1MFt+ t (i) The Eviews output from estimating this equation with data on average female verbal and math scores between the years 1967 and 1990 is given below: Dependent Variable: VF Method: Least Squares Sample: 1967 1990 Included observations: 24 Variable C MF R-squared Adjusted R-squared S.E. of regression Sum squared resid Log likelihood Durbin-Watson stat Coefficient -336.0593 1.698514 0.797654 0.788457 8.011268 1411.969 -82.95077 0.272452 Std. Error 82.75977 0.182389 t-Statistic -4.060660 9.312613 Prob. 0.0005 0.0000 434.5000 17.41813 7.079230 7.177402 86.72476 0.000000
Mean dependent var S.D. dependent var Akaike info criterion Schwarz criterion F-statistic Prob(F-statistic)
The estimated counterpart of equation (2) is:
VFt = 0 + 1 MFt + et OR
VF t = 336.06 + 1.6985MFt
The above regression shows that the slope of the estimated regression line is 1.6985. The interpretation of this coefficient is that if the average math score goes up by one point, the average verbal score for females goes up by about 1.70 points. This value is slightly higher than that obtained for the male students. Again, it seems that the two variables are positively related. The intercept of -336 has no particular practical interpretation.
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(ii) Step 1: H0: 1 = 0 Ha: 1 > 0 In words: To test that the average math score has absolutely no power in explaining the variation in average verbal score for females we test whether the estimated counterpart of 1 is equal to zero. My alternative hypothesis reflects my expectation that if a female is good at math, her verbal aptitude is at least at good. That is, I expect a positive relationship. (My expectations here are formulated strictly for illustration purposes. What I really think about male and female math/verbal abilities is not necessarily reflected in this answer key just thought Id make this clear.) My expectation requires a one-sided t-test. Step 2: The test statistic is t =
1 s 1
=
1.698514 = 9.31259 t 22 0.182389
Step 3: Degrees of freedom are 22. The critical values for a one-sided test at the 10%, 5%, 1% levels of significance are tc = 1.321, tc = 1.717, and tc = 2.508 respectively. Step 4: Reject H0 if t > tc. Since t = 9.3126 is greater than all three critical values, I reject the null that hypothesis 1 is zero at the 10%, the 5%, and the 1% level of significance. (iv) To obtain the verbal score prediction for 1991 we use our estimated equation
VF t = 336.06 + 1.6985MFt and plug the value of 425 for MFt. Based on this
calculation, I predict that the average verbal score for females in 1991 is going to be approximately 385.80.
(iv) We have 22 degrees of freedom, 1 = 1.6985 , SE( 1 ) = 0.182389, a 99% two-sided critical t value for 22 d.f. is tc = 2.819. Then we compute:
Confidence interval
= 1 tc * SE( 1 ) = 1.6985 0.514154591 = {1.1843; 2.2127} In words, I expect that 99% of the time the true coefficient will fall between 1.1843 and 2.2127.
Problem No. 2
Violating the principle that the null hypothesis contains that which we do not expect to be true is a concern for part (A) but not for parts (B) and (C) in exercise 6. In part A, we most probably expect that the coefficient is 160. After all, if our expectation was something else, why did we pick 160? This is an illustration of the point made on p. 136 and footnote 9. To the extent that we feel that the hypothesized value of 160 is theoretically correct, we will also violate the normal practice of using the null hypothesis to state that which we believe to be untrue. In parts (B) and (C) on the other hand, it seems unlikely that we would expect the coefficients to be zero.
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Problem No. 3
(a) For all three, H0: 0, HA: > 0, and the critical 5% one-sided t-value for 24 degrees of freedom is 1.711. For LOT, we can reject H0 because +7.0 > 1.711 and 7.0 is positive. For BED, we cannot reject H0 because +1.0 < 1.711 even though 1.0 is positive. For BEACH, we can reject H0 because +10.0 > 1.711 and 10.0 is positive. (b) H0: 0, HA: < 0, and the critical 10% one-sided t-value for 24 degrees of freedom is 1.318, so we reject H0 because 2.0 > 1.318 and 2.0 is negative. (c) H0: = 0, HA: 0, and the critical 5% two-sided t-value for 24 degrees of freedom is 2.064, so we cannot reject H0 because 1.0 < 2.064. Note that we dont check the sign because the test is two-sided and both signs are in the alternative hypothesis. (d) The main problems are that the coefficients of BED and FIRE are insignificantly different from zero. (e) Given that we werent sure what sign to expect for the coefficient of FIRE, the insignificant coefficient for BED is the most worrisome. (f) It is possible that the dataset is unrepresentative, or that theres an omitted variable causing bias in the estimated coefficient of BED. Having said that, the most likely answer is that BED is an irrelevant variable if LOT also is in the equation. Beach houses on large lots tend to have more bedrooms than beach houses on small lots, so BED might be irrelevant if LOT is included.
Problem No. 4
(a) For the t-tests:
P M Coefficient: Hypoth. Sign: + + t-value: 5.8 6.3 tC = 1.671 reject reject (5% one-sided with 60 d.f., as close to 73 as Table B1 goes)
F-test: first compute the R2 measure:
S
1.0 do not Reject
T
3.3 reject
2 n K 1 78 4 1 R 2 = 1 (1 R ) * = 1 (1 0.682) * 78 1 = 0.6985 n 1
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STEP 1: STEP 2: STEP 3: STEP 4:
H0: P =M = S =T = 0, H1: H0 is not true.
R2 / K The test statistic is F = = 3.8434 F4,73 . (1 R 2 ) /(n K 1)
The critical values for the 5% level of significance Table B-2 is 2.53 Reject H0 if F Fc; At the 5% we reject the null hypothesis that P =M = S =T = 0 and conclude that at least one of the coefficients is not equal to zero. The model is statistically significant overall.
(b) No. We still agree with the authors original expectations despite the contrary result. (c) Keynes point is well taken; empirical results will indeed allow an econometrician to discover a theoretical mistake now and then. Unfortunately, far too many beginning researchers use this loophole to change expectations to get right signs. The bottom line is that a lot of thinking and analysis should be undertaken before one changes one's expectations. (d) Holding all other included explanatory variables constant, an increase in winning percentage of 150 points will increase revenues by $7,965,000 ($53.1 times 150 times 1000) and thus it would be profitable for this team to hire a $4,000,000 free agent who can raise its winning percentage to 500 from 350.
Problem No. 5
(a) All the expected signs for the coefficients are positive, so: H0: 0, HA: > 0. (Some students will see the Wall Street Journal as a competitor and will hypothesize a negative coefficient for J, but the authors intended the Journals circulation as a measure of what the Posts circulation would have been without Watergate.) (b) Coefficient: Hypoth. Sign: t-value: tC = 1.717 (5% one-sided with 22 d.f.)
J
+ 14.27 reject
S
+ 6.07 reject
W
+ 1.31 do not reject
(c) Assuming that the specification is correct, Watergate had a positive but statistically insignificant effect on the Posts circulation.
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Problem No. 6
(a) Coefficient: Hypoth. Sign: t-value: tC = 1.314 (10% one-sided with 27 d.f.)
C
+ 4.0 reject
E
+ 4.0 reject
M
+ 2.0 do not reject
The problem with the coefficient of M is that it is significant in the unexpected direction, one indicator of a possible omitted variable. (b) The coefficient of M is unexpectedly negative, so were looking for a variable the omission of which would cause negative bias in the estimate of M . We thus need a variable that is negatively correlated with meat consumption with a positive expected coefficient or a variable that is positively correlated with meat consumption with a negative expected coefficient. So: Possible Omitted Variable B F W R H O
*
Expected Sign of + + +*
Correlation with M +* + + +
Direction of Bias + + + + +
indicates a weak expected sign or correlation.
(c) The only one of the above variables that seems likely to have caused any of the observed
negative bias in M is H, and that variable is a good proxy for the obvious omission of the general level of quality of cardiac health care, so wed choose to add it. We wouldnt overly penalize a student who chose to add another variable as long as the alternative was an annual aggregate variable. (Adding disaggregate variables would not be appropriate.)
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