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ECMT: Econometric Applications Problem

Set 3 Solutions

Semester 2 2022

Question 1.

(i) Test:

H0  : β 1  = β2  = β3  = β4  = 0

H1  : H0  is false

Note this is a test of the overall significance of the model – and hence the R2   =  0 for the restricted model

Test Statistic:

R2/q                

F =                                               Fq,n  1

0.0395/4       

=

(1 0.0395) /137

0.0099

= 0.0070

= 1.4085

Rejection Rule: Reject H0  in favour of H1  if F > c, where c is the critical value for the F distribution with df = (4, 137) and a 5% significance level.

In our case, F = 1.4085 and c = 2.37.

Decision: Since F  c we do not reject the null in favour of the alternative at the 5% signifi- cance level, all else equal.

Conclusion: The explanatory variables in the model are jointly insignificant at the 5% level.

(ii) Test:

H0  : β 1  = β2  = β3  = β4  = 0

H1  : H0  is false

Test Statistic:

R2/q                

F =  (1 R2) / (n 1) Fq,n 1

0.033/4       

=

(1 0.033) /137

0.0083

= 0.0071

= 1.176

Rejection Rule: Reject H0  in favour of H1  if F > c, where c is the critical value for the F distribution with df = (4, 137) and a 5% significance level.

In our case, F = 1.176 and c = 2.37.

Decision: Since F  c we do not reject the null in favour of the alternative at the 5% signifi- cance level, all else equal.

Conclusion: Again, we find that the explanatory variables in this model are jointly insignif- icant at the 5% level.

(iii) The absolute value of the t-statistics for the individual coefficients are:

β         |t|

β 1

β2

β3

β4

1.611 0.863 1.407 1.147

The critical value for a 5% significance level and 2-sided alternative is t(0 .025 , 137)    =  1 .96.

Therefore no explanatory variable is individually significant at the 5% level.

(iv) Overall, there is very weak evidence for the predictability of stock market returns. The F statis- tics are insignificant in both models, and there are no significant individual t-statistics (at the 5% level against a two-sided alternative) in the second model. (Incidentally, the same is also true for the first model). Further, less than 4% of the variation in return is explained by the independent variables.

Question 2. Computer Exercise: Explaining the Salary of MBA Graduates

(i) Descriptive statistics for the sample:

salary

testsc

WAM

libsize

rank

mean

139,750

10.587

3.31

354.93

79.54

min

124,900

9.67

2.73

124.00

1.00

max

179,000

11.40

3.82

1745.00

174.00

(ii) The parameter β4 measures the effect on expected log (salary) of a graduate who attended an

MBA program that is ranked one place lower than a comparable MBA program, while holding testsc, WAM and log (libsize) constant.

We expect β4   ≤ 0 as graduates from more prestigious programs (programs that are ranked more highly – so have a lower value of rank), cet. par., are expected to receive higher salaries. The coefficient on libsize is the elasticity of salary w.r.t. libsize, holding testsc, WAM and rank constant. That is, the coefficient β3  measures the percentage change in salary given a 1% change in the size of the library holdings, other things equal.

(iii) Estimated model:

log (salary) = 11.000 + 0.050 testsc + 0.048 WAM + 0.040 log (libsize) − 0.0009 rank

(0.143)   (0.017)             (0.023)              (0.009)                         (0.0001)

n = 250,     R2  = 0.7927,     SSR = 0.40058

(iv) Test:

。0  : β4  = 0

。1  : β4  < 0

Test Statistic:

t =                 =                      = 10.40

Rejection Rule: Reject 。0  in favour of 。1  if t < −c, where t is the t-statistic and c is the critical value for the t distribution with df  =  245 and a 5% significance level (a different significance level could be used).

In our case, t = −10.40 and c = −1.645 (from Table G.2 with df = ∞).

Decision: Since −t < −c we reject the null in favour of the alternative at the 5% significance level.

Conclusion: The rank on the MBA program, holding testsc, WAM and rank constant, has a significant effect on the median starting salary.

(v) Test:

。0  : β 1  = β2  = 0

。1  : β 1   0 or β2   0

Test Statistic:

F =         (Ru(2)r  Rr(2))/q        Fq,n 1

=

0.0141

= 0.00085

= 16.674

Rejection Rule: Reject H0  in favour of H1  if F  > c, where c is the critical value for the F distribution with df = (2, ∞) and a 5% significance level.

In our case, F = 16.674 and c = 3.00.

Decision: Since F  > c we reject the null in favour of the alternative at the 5% significance level, all else equal.

Conclusion: The two features of the graduate class testsc and WAM – are jointly signifi- cant in explaining median starting salary, holding libsize and rank constant.

(vi) Test:

H0  : β 1  = β2

H1  : β 1   β2

Rewrite the hypotheses as: H0  : β 1  − β2  = 0, H1  : β 1  − β2   0.

Let θ = β 1  − β2 . Then equivalent to these hypotheses is: H0  : θ = 0, H1  : θ  0.

This can be tested directly by estimating a transformed population model and doing a t-test on an individual parameter. To see this, from θ = β 1  − β2 we have β 1  = θ + β2 . Plug this expression for β1 into the population model:

log (salary) = β0 + (θ + β2 ) testsc + β2WAM + β3 log (libsize) + β4 rank + u       = β0 + θ testsc + β2 testsc + β2WAM + β3 log (libsize) + β4 rank + u = β0 + θ testsc + β2  (testsc + WAM) + β3 log (libsize) + β4 rank + u

Estimate this transformed testsc.

Test:

population model and test the significance of the coefficient on

H0  : θ = 0

H1  : θ  0

Test Statistics:

|t| = se (  )θˆ = '  0.03611   ' = 0.05769

Rejection Rule: Reject H0  in favour of H1  if |t| > c, where t is the t-statistic and c is the critical value for the t-distribution with df = n − k − 1 = 245 and a 5% significance level (a different significance level could be used).                                                                         In our case, |t| = 0.05769 and c = 1.96 (from Table G.2 with df = ∞).

Decision: Since |t|  c we do not reject the null in favour of the alternative at the 5% signifi- cance level, all else equal.

Conclusion:  Therefore testsc and WAM have the same effect on median starting salary, other things equal.