262 Bs (e) Test the hypothesis Ho: Consider the wage equation . 6.4 + 6(EDUC × E

Question

262 Bs (e) Test the hypothesis Ho: Consider the wage equation . 6.4 + 6(EDUC × EXPER) + 7HRSWK + e where the explanatory variables are years of education, years of experience and hours worked per week. Estimation results for this equation, and for modified versions of it obtained by dropping some of the variables, are displayed in Table 6.4. These results are from the 1000 observations in the file cps4c_small.dat (a) Using an approximate 5% critical value of tc-2, what coefficient estimates are not significantly different from zero? (b) What restriction on the coefficients of Eqn (A) gives Eqn (B)? Use an F-test to test this restriction. Show how the same result can be obtained using a t-test.

Explanation / Answer

a) Using an approximate 5% critical value of tc= 2, what coeffcient estimates are not significantly different from zero?

Ans:- use of the table 6.4 calculat t statistics and the formula of t-statistics are as follows;

t= bk/ se(bk)

I only show how to calculate one value of t

In the table 6.4 equation A the intercept C ; Coefficient value = 1.055 and standrad error = 0.266

t= bk/ se(bk) for C = 1.055/0.266= 3.9661 ( approx. 3.97) like that we will calculate other variable equation wise and get the below table of T value

According to the question we reject the null hypothesis if t value is greater than tc equal to 2 . The above table explain that In the equation A EDUC & EDUC2 is statistically insignificant different from zero but there is some exception in equ B that EDUC2 is statistically signficant with 5% level .EXPER*EDUC is not significant in the Equ 1

(b)What restriction on the coeffcients of Eqn (A) gives Eqn (B)? Use an F-test to test this restriction. Show how the same result can be obtained using a t-test.

Ans:- If you see the Table 6.4 it is easy to understand the restriction on the coefficient of Eqn(A) gives Eqn(B) is beta 6 = 0 . so F test value the null hypothesis H0: ß6 = 0 against the alternative hypothesis ß6 is not equal to 0 . Calculate the restricted and unrestricted sum of square are as follows;

F= ((SSER-SSEU)/J)/(SSEU/N-K) SSER & SSEU, N, K,J are already given on the table 6.4

= ((222.6674222.4166) /1)/(222.4166 / 993) = 1.120

The P value = 0.290 so 5% level of significant F( 0.95, 1, 993)= 3.851 ( we can get from F table)

so F value is less than the critical value or p value is less than 0.05 then we reject the null hypothesis

Find out the same result of both t-test & f-test below are the explanation

So t value for testing the null hypothesis H0: ß6 = 0 against the alternative hypothesis ß6 is not equal to 0 is -1.058(approx 1.06) which is given on the above t test table .and 5% level of significant the actual value is less than the critical value , t( 0.95, 1, 993)= 1.962 . So the t test give the same result with f test.

Squr root of 1. 120= 1.058 and Squre root of 3.851 = 1.962

(c) What restrictions on the coefficients of Eqn (A) give Eqn (C)? Use an F-test to test these restrictions. What question would you be trying to answer by performing this test?

Ans:- If you see the Table 6.4 it is easy to understand the restriction on the coefficient of Eqn(A) gives Eqn(C) is beta 4,5,6 = 0 . so F test value the null hypothesis H0: ß4 = 0,ß5 = 0ß6 = 0 against the alternative hypothesis at least one of the ß4,ß5 ,ß6 is non zero  . Calculate the restricted and unrestricted sum of square are as follows;

F= ((SSER-SSEU)/J)/(SSEU/N-K) SSER & SSEU, N, K,J are already given on the table 6.4

= ((233.8317-222.4166) /3)/(222.4166 / 993) = 16.988 and the p value is 0.0000

The critical value of 5% lelvel significant F( 0.95, 3, 993)= 2.614 so F value is greater than critical value and we reject the null hypothesis and conclude that at least one of the coefficient is non zero.

(d) What restrictions on the coefficients of Eqn (B) give Eqn (D)? Use an F-test to test these restrictions. What question would you be trying to answer by performing this test?

Ans:- If you see the Table 6.4 it is easy to understand the restriction on the coefficient of Eqn(B) gives Eqn(D) is beta 2,3,6 = 0 . so F test value the null hypothesis H0: ß2 = 0,ß3 = 0ß6 = 0 against the alternative hypothesis at least one of the ß2,ß3 ,ß6 is non zero  . Calculate the restricted and unrestricted sum of square are as follows;

F= ((SSER-SSEU)/J)/(SSEU/N-K) SSER & SSEU, N, K,J are already given on the table 6.4

= ((280.5061-222.6674) /3)/(222.6674 / 993) = 262.86and the p value is 0.0000

The critical value of 5% lelvel significant F( 0.95, 3, 993)= 2.621 so F value is greater than critical value and we reject the null hypothesis and conclude that at least one of the coefficient is non zero.

Variables t- values Eq(A) Eq(B) Eq(C) Eq(D) Eq( E ) C ß1 3.97* 6.59* 8.38* 23.82* 9.42* EDUC ß2 1.26 0.84 1.04 15.9 EDUC2 ß3 1.89 2.12 1.73 EXPER ß4 4.58* 6.28 5.17* 6.11* EXPER2 ß5 –5.38* –5.31* –4.90* –5.13* EXPER* EDUC ß6 –1.06 HRSWK ß7 8.34* 8.43* 9.87* 10.11* 8.71*

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