You are given the following economic model: log( wage) = 0.478 + 0.085edu + 0.05

Question

You are given the following economic model: log( wage) = 0.478 + 0.085edu + 0.059ten - 0.058female - 0.011 edu. female - 0.021 female Jen - 0.0012ten^2 Std errors With all the variables described as follows: log(wage) = log of average hourly wage; female is a dummy variable equal to 1 if the observed person is a female, and 0 if male; edu.female is an interaction variable equal to education *female; edu is the number of years of schooling; ten is the number of years of tenure female.ten is an interaction variable equal to female*ten ten^2 = tenure*tenure. Is the parameter of the variable ten^2 statistically significant at 5%? Provide a possible reason for the presence of the variable ten^2 in the model estimated above What is the estimated return to an additional tenure year for a female worker? For a male worker? What is the optimal number of tenure years for a female worker? For a male worker? Calculate the estimated effect of the 10^th year of tenure on a female's wage, and on a male's wage. From the estimated equation above, provide an appropriate interpretation for each coefficient associated to the following variables: female, education, and female.edu ? Is education more valuable for male or for female workers? What about tenure? If a male and a female have 10 years of tenure each, how many years of education would it take for their wages to be the same?

Explanation / Answer

a) Yes the variable ten square is statistically significant at 5%significance level. Since 2xstd error is less than 5.

B) with the increase in tenure log of wage is increasing but this relation cannot be captured linearly so. ten square is present in the model it shows with 1 unit increase in tenure log of wage will increase by less than 0.059. it will increase by (0.059-0.0012). with the increase in tenure period this less will be more.

C) for female worker it must be(+0.059 -0.011-0.021-0.0012)=(0.059-0.0332)=0.0258

for male worker it will be (0.059-0.0012)=0.0588

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