Use the regression results in columns: (a) Using the regression results in colum

Question

Use the regression results in columns: (a) Using the regression results in column (1), is the college-high school earnings difference estimated from this regression statistically significant at the 5% of significance level ( = 0.05)? Use a 95% confidence interval of the difference, where you can use the standard normal approximation for a critical value (i.e. tcrit 1.96). (Hint: think about the meaning of dummy variable.) (b) Using the regression results in column (2), is age a determinant of earnings? Use an appropriate statistical test of age at the 5% significance level ( = 0.05). (c) Using the regression results in column (2), Sally is a 29-year-old female college graduate. Sophie is a 32-year-old female college graduate. Using the standard normal approximation (i.e. tcrit 1.96), construct a 95% confidence interval for the expected difference between their earnings. (Hint: think about what makes them different and the meaning of dummy variable.) (d) Using the regression results in column (3), do there appear to be important regional differences? Assuming the variance of errors is homoskedastic, use an appropriate hypothesis test to explain your answer. (Hint: you have multiple parameters in your null hypothesis.) (e) The regression shown in column (2) was estimated again, this time using data from 1992 (inflation-adjusted to 2012 dollars). The results are: +8.66 (.33) College+4.24 (.29) F emale .65 (.05) Age, n = 7440, SER = 9.57, adj.R2 = 0.21, Comparing this regression to the regression for 2012 shown in col- umn (2), was there a statistically significant change in the coefficient on College? That is, test for H0 : college,2012 college,1992 = 0.
Dependent Variable: average hourly earnings (AHE) Regressor College (X1) Female (X2) Age (X3) Northeast (X4) Midwest (Xs) South Xe) Intercept 8.31 8.32 8.34 (0.23) (0.22) (0.22) 3.85 -3.81 3.8o (0.23) (0.22) 0.22) 0.51 0.52 (0.04) (0.04) 0.18 (0.36) (0.31) 0.43 (0.30) 17.02 87 2.05 Summary Statistics SER R2 Adj. R2 (or R2) 9.79 9.68 9.67 0.162 0.180 0.182 7440 7440 7440 Note: The standard errors are in parentheses along with the estimates.

Explanation / Answer

(a) Using the regression results in column (1), is the college-high school earnings difference estimated from this regression statistically significant at the 5% of significance level ( = 0.05)? Use a 95% confidence interval of the difference, whereyou can use the standard normal approximation for a critical value.

Answer. Here college high school eaarnings difference is dependent on the dummy variable college (x1) r where college = 1 and highschool = 0

Here coefficient for dummy variable (college) is = 8.31

standard error of the coefficient = 0.23

test statistic t = 8.31/0.23 = 36.13 > 0.05

so we can reject the null hypothesis and can conclude that the college-high school earnings difference estimated from this regression statistically significant at the 5% of significance level.

95% confidence interval = 1 +- Z0.05 se (1) = 8.31 +- 1.96 * 0.23 = (7.86, 8.76)

(b) Using the regression results in column (2), is age a determinant of earnings? Use an appropriate statistical test of age at the 5% significance level ( = 0.05).

Answer : Here test statistic

t = Coefficient for age/ standard error for the age = 0.51/0.04 = 12.75

Here t (critical) = 1.96 as n is very high.

so here t > t (critical) so we can reject the null hypothesis and can conclude that age is a determinant of earnings.

(c) Using the regression results in column (2), Sally is a 29-year-old female college graduate. Sophie is a 32-year-old female college graduate. Using the standard normal approximation (i.e. tcrit 1.96), construct a 95% confidence interval for the expected difference between their earnings.

Answer : Earning for Sally [College = yes, Female = yes ; Age = 29] let say y1

Earning for Sophie [College = yes, Female = yes; Age=32] let say y2

Here we can say that the only difference here is age only.

so here E(y1 - y2) = 3* coefficient of Age= 3 * 0.51 = 1.53

standard error of difference sed= sqrt [3 * se2] = sqrt [3 * 0.042] = 0.07

95% confidence interval for difference = E(y1 - y2) +- t0.05 * sed = 1.53 +- 1.96 * 0.07 = (1.394, 1.666)

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.