Question 10 Suppose you believe that wages depend on education. You gather data

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Question 10 Suppose you believe that wages depend on education. You gather data for 526 people and estimate the following regression Source df MS 526 Number of obs FC 1, 524)103.36 Prob > F R-squared Adj R-squared0.1632 Root MSE -0.0000 Modl 1179.732041 1179.73204 524 11.4135158 Residual l Total 7160.41429525 13.6388844 3.3784 wage I Coef Std. Err [95% Conf. Interval] educ .5413593 053248 0.000 -cons.9048516 -2.250472 4407687 1. What is the interpretation of the estimated coefficients? Does the interpretation of the intercept make sense to you? 2. What is the t-statistic of the coefficient on education1? 3. Find a 95% confidence interval for the coefficient on education 4. Test whether the coefficient on education is greater than 0.5, at a 5% level of significance 5. What is the estimated standard error of the intercept? 6. What is the p-value reported for the constant2? 7. What is the sum of squared residuals SSR)? 8. What is the R2 of the regression? What does it mean? Remember that the t-statistic that is commonly reported by statistical softwares is the one computed under the null hypothesis Ho 0. Remember that the p-value that is commonly reported by statistical softwares is the one computed for the two-sided test H0 : ? =0; H0 : ??0

Explanation / Answer

1.
Interpretation of slope - Wage is increased by    0.5413593 unit for each unit increase in education.
Interpretation of intercept - Wage is -0.9048516 when education is 0.
The Interpretation of intercept does not make sense, as the wage cannot be negative.

2.
t statistic = Estimated Coeff / Standard error = 0.5413593 / 0.053248 = 10.16675368

3.
Degree of freedom of residual = 524
t value at 95% confidence interval and DF = 524 is 1.96
95% confidence interval is
(0.5413593 - 1.96 * 0.053248, 0.5413593 + 1.96 * 0.053248)
(0.43699322, 0.64572538)

4.
H0: b1 = 0.5
Ha: b1 > 0.5

Test statistic t = (Estimated Coeff - Hypothesized value) / Standard error
= (0.5413593 - 0.5) / 0.053248
= 0.7767

P-value = P[t > 0.7767] for DF = 524 is 0.2188
As p-value is greater than significance level of 0.05, we fail to reject the null hypothesis and conclude that there is no significant evidence that the coefficient is greater than 0.5.

5.
Estimated standard error of the intercept = Coeff / t = -0.9048516 / -1.32 = 0.6854936364

6.
For two tail test,
P-value = 2 * P[t < -1.32] for DF = 524 is 2 * 0.09365951827 = 0.1873190365

7.
SSR = SS(total) - SS(model) = 7160.41429 - 1179.73204 = 5980.68225

8.
R-sq = SS(model) / SS(total) = 1179.73204 / 7160.41429 = 0.1647575115
This means that the model with education as predictor explains 16.475% of variation in wages.

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