Question

Consider the relationship between hourly wage rate and education attainment. A random sample of 21 male workers was collected to estimate the following model Y; = Bo + B1X; + uj, for i = 1,..., 21. Here, Y; is the logarithm of hourly wage rate, log(wage), for the i-th worker. Xi is the education level, husedu, of the i-th worker, which is measured as the years of schooling, and uị is the error term for the i-th worker. The ordinary least squares (OLS) estimation of the model is reported in the table below. The variable ones OLS Regression Results ============================================================================== Dep. Variable: Model: Method: Date: Time: No. Observations: Df Residuals: Df Model: Covariance Type: huswage OLS Least Squares Wed, 08 Apr 2020 00:13:51 21 19 R-squared: Adj. R-squared: F-statistic: Prob (F-statistic): Log-Likelihood: AIC: BIC: 0.525 0.499 20.96 0.000205 6.6582 -9.316 - 7.227 nonrobust = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = coef std err t P> [0.025 0.975] ones huseduc 1.1466 0.0736 0.207 0.016 5.546 4.578 0.000 0.000 0.714 0.040 ======== 1 .579 0.107 ========== EEEEE = = = = = = = = = = = = 18. (3 points) According to the estimates , what is the predicted value of the logarithm of hourly wage for a male worker with 12 years of schooling? 19. (3 points) The standard error of is selối) = 0.016 as shown in the OLS results. Test the hypothesis He: B1 = 0.03 Hi : B1 +0.03. Use the 0.01 significance level. 20. (3 points) Construct the 95% confidence interval for the population parameter ß1. 21. (3 points) The standard error of the regression is 0.1853 (not shown in the results table), and the R-squared is 0.525. Use these values to obtain the total sum of squares of the dependent variable.

Answer 1

solution:

Given data:

18.

The regression line is:

, thus for a male worker with 12 years of schooling, the predicted value of the log of hourly wage is:

log(wage;) = 1.1466+0.0736 * huseduc

log(wage) = 1.1466 +0.0736 * 12 = 2.0298

19.

Here we are testing the hypothesis that:

. To the test the above hypothesis the test statistic is given by:

HO: B1 = 0.03 vs H:81 0.03

, which follows a t distribution with n-2= 21-2 = 19 degrees of freedom. Thus the test statistic is:

_81-0.03 se(81)

0.0736 -0.03 ta =2.725 0.016
. The critical value at a significance level 0.05 is given by:
土10.975.19 = ±2.093
. Here test statistic lies beyond the critical value thus we reject the null hypothesis. i.e. there is enough evidence to support the claim that
¥ 0.03
.

20)

we need to find the 95% Confidence interval for ttge population parameter $\beta_1$

And the confidence interval is given by:

B1 tdf=n-2,0/2) SE(B1)

where  
tdf=n-2,0/2)
is the critical value of t distribution with n-2 df and at $\alpha$ level of significance.

n = sample size

$\hat{\beta_1}$ = 0.0736

= 0.016

SE(81)

n = 21

= 2.093024

t(21,0.05/2)

It is calculated in R

code is attahced herewith.

> at (0.975, 19)

[1] 2 .093024

So 95% CI is given by:

  

0.0736 +2.093024 * 0.016

(0.04011162, 0.1070884)

As we can aslo verify this from the above given output.

21)

Here we need to find the Total Sum of Squares

As we know that

TSS = SSR + SSE

where

TSS = Total sum of squares

SSR = Sum of squares due to regression

SSE = Sum of squares due to Error

and we also know that

standard error = VMSE

and

SSESSE MSE = → SSE = MSE* 19 = 0.1853 * 19 = 3.5207 n - 2 1970

Similary we are also given with R - square = 0.525

and we know that

SSR SSE SSE 3.5207 R² = 0.525 = TSS = T SS = 1 TSS-1 1 - R2 1 - 595 = 1.412

So TSS = 7.412

please give me thumb up