1. Empirical Exercise (chapter 4, 4.3)
a. Edi =13.93373-0.064855 Disti
(0.0678954) (0.0240502)
The intercept β0 is 13.93373. The estimated coefficient of Distance (β1) is -0.64885. The regression predicts that if colleges are built 10 miles closer to the high schools, years of completed education increase by 0.64855 years.
b. Prediction for Bob: Ed = 13.93373-0.064855*2= 13.80402 years
If Bob lived 10 miles from college, then the result would be as following: Ed = 13.93373-0.064855*1= 13.868875 years
c. No, the distance for college does not explain a large fraction of the variance, because the estimated R2 is very low, only 0.0064. Thus, it means that distance merely does explain the smallest percentage.
d. The standard error of the regression is 1.8022, and it is measured in years.
2. Empirical Exercise (chapter 5, 5.3)
- H0: β1=0
HA: β1≠0
I reject the null hypothesis, if the | tk | >tc.
The critical t-value for two-sided 10% confidence level is 1.711. tk is (β1 /SE) = -0.0648552/ 0.0240502 = - 2.7. I reject the null hypothesis because |-2.7|> 1.645
The critical t-value for two-sided 5% confidence level is 1.960. tk is (β1 /SE) = -0.0648552/ 0.0240502 = - 2.7. I reject the null hypothesis because |-2.7|> 1.960
The critical t-value for two-sided 1% confidence level is 2.576. tk is (β1 /SE) = -0.0648552/ 0.0240502 = - 2.7. I reject the null hypothesis because |-2.7|> 2.576
The p-value associated is 0.007.
- Construct a 95% confidence interval for the slope coefficient.
Confidence intervals for β1: -0.0648552 ± 1.960*0.0240502 =-0.0648552± 0.047138392
Confidence interval is between -0.017716808 and - 0.111993592.
- Run the regression using data only on females and repeat (b).
Ed_femalei =13.91252-0.0663379Disti
(0.0936246) (0.0335904)
R2= 0.0064
Confidence intervals for β1: -.0663379± 1.960*0.0335904=-.0663379 ± 0.065837184
Confidence interval is between - 0.000500716 and - 0.132175084
- Run the regression using data only on males and repeat (b).
Ed_malei =13.9577-0.0632007Disti
(0.0988613) (0.0345317)
R2= 0.0062
Confidence intervals for β1: -0.0632007 ± 1.960*0.0345317= -0.0632007± 0.067682132
Confidence interval is between 0.004481432 and – 0.130882832.
The effect of professor beauty on student evaluations
3. Empirical Exercise (Chapter 4, 4.2)
- Construct a scatter plot of average course evaluation (Course_Eval) on the professor's beauty (Beauty). Does there appear to be a relationship between the variables?
The scatter plot reveals no relationship between course_eval and beauty.
Course_Evali= 3.998272+ 0.1330014Beautyi
(0. 0253493) (0 .0321775)
R2= 0.0357
The intercept is 3.998272, and the estimated slope is 0.133.
The estimated intercept is equal to the sample mean, because the intercept is the mean of the dependent variable minus the estimated slope, which is 0.133 multiplied by the mean of the independent variable, which is 0.
- 1 standard deviation above the mean is equal to 0 .0321775 + 0= 0.0321775.
If the Professor Watson has an average value of Beauty, then it is 3.998272.
Evaluation for Professor Stock is 3.998272+0.133001*0.0321775=4.1634505
- The estimated effect is small because an increase in the standard deviation of Beauty will increase the standard deviation of the Course_Eval by 0.104937, which is insignificantly small.
- R2=0.0357, which means that only 3.57% of the variance is explained.
4. Empirical Exercise (Chapter 5, 5.2)
Course_Evali= 3.998272+ 0.1330014Beautyi
(0. 0253493) (0 .0321775)
R2= 0.0357
H0: β1=0
HA: β1≠0
I reject the null hypothesis, if the | tk | >tc.
The critical t-value for two-sided 10% confidence level is 1.711. tk is (β1 /SE) = 0.1330014/0.0321775 = 4.13. I reject the null hypothesis because |4.13|> 1.645
The critical t-value for two-sided 5% confidence level is 1.960. tk is (β1 /SE) = 0.1330014/0.0321775 = 4.13. I reject the null hypothesis because |4.13|> 1.960
The critical t-value for two-sided 1% confidence level is 2.576. tk is (β1 /SE) = 0.1330014/0.0321775 = 4.13. I reject the null hypothesis because |4.13|> 2.576
The p-value associated is 0.000.
