Answer:
X is the GPA
Y is the Salary
Standard deviation of X is 0.4
Standard deviation of Y is 8500
E(X)=2.9
E(Y)=47200
We are given that The correlation between the two variables was r = 0.72
a)[tex]y = a+bx[/tex]
[tex]b = \frac{\sum(x_i-\bar{x})(y_i-\bar{y})}{\sum(x_i-\bar{x})^2} = \frac{r \times \sqrt{var(X) \times Var(Y)}}{Var(X)} = \frac{0.72 \times \sqrt{0.4^2 \times 8500^2}}{0.4^2} = 15300[/tex]
[tex]a=y-bx = 47200-(15300 \times 29) = 2830[/tex]
So, slope = 15300
Intercept = 2830
So, equation : [tex]y = 2830+15300x[/tex]
b) Your brother just graduated from that college with a GPA of 3.30. He tells you that based on this model the residual for his pay is -$1880. What salary is he earning?
[tex]y = 2830+15300 \times 3.3 = 53320[/tex]
Observed salary = Residual + predicted = -1860+53320 = 51440
c)) What proportion of the variation in salaries is explained by variation in GPA?
The proportion of the variation in salaries is explained by variation in GPA = [tex]r^2 = (0.72)^2 =0.5184[/tex]