Question

A college job placement office collected data about students’ GPAs and the salaries they earned in their first jobs after graduation. The mean GPA was 2.9 with a standard deviation of 0.4. Starting salaries had a mean of $47200 with a SD of $8500. The correlation between the two variables was r = 0.72. The association appeared to be linear in the scatterplot. a) Write an equation of the model that can predict salary based on GPA and find the slope and intercept. 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? c) What proportion of the variation in salaries is explained by variation in GPA?

Answer 1

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)
`y = a+bx`

`b = \frac{\sum(x_i-\bar{x})(y_i-\bar{y})}{\sum(x_i-\bar{x})^2} = (r * √(var(X) * Var(Y)))/(Var(X)) =  (0.72 * √(0.4^2 * 8500^2))/(0.4^2) = 15300`

`a=y-bx = 47200-(15300 * 29) = 2830`

So, slope = 15300

Intercept = 2830

So, equation :
`y = 2830+15300x`

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?

`y = 2830+15300 * 3.3 = 53320`

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 =
`r^2 = (0.72)^2 =0.5184`