Question

Call: lm (formula = GPA ~ HSGPA + SAT, data = GPA) Residuals: 0.97894 -0.27639 0.02867 0.30133 0.87956 Coefficients:l 3Q Min 10 Median Max Estimate Std. Error t value Pr (>ItI) (Intercept) 0.6351217 0.2955033 2.149 0.03272 HSGPA SATV 0.4975320 0.0750569 6.629 2.66e-10 0.0012283 0.0003373 3.641 0.00034 Residual standard error 0.4061 on 216 degrees of freedom Multiple R-squared: 0.246, Adjusted R-squared: 0.239 F-statistic: 35.23 on 2 and 216 DF, p-value: 5.711e-14

1.) Interpret the estimate of the intercept Bo

2.) Estimate the expected GPA for a student with a HS GPA of 3.5 and a verbal SAT score of 600, and construct a 95 percent confidence interval for the expected GPA.

3.) Predict the GPA for a new student with a HS GPA of 3.5 and a verbal SAT score of 600, and construct a 95 percent prediction interval for the new student’s GPA.

Answer 1

1)

The Summary of multiple regression states:

Bo = 0.635; B1 = 0.497; B2= 0.001

GPA ~ HSGPA + SATV

2)

GPA = 0.635 + 0.497 * HSGPA + 0.001*SATV

given HSGPA = 3.5 and SATV = 600

GPA = 0.635+1.739+0.6 = 2.974

95%confidence iterval

predict(data, GPA, interval="confidence")

*data <- lm(GPA ~ HSGPA + SATV)

3)

GPA = 0.635 + 0.497 * HSGPA + 0.001*SATV

given HSGPA = 3.5 and SATV = 600

GPA = 0.635+1.739+0.6 = 2.974

predict(data, GPA, interval="confidence")

*data <- lm(GPA ~ HSGPA + SATV)