Question

Suppose I want to estimate the effect of exercising on GPA. I collect a random sample of data. For each individual, I collected the current GPA (variable GPA measured 0-4) and an average number of hours spent exercising during a week (variable excer in hours).

a) Write down the population model.(2pts)

b) Suppose the sample covariance between GPA and exercising is 100 and suppose that the variance of the variable excer is 200. Also, it was calculated that the average GPA in the sample is 3.0 while the average number of hours individuals spent exercising per week is 2 hours. Write down the estimated regression. (5pts)

c) Given ??(? 0 ̂ ) = 0.2 ??? ??(? 1 ̂ ) = 0.1, construct 95% level confidence intervals for both the intercept and the slope.(5pts) 6

d) Interpret the coefficients. (5pts)

e) What is the predicted value of the GPA for an individual who spends 3 hours exercising per week? What is the residual for this individual given that you know that the individual’s GPA is 2.5? Is the estimated regression under- or over-predicating GPA for this individual? (5pts) 7

f) Is there evidence that the true effect of exercising on GPA is 0.55?(2pts)

Answer 1

For shorthand, we use the following variable names: current GPA = cgpa, and the average number of hours spent exercising during a week = excer.

a) Population model: cgpa = B₀ + B₁excer + u ; where u is the population error term

b) We can estimate B₀ and B₁ using the following formula:

B^{^}₁= (Sample covariance of cgpa and excer)/(Sample variance of excer) = 100/200 = 0.5

B^{^}₀ = mean(cgpa) - B^{^}₁.mean(excer) = 3 - 0.5*2 = 3 - 1 = 2

Therefore, we have the estimated equation:

c) Confidence interval formula: CI = B^{^}_j +/- c.se(B^{^}_j) ; where c is the critical value at the level of significance (at 95%, c=1.96)

CI for B^{^}₀ = 2 + 1.96*0.2 and 2 - 1.96*0.2 = 2.392 and 1.608

Hence CI for B^{^}₀ is [1.608, 2.392]

CI for B^{^}₁ = 0.5 + 1.96*0.1 and 0.5 - 1.96*0.1 = 0.696 and 0.304

Hence CI for B^{^}₁ is [0.304, 0.696]

d) The coefficient on B^{^}₁ shows that a one hour increase in exercising in the week leads, on average, to a 0.5 point increase in the GPA. Alternatively, it would mean that a 2 hour increase woule lead to a 1 point increase in the GPA. This also shows that we might want to add a quadratic term of hours exercised in the week to see at what point will this average increase in GPA change direction or stop.

The coefficient on B^{^}₀ , of little interest here, means that the expected value of current GPA, when hours exercised in the week is zero, is 2.0.

e) E(cgpa | excer = 3) = 2 + 0.5(3) = 2 + 1.5 = 3.5

Actual GPA = 2.5. Hence, the model overpredicts the GPA of the individual by one point, that is the residual is negative (hence, overprediction).