Question

1. (Based on Stock & Watson “Introduction to Econometrics” 6th ed., Exercise 4.1 and 5.1.) Suppose that a researcher, using data on class size (CS) and average test scores from 100 third-grade classes, estimate the simple linear regression: TestScore d = 520.4−5.82 × CS, n = 100, R^2 = 0.08. (20.4) (2.21)

(d) Name one factor in the error term and discuss its correlation with class size and average test score.

(e) Construct 95% and 90% confidence intervals for β1.

(f) Calculate the p-value for the two-sided test of the null hypotheis H0 : β1 = 0. Do you reject the null hypothesis at the 95% level? At the 1% level?

(g) Calculate the p-value for the hypotheses H0 : β1 = −5.6 v.s. H1 : β1 6= −5.6. Without doing any additional calculations, determine whether −5.6 is contained in the 95% confidence interval for β1.

(h) Construct a 99% confidence interval for β0.

Answer 1

= 0.08 = 0.283 The correlation with class size and average test score is 0.283. CI = bt 98.0.025) SE (6) =-5.82 1.984x2.21 =-5.824.3846 =(-5.82-4.3846.-5.82+4.3846) =(-10.2046.-1.4354) The 95% confidence interval for the slope is -10.2046<B<-1.4354.

CI = 5 798,0.05) * SE (6) =-5.82 1.661x 2.21 =-5.82 =(-5.82-3.6708.-5.82 +3.6708) =(-9.4908.-2.1492) The 95% confidence interval for the slope is -9.4908<A<-2.1492 SE(5) --5.82-0 = 221 = -2.63 The p-value is 0.0099. Reject the null hypothesis at the 95% level and at the 1% level.

t-5-6 SE (6) -5.82-(-5.6) 2.21 -5.82 +5.6 2.21 =-0.10 The p-value is 0.9205. Do not reject the null hypothesis at the 95% level. Yes, -5.6 is contained in the 95% confidence interval for A. (h)

CI = b, 88.000 SE(b) = 520.4+2.627 x 20.4 = 520.4+53.5908 =(520.4-53.5908,520.4+53.5908) = ( 466.8092,573.9908) The 99% confidence interval for the intercept is 466.8092 <B, < 573.9908.