Question

Using output given below,

(a) Set-up simultaneous 95% confidence intervals for Bo and B1. (b) Set-up a 95% confidence interval for u{5.1}= expected vo at a speed of 5.1. (c) What is the covariance between bo and by? What does it mean?

Answer 1

$(a)~\\ 95\%~C.I.~for~\beta_0~is\\ \left(\hat{\beta}_0-t_{0.025,4}se(\hat{\beta}_0),~\hat{\beta}_0+t_{0.025,4}se(\hat{\beta}_0) \right )=(-65.1282,~ -19.2814)\\ where,~\hat{\beta}_0=-42.20476,~se(\hat{\beta}_0)=8.25652,~t_{0.025,4}=2.7764\\ 95\%~C.I.~for~\beta_1~is\\ \left(\hat{\beta}_1-t_{0.025,4}se(\hat{\beta}_1),~\hat{\beta}_1+t_{0.025,4}se(\hat{\beta}_1) \right )=(10.7073,~20.2070)\\ where,~\hat{\beta}_1=15.45714,~se(\hat{\beta}_1)=1.71079,~t_{0.025,4}=2.7764.\\ (b)~\bar{x}=sample~mean~of~speed=(3.5+4.0+4.5+5+5.5+6)/6\\ =4.75\\ \hat{\mu}\{5.1\}=-42.20476+15.45714*5.1=36.6267\\ 95\%~C.I.~for~\mu\{5.1\}~is\\ \left(\hat{\mu}\{5.1\}-t_{0.025,4}\sqrt{\frac{MSE}{4}+(se(\hat{\beta}_1))^2(5.1-\bar{x})^2},~\hat{\mu}\{5.1\}+t_{0.025,4}\sqrt{\frac{MSE}{4}+(se(\hat{\beta}_1))^2(5.1-\bar{x})^2} \right )\\ =(31.3884 ,~41.8650).\\$

$(c)~Cov(b_0,b_1)=-\bar{x}(se(\hat{\beta}_1))^2=-4.75*(1.71079)^2=-13.9023\\ b_0~and~b_1~are~negatively~correlated~if~\bar{x}>0\\ b_0~and~b_1~are~positively~correlated~if~\bar{x}<0\\ b_0~and~b_1~are~uncorrelated~if~\bar{x}=0\\$