Confidence intervals and α

Problem

Setup: a linear regression model with a random sample, the Gauss-Markov assumptions hold and the errors are normally distributed,

\[y_i=β_1+β_2x_i+ε_i i=1, \ldots ,n\]

Show that the 95% confidence interval for \(β_2\) is a subset of the 99% confidence interval for \(β_2\) .

Hint: Show that if \(c\) is in the 95% C.I. for \(β_2\) then \(c\) must be in the 99% C.I. for \(β_2\) .

Solution

If \(c\) is in the 95% C.I. for \(β_2\) then

\[-SE\left( b_2) \right.⋅t_{0.05/2,n-2}<c<SE\left( b_2) \right.⋅t_{0.05/2,n-2}\]

Since \(t_{0.01/2,n-2}>t_{0.05/2,n-2}\) we have \(-t_{0.01/2,n-2}<-t_{0.05/2,n-2}\) . Thus,

\(SE\left( b_2) \right.⋅t_{0.01/2,n-2}>SE\left( b_2) \right.⋅t_{0.05/2,n-2}\) and \(-SE\left( b_2) \right.⋅t_{0.01/2,n-2}<-SE\left( b_2) \right.⋅t_{0.05/2,n-2}\) . Therefore,

\[-SE\left( b_2) \right.⋅t_{0.01/2,n-2}<c<SE\left( b_2) \right.⋅t_{0.01/2,n-2}\]

The argument is sort of similar to stating that if \(-2<c<2\) then it must follow that \(-3<c<3\) .