A 1 \\\\alpha confidence interval for the mean\ response at a predictor value xi is:\ Yi \\\\pm t\\\\alpha /2\ np1\ \ \\\\sigma 2x\ i(xTx)1xi\ Here youll derive this formula. The following steps walk you through the derivation; each step can be\ answered in just 1-3 lines (usually 1 line).\ i. Write the distribution of Yi E[ Yi].\ ii. Let Z = c\ ( Yi E[ Yi]\ )\ ; find the value of c such that Z N (0, 1). (You do not need to prove that\ Z N (0, 1) for the value of c you identify; just find and state the correct c.)\ iii. Let W = np1\ \\\\sigma 2 \\\\sigma 2 and \\\ u = n p 1. Write an expression for 1W/\\\ u .\ iv. Let T = ZW/\\\ u . Write T in terms of \\\\sigma 2, x\ i(xTx)1xi, Yi, and E[ Yi]. What is its distribution?\ v. Note that if T t\\\ u :\ 1 \\\\alpha = P\ (\ t\\\\alpha /2\ \\\ u <= T <= t\\\\alpha /2\ \\\ u \ )\ Substitute the proper quantities for \\\ u and T from above and rearrange to show that:\ 1 \\\\alpha = P\ (\ Yi t\\\\alpha /2\ np1\ \ \\\\sigma 2xi(xTx)1xi <= E[ Yi] <= Yi + t\\\\alpha /2\ np1\ \ \\\\sigma 2xi(xTx)1xi\ )\ 3