Hi, I do not understand the formula for the width of the confidence interval used here: 2 * t_2.5%;2n-2 * sqrt(2/n) * pool variance So normally when we have seen width questions in this chapter it is referring to 2t_(a/2 %; n -1) * sigma/n. i do not understand here why we use: sqrt(2/n) * pool variance Additoinally, why do we use 2n - 2 degrees of freedom in paired values?
Hi Rayhan I think you are talking about Q9.10 in the Course Notes. You haven't given us a full reference, so i'm sort of guessing here. Here we are using the 2-sample t result, which is given at the top of p23 of the Tables. Using this result, a 95% CI for mu1-mu2 is given by: x1(bar) - x2(bar) +/- upper 2.5% pt of t(n1+n2-2) * sqrt(pooled var *(1/n1 + 1/n2)) In this question n1 = n2. Let's call this n for simplicity, in which case the formula above can be written as: x1(bar) - x2(bar) +/- upper 2.5% pt of t(2n-2) * sqrt(pooled var *(2/n)) Does this help? Julie
