T-distribution problem I am working through the exercises of Martin Sternstein's Statistics (Barron's College Review Series) ... struggling with Exercise 4 in Chapter 14.
Given are the weight lifting strengths of 9 people before/after 8 weeks of weight lifting exercises: Before: 3,3.5,4,6,7,8,8.5,12.5,15. After: 7,17,19,12,19,22,28,20,28.
Asked is to construct the 95% CI a) before training strength, b) after training strength and c) strength gain.
The answers are given at the back of the book as follows: a) 7.5 +- 2.262(4.085/sqrt(9)) b) 19.11 +- 2.262(6.791/sqrt(9)) and c) 11.61 +- 2.262(4.891/sqrt(9)).
My struggles:
a) given n=9, I thought df=8 and looked up 2.5% (due to two-sided) and got critical value of 2.306, but the book's answer uses 2.262 which corresponds to df=9. Anyone know why?
b) I understand both 4.085/sqrt(9) and 6.791/sqrt(9) for a) and b) but not the 4.891/sqrt(9). I had sqrt(4.085^2/9 + 6.79^2/9). Any ideas?

It seems you may still be puzzled by the CI for gain.
You have paired data. You need to start by getting the nine gains, and then
do a one-sample CI using those nine values.
Here is a bit of R code that computes the CI directly
and then again as part of the t.test procedure in R.
g = a-b
g
[1]  4.0 13.5 15.0  6.0 12.0 14.0 19.5  7.5 13.0

mean(g); sd(g)
[1] 11.61111
[1] 4.891177
qt(c(.025,.975),8)
[1] -2.306004  2.306004

CI = mean(d) + qt(c(.025,.975), 8)* sd(g)/3;  CI
[1]  7.85142 15.37080

t.test(g)$conf.int
[1]  7.85142 15.37080
attr(,"conf.level")
[1] 0.95

