# Problem with paired t-test t value

I sucessfully find the standard error of difference, but can't get the t value right.

According to this calculator , I am supposed to get a sd of 0.092, which I do.

But for some reason, when I do $t = \bar{d} / (sd / \sqrt{11}$ I get 11 instead of 3.xxx

Here is the answer, but I can't figure what I've done wrong about the t calculation.

Answer: Let μ = μ1 − μ2 = μwithout − μwith = the difference between the average completion times of the puzzle. Hypotheses are: H0 :μ = 0 vs Ha: μ < 0.

Using paired t-procedure, t =-3.49 and p-value < 0.005 (less than α = 0.01) Conclusion: The data provide sufficient evidence to indicate that it requires more time, on average, to complete the puzzle after consuming alcohol.

• your sd is wrong. What you have there is variance. – Glen_b Apr 24 '14 at 1:38
• The calculator was NOT telling you that the sd was 0.92, it actually tells you that something else is about 0.92 (the standard error of the difference). – Glen_b Apr 24 '14 at 1:46

Perhaps you are dividing by n and not root(n)

> a<-c(7.1,6.3,6.8,8.4,6.9,8.5,7.3,7.7,8.1,7.4,6.6)
> b<-c(7.4,6.2,6.6,9.3,7.2,8.8,7.6,7.9,8.7,7.9,7)


if you divide by the correct root n you get:

> (mean(a)-mean(b))/(sd(a-b)/sqrt(length(a)))
[1] -3.486083


if you divide by just n you get

> (mean(a)-mean(b))/(sd(a-b)/(length(a)))
[1] -11.56203


also note that sd(a-b) is not 0.092

> sd(a-b)
[1] 0.302715


The Variance is 0.092

> var(a-b)
[1] 0.09163636