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I have two samples d1 and d2:

d1 <- c(18,21,16,22,19,24,17,21,23,18,14,16,16,19,18,20,12,22,15,17)
d2 <- c(22,25,17,24,16,29,20,23,19,20,15,15,18,26,18,24,18,25,19,16)+5
d  <- d2-d1

In order to perform a paired two-tail t-test I use the following code on the differences d:

standard_error <- function(vals){
  return(sd(vals) / sqrt(length(vals)))
}

t_test <- function(d){
  t <- mean(d) / standard_error(d)
  return(2*pt(abs(t), length(d)-1, lower=F))
}

This gives me that the p-value is 9.37919e-10, so quite a significant result. As an alternative I tried using t.test-function in R:

t.test(d1, d2, paired=T)

This gives me the same as above, so everything is OK here.

However, now I want to perform a one-tailed t-test for if the mean of d2 is larger than that of d1. Is it correct that I can achieve this by simply returning pt(abs(t) in the t_test-function above, i.e., divide it by 2?

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    $\begingroup$ @user2974951 As a general rule it can't be correct, because that would never give a p-value above 0.5 $\endgroup$
    – Glen_b
    Commented Sep 27, 2018 at 16:07
  • $\begingroup$ @Glen_b That is a good point. What is the correct (i.e., most general way) to do it? $\endgroup$
    – N08
    Commented Sep 27, 2018 at 16:48
  • $\begingroup$ What happens when there's a one-tailed hypothesis test and the results are in the other tail? ... the answer there seems to cover your question so this may in effect be a duplicate $\endgroup$
    – Glen_b
    Commented Sep 29, 2018 at 6:28

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