I determined some parameter in a group of subjects under two conditions, such that each subject was tested in both conditions. I want to know if the condition has an effect on the parameter. My design is obviously paired. However, previous experience with this experiment makes me believe that the subject's response is independent during each measurement. Thus, I assume that subjects are independent and not paired. Accordingly, I decide to analyze the data with an unpaired t-test. But out of curiosity, I also run a paired t-test, with the following results:

P value for the paired T-test is 0.08 (no difference). P value for the unpaired T-test is 0.04 (significant difference).

I am confused because I thought that a paired test should always give lower p-value (or equal in the worst case) than an unpaired test, by eliminating between-subject variability and increasing power.


1) When can a paired test produce higher p-value that an unpaired test? What does my result mean in terms of sources of variation? It seems that pairing the subjects not only does not eliminate between-subject variability, but adds some kind of a new variability.

2) Can this result tell me something about my assumption that subjects are independent? Can it invalidate this assumption?

3) In summary, is it always a matter of choice to pair or not to pair subjects during analysis? Or is it formally incorrect to use an unpaired test when subjects are logically paired?

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    $\begingroup$ "Eliminating between-subject variability" usually is impossible; surely you mean to refer to reducing that variability. But a reduction occurs only when the pairs are positively correlated--not when they are negatively correlated! $\endgroup$
    – whuber
    Commented Oct 29, 2014 at 20:17
  • $\begingroup$ @whuber, I might describe a paired t-test as analyzing 'scores conditional on the subject'. In that context it seems reasonable to me to say that the b-s variability has been eliminated--it's certainly still there in the data but not in the data as analyzed. Re: -cor, that seems to be the answer, why not make it official? $\endgroup$ Commented Oct 29, 2014 at 20:33
  • $\begingroup$ @gung This terminology always makes me think for a minute: isn't "between-subjects" variability the component attributable to variation from one subject to another, while "within-subjects" is the component specific to each subject? Regardless, why would it be valid to suppose that either one can be eliminated altogether? I guess I don't understand the distinction you are making between being present "in the data" and being present in the "data as analyzed." BTW, correlation is not the entire story: it has to be balanced against the effect of degrees of freedom. $\endgroup$
    – whuber
    Commented Oct 29, 2014 at 20:44

1 Answer 1

  1. As @whuber says in the comment above, when the measures are negatively correlated, the p-value can be lower in the unpaired test than in the paired test. Here's an example where there is a difference:

    s  <- matrix(c(1, -0.8, -0.8, 1), 2)
    df <- mvrnorm(n=100, mu=c(0, 0.3), Sigma=s, empirical=TRUE)
    t.test(df[,1], df[, 2], paired=FALSE)
    t.test(df[,1], df[, 2], paired=TRUE)

    The first test (unpaired) gives p=0.035, the second gives p=0.117.

  2. Yes, this is a design issue. This book chapter discusses it: Keren, G. (2014). Between-or within-subjects design: A methodological dilemma. A Handbook for Data Analysis in the Behaviorial Sciences: Volume 1: Methodological Issues Volume 2: Statistical Issues, 257, which you can read some of on Google books.

  3. Hmmm... I'm not sure. I'd do a simulation to find out the effect on the type I error rate. How this affects your power is a separate issue that I haven't looked into here. Slight adaptation of my previous code:

    paired   <- rep(NA, 1000)
    unpaired <- rep(NA, 1000)
    for(i in 1:1000){
          df          <- mvrnorm(n=100, mu=c(0, 0), Sigma=s, empirical=FALSE)
          unpaired[i] <- t.test(df[,1], df[, 2], paired=FALSE)$p.value
          paired[i]   <- t.test(df[,1], df[, 2], paired=TRUE )$p.value
    sum(paired < 0.05)
    sum(unpaired < 0.05)


    > sum(paired < 0.05)
    [1] 46
    > sum(unpaired < 0.05)
    [1] 137

Well look at that. If you treat them as unpaired, your type I error rate rockets. You need to treat them as paired to get the right answer. I believe (it's a long time since I've read it) that this is one of the issues Keren talks about in that chapter. If you're going to have data that might be negative correlated (e.g. amount of soup and amount of burgers someone eats) you'll have more power with an unpaired design.

  • $\begingroup$ I beg to disagree with your interpretation of my comment: for the p-values to be reversed, it is necessary for there to be negative covariance; but this is not sufficient. The negative covariance has to overcome the loss of degrees of freedom in the paired t-test. $\endgroup$
    – whuber
    Commented Oct 29, 2014 at 21:05
  • $\begingroup$ What should we make of your proposed answer to #1 in light of your proposed answer to #3? $\endgroup$ Commented Oct 29, 2014 at 21:26
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    $\begingroup$ @JeremyMiles Thanks for the detailed answer. I read the chapter from the book, but I couldn't find a more or less direct answer to my question. The core of my question is "can you treat paired subjects as unpaired", while the chapter just compares true within-subject and between-subject designs in general. $\endgroup$
    – Viktor
    Commented Oct 31, 2014 at 15:20
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    $\begingroup$ Regarding your answer, let me just confirm if I understand it correctly. Your conclusions seems to be: 1) My results show that the measurements of one subject in two different conditions cannot be considered independent, because they are (negatively) correlated; 2) thus, paired test must be used, and provides "more true" measure of the treatment effect. Is that correct? $\endgroup$
    – Viktor
    Commented Oct 31, 2014 at 15:21
  • 1
    $\begingroup$ Yes, I think you need to treat them as paired. $\endgroup$ Commented Oct 31, 2014 at 16:51

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