A review problem for one of my daughter's statistics classes: The arm span and height of 10 individuals is measured in order to test the hypothesis that arm-span equals height in humans. Perform an hypothesis test to determine if there is a significant difference between the mean of the two groups.

I am not sure if this is a match pairs data set and therefore not sure if one should perform a single sample t-test on the difference (for a matched pair) or a 2-sample t-test for the difference between the means of two independent data sets.

I am inclined to believe that this is NOT a matched pairs data set even though the two measurements are taken from the same person. Just because there is a high correlation between the two does not mean that they are dependent.

I understand that if you take measurements from a single group on two occasions (e.g., weight now and 6 months from now) it is a matched pairs data set. But, I'm looking for a general description of "matched pairs" that would help us distinguish this (and similar cases) in the future.

  • $\begingroup$ t-test cannot be used in this situation. simple regression is needed. $\endgroup$ – user158565 Dec 9 '18 at 22:05
  • $\begingroup$ @user158565 you should explain why (in some detail) you believe this to be the case, perhaps in an answer if there's insufficient space in a comment. $\endgroup$ – Glen_b -Reinstate Monica Dec 9 '18 at 22:33
  • 2
    $\begingroup$ I'd have marked this as a duplicate (e.g. of this) if not for the 'matched' terminology issue and the potential discussion with comparing two quite different quantities that user158565 is raising. $\endgroup$ – Glen_b -Reinstate Monica Dec 9 '18 at 23:20

Just because there is a high correlation between the two does not mean that they are dependent.

Any non-zero correlation in the population is proof of dependence, but you don't need to look at the correlation in the sample to infer that this is matched (on person).

I'm looking for a general description of "matched pairs"

I guess that terminology may differ across books, but to my mind a matched pair is not what you have here. (Matched pairs are where you identify similar subjects/experimental units and associate them together e.g. matching on size, age and gender -- see the first paragraph here - that is, take an action to identify pairs that will be similar on any variables of importance in order that they can be used as blocks). Some definitions of 'matched pairs' don't seem to follow this convention, though (a pity because this is a useful distinction for which they'll now need a new term).

You have paired data, certainly, but they're naturally occurring pairs. The fact that they're both from the same person is the giveaway. [If your book would call values from twins a 'matched pair' then this would be a matched pair of values.]

Paired of values will tend to be more alike within a specific pair than across them. In this case the "person" is giving you the specific (height, arm-length) pair.

People vary in general size, so if height and arm-length tend to be similar (whether or not the null is actually true) they will tend to be more similar within one person than they would be across people (person A's height will be less similar to person-B's arm-length than to her own).

These inherent 'pairs' are what is giving you 'paired data'.

| cite | improve this answer | |

You have paired data, but a paired t-test is not very suitable (unless the question explictly asks you to compare means but that is not the same as "arm span equals height")

  • You have paired data when two variables are measured in the same person/unit each case/measurement. E.g. you can say you measured ten pairs of (armspan,height).

    If you wish to test $\mu_X = a + \mu_Y$ then you can also test $\mu_{X-Y} =a $ which often is more precise since there is a smaller variance. E.g. the variables X and Y have large variance due to smaller and larger people. But the difference X-Y has smaller variance because within a large/small person X and Y are both large/small and you eliminate some of the variation by taking the difference.

  • Like mentioned in the comments a t-test may not be suitable. A t-test compares equality of the means. But possibly you wish to prove the linear relation X=Y. E.g if you have the following data you have equal means of X and Y but not really equal X and Y


  X   Y
  1   5
  3   6
  5   7
  7   8
  9   9
 11  10
 13  11
 15  12
 17  13
| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.