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Participants (N=14) filled in two questionnaires at two points of time, and I want to see the correlation between the variables.

Is this sample size sufficient?

Also, should I have reverse coded the items before running the correlation analysis?

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  • $\begingroup$ Are you interested in the correlation between each variables/items or a composite score (e.g., sum or mean score) derived from them? $\endgroup$ – chl Oct 16 '12 at 21:46
  • $\begingroup$ I'm looking at the correlation between each item. $\endgroup$ – Julie Oct 16 '12 at 21:52
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    $\begingroup$ "Is this sample size sufficient?" ... sufficient to do what, exactly? You can compute correlations with fewer data points. And why would you reverse-code anything? $\endgroup$ – Glen_b -Reinstate Monica Oct 16 '12 at 21:54
  • $\begingroup$ Then reverse scoring doesn't matter, but we'll probably want to know how many items you have. $\endgroup$ – chl Oct 16 '12 at 21:54
  • $\begingroup$ I meant is the sample size sufficient for correlation analysis? $\endgroup$ – Julie Oct 16 '12 at 21:59
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Correlations are valid with fewer than 14 points, but they become more accurate with more data. Exactly how much more accurate depends on various things, but if the data are bivariate normal, the standard error is approximately

$SE(r) = \frac{(1-\rho^2)}{\sqrt{n-1}}$

However, with 14 points, the data clearly aren't going to be exactly bivariate normal. Still the denominator term should give you a rough idea: Accuracy goes up as the square root of N; with 4 times as many subjects, it's twice as accurate. And it's more accurate the higher the population correlation.

There is no reason to reverse code, that will just change the sign of the correlation.

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  • $\begingroup$ Peter, what does "However, with 16 items, the data clearly aren't going to be exactly bivariate normal." mean? This seems to suggest the joint distribution of the data might depend on the sample size... (as a secondary point, the wording is confusing since clearly a 16-dimensional vector can't be bivariate normal. I assume what you meant was that each pair of variables is bivariate normal, in which cases it suffices to say the data is multivariate normal) $\endgroup$ – Macro Sep 11 '13 at 14:52
  • $\begingroup$ @Macro The formula I gave for the SE depends on bivariate normality. I guess I should say 16 subjects instead of items? Participants? $\endgroup$ – Peter Flom Sep 11 '13 at 15:06
  • $\begingroup$ Hi Peter- it looks like 14 is the sample size and, in the last comment to the main question, there are 16 observations made (8 pre, 8 post). This seemed to be what was reflected in your post. (just saw how old this post was - sorry for digging this uP!!) $\endgroup$ – Macro Sep 11 '13 at 15:29
  • $\begingroup$ Oh, OK. So 16 should be 14, I'll fix that. $\endgroup$ – Peter Flom Sep 11 '13 at 15:35
  • $\begingroup$ Peter, my previous comments still apply to your modified statement: However, with 14 points, the data clearly aren't going to be exactly bivariate normal. As I said before, this wording makes it seem as though the joint distribution depends on the sample size. (again, very secondarily, the data isn't bivariate). $\endgroup$ – Macro Sep 11 '13 at 16:09

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