A magazine wants to know if the online viewers are more than its printed copy buyers.

Which statistics test is the best to use, the t-test or Wilcoxon signed rank or Mann–Whitney U?

If I have less than $30$ days, is it correct that I should not use t-test since it is not a normal distribution?

$$\begin{array}{c|c|c|} \text{Days} & \text{Copy} & \text{Online}\\ \hline \text{1} & 3k & 5k \\ \hline \text{2} & 5k & 3k \\ \hline \text{3} & 6k & 9k \\ \hline \end{array}$$ $$...$$

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    $\begingroup$ A paired or repeated-measure t-test may be a good choice since "copy" and "online" give the measurements on the same magazine. $\endgroup$
    – JACKY88
    Jan 1, 2013 at 1:11
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    $\begingroup$ What data does the magazine have? If, for instance, we interpret the "$k$"s in the table as thousands and take the table to be a complete census of purchasers, then over these three days there were 14,000 copy buyers. This was less than the 17,000 online buyers: you don't need a test to determine that! $\endgroup$
    – whuber
    Jan 1, 2013 at 3:50
  • $\begingroup$ @whuber you are right , 17k > 14k , but I am looking for a significance not by chance $\endgroup$
    – tnaser
    Jan 7, 2013 at 16:14

1 Answer 1


Your data are paired, so the Mann-Whitney U test (which isn't) is immediately out.

We can easily argue for normality (the counts are large, if "$k$" means thousands, so near-normality would be reasonable), but the actual problem is that since for count data we expect the variance to be related to the mean, unless we assume that the mean isn't varying across days, we don't have constant variance, which might suggest some caution with the t-test; it will still have decent power, but its type I error rate may be somewhat affected.

You can argue for the Wilcoxon signed rank test, but there are several issues there that need to be thought about.

Note that there's also the sign test.

One could easily argue for a test that's appropriate for count data. So, for example, you could do a form of proportion test that takes account of the apparent one-tailedness of the hypothesis. (If you do a two-tailed test, you could also do it as a chi-square.)

Finally, one could also fit a Poisson, quasi-Poisson or Negative Binomial GLM.


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