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I'm attempting to compute statistical significance for two data sets. I have the mean and the number of data points, but I don't think I can compute std deviation, because I don't have the individual data points.

Under previous teacher, 1,072 tests given with average score of 27.7 correct.

New teacher, 179 tests given with average score of 28.6.

Seems self-evident to me that combination of small uptick and tiny sample size means that conclusion "new teacher responsible for improvement" is untenable, but I'd like to be able to say that the improvement is not statistically significant.

All help appreciated.

Matt
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    It might matter whether it's $27.7$ out of $100$ or $27.7$ out of $30$. In the latter case, the standard deviation couldn't be very big by comparison to what it might be in the former case. – Michael Hardy Jul 10 '15 at 04:13
  • Maybe some non-parametric test? – Gary. Jul 10 '15 at 04:18
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    I don't see how you could answer this question with no information about the variability in the data. That's the whole idea behind significance testing, comparing "natural" variation to potentially systematic variation. For all we know the score was always exactly $28.6$ for the new teacher and always $27.7$ for the old. – dsaxton Jul 10 '15 at 04:55
  • Following up on Comment by @MichaelHardy: I assume the two teachers used the same type of test, otherwise comparison is inappropriate. Do you know anything more about the type of tests given. If so many of them were given by two teachers, there must be some information. Knowing or being able to deduce something about the population SD might be enough. NO test of significance is possible without some information about variability. (Nonparametric tests are not "magical", they simply don't require data to be normal, or from other known distribution family.) – BruceET Jul 10 '15 at 16:47
  • @Bruce Trumbo: But you can do, i.e., a contingency table without (th magic of ) knowing neither the pop $\sigma$ nor the standard error. – Gary. Jul 10 '15 at 23:32
  • @Gary: Even a 2x2 contingency table requires some additional info about data to find counts. For example, it might be logical to have bins above and below 28.2 for each teacher, but I don't see how to get the four cell counts for that. And averages may not be even close to medians.. – BruceET Jul 11 '15 at 01:29

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