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I performed an independent groups t-test and got a t value of 4.48. Each group is of 26 and I used a criterion of 0.05 (just because...).

Before doing the t-test I had to transform my values because both groups weren't of equal variance (found by calculating variance and S.D.). To transform, i used log10. Put values in histogram and it isn't a t-curve. What did I do wrong? Why don't I have a proper t distribution?

LostStatsStudent
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    Note that a t-test applied to the logarithms addresses a *different hypothesis*: it is concerned with whether the two populations have different geometric means, rather than whether they have different arithmetic means. But in no case does the t-test assume the population (or the sample) follows a t distribution! – whuber Dec 01 '14 at 16:52

2 Answers2

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The assumptions of t-test is that if two population are compared they:

  • both should follow normal distribution but this is not a problem with big sample size,
  • both should have the same variance - for unequal variances Welch t-test could be used.

They do not have to follow the t-distribution.

See also this discussion on t-tests for data with unequal variances and on log transformations of data.

Community
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Tim
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    Lack of normality is not necessarily a problem, particularly if the sample sizes are large. If variances are suspected to be unequal a good alternative to the standard t test is the Welch t test which corrects the degrees of freedom to allow for the unequal variances. – Silverfish Dec 01 '14 at 12:09
  • Yes, but in this case sample sizes are small. – Tim Dec 01 '14 at 12:10
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    Yes, though with a total sample size of over 50 in this case, that's still fairly forgiving of small deviations from normality. My main point was that your statement as it is currently written seemed too sweeping - it isn't true in general that we demand equal variances and exact normal distributions before we can countenance a t-test. – Silverfish Dec 01 '14 at 12:15
  • Related discussion: http://stats.stackexchange.com/q/9573/22228 – Silverfish Dec 01 '14 at 12:24
  • So I changed the wording that it is more clear. – Tim Dec 01 '14 at 12:24
  • I understand that Welch's t test may be the better solution, but is transforming my data with log10 just as good? Or is Welch's t test the only way that, in my case, I will obtain a t distribution? @Tim – LostStatsStudent Dec 01 '14 at 12:25
  • @Silverfish I understand that Welch's t test may be the better solution, but is transforming my data with log10 just as good? Or is Welch's t test the only way that, in my case, I will obtain a t distribution? – LostStatsStudent Dec 01 '14 at 12:30
  • The appropriate way of dealing with unequal variances is Welch t-test. – Tim Dec 01 '14 at 12:32
  • The adequacy of the approximation of the sampling distribution of the test statistic is *far less important* than ensuring that the t-test has proper calibration. Proper calibration of the t-test is had for *nearly any* sample size. Simply put: there are virtually no issues with sample size in using the t-test as a measure of association. Every other inconsistency with the approximation can just be chalked up to subpar power, but that may be preferable to making large and fallible assumptions about the data. – AdamO Dec 01 '14 at 14:05
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You should be using the t-test with an unequal variance assumption. Roughly put, this should handle virtually any of the discrepancies in the data and give you a well powered test of mean difference between the two groups.

We do not expect the data themselves to appear to have a t-distribution, but we expect that the sampling distribution of the test statistic will have a t-distribution. In the words of Student himself in the first line of his article leading to the eponymous test, "Any experiment may be regarded as forming an individual of a population of experiments which might be performed under the same conditions." You do not observe this population, you only make inference to its distribution based on the nature of the statistic you compute.

AdamO
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