I have to group $A$ ($\bar X_1 =20, N_1=10$) and group $B$ ($\bar X_2 = 25, N_2=15$). Is there a way I can compare these two groups and get some statistical significance? I was thinking about using the $t$-test, but the standard deviation is not provided.

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    $\begingroup$ I don't suppose $\bar X_{1\&2}$ are mean counts of Poisson-distributed variables...? $\endgroup$ Jun 6 '14 at 22:03
  • $\begingroup$ It is payment received for group of people for doing same things for example . Doing laundry. If you have 10 workers doing laundry in 10 different places. You have there average earnings and you want compare if their incomes are statistical significant. It got difficult for me because some of them have very large discrepancy in average payment ( $10 vs $100 but 0 STDEV) $\endgroup$ Jun 6 '14 at 22:18
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    $\begingroup$ @Nick That was a great question to ask. A lot of people tend to give up on this sort of question much too easily, while your response covers a case that often crops up, and for which there is a practical solution. $\endgroup$
    – Glen_b
    Jun 7 '14 at 5:35
  • $\begingroup$ @statlearner You say something about standard deviations in your comment - if you have the standard deviations (even if they're zero as you suggest in your comment), why wouldn't you mention that? Is there anything you do know about these particular samples that you haven't said? $\endgroup$
    – Glen_b
    Jun 7 '14 at 5:36

I ended up with what I think is a practically useful suggestion (1(b)), and some other ideas that might be pursued if that's not suitable.

1) We can make a parametric assumption (though we cannot check it for reasonableness). We might, for example, assume that the payments are exponentially distributed - we can certainly base a test off that if we know the means and sample sizes. The sensitivity to that assumption may be an issue though. At least it's something. (If we're prepared to assume a shape - which might be garnered from external data of similar type, say - we might even be able to use a gamma rather than an exponential.)

Now I don't know that an exponential distribution will be a great approximation, but I'll give a suitable statistic for that case, and then suggest something that may work better - better enough as to perhaps be actually useful.

a) Exponential assumption. If we assume both samples are exponential, that all observations are independent, and that under the null they have the same mean, then

$$\frac{\bar x_1}{\bar x_2}\sim F_{2n_1,2n_2}$$

So you can do a (two-tailed) F-test of that hypothesis

(Indeed, with your sample, this would be $F_{30,20}=1.25$, and this would not lead to rejection at typical significance levels; but samples fifteen times as large would lead to rejection at the 5% level).

b) Here's what I think is the the better choice. If there's a minimum payment (such as in places with a minimum wage and on tasks where you can reasonably know or guess the time or minimum time), then a shifted exponential would probably be a pretty good approximation. If you know the minimum, $m$, say (and if it's a legislated amount, you probably do), then you can instead do:

$$\frac{\bar x_1-m}{\bar x_2-m}\sim F_{2n_1,2n_2}$$

And again the test is two-tailed.

(For example, if $m$, the minimum payment for this task, was say $17.50$, then for the data you have, the statistic would be significant at the 5% level.)

Note that equality of the original means implies equality of the mean above the minimum, so this still speaks to the original hypothesis. If you have the information to pursue this option, I suggest trying it; in the right circumstances, it may be possible to justify it as a reasonable approximation to the actual distribution.

c) Rather than exponential as in (a), we might assume a gamma distribution with some other shape parameter (exponential is gamma with shape parameter 1).

For example, if we assumed the shape parameter were 3 rather than 1, the test would work the same way but the degrees of freedom for the $F$ statistics (numerator and denominator) would also be multiplied by 3. So:

(i) this approach could accommodate any external shape information (such as for similar samples).

Let's say that with similar samples we concluded that the shape parameter was typically around 10. Then (for your data), the test statistic would still be 1.25 but the df would be around $(300,200)$ ... but the result would still not be significant at the 5% level.

(ii) this opens up the possibility of considering some worst cases - with a given statistic you could compute the smallest shape parameter for a gamma distribution that would lead to rejection. With a large ratio of means and enough observations, it would sometimes be the case that only very heavily skew payment distributions should the test fail to reject.

d) of course there are other possible parametric assumptions than the exponential/gamma; similar approaches to the above may be fruitful.

2) Even without such an assumption, there's at least something here we can work with: We know the data are positive (payment received for doing work; presumably zero is excluded, but if the payment can be zero we need also to allow for that). That does tell us something, since it places some sort of limits on where the data can be relative to the mean.

So for example, the Markov inequality tells us that for any non-negative $X$ and $t>0$,

$$\text{P}(X \geq t) \leq \frac{\text{E}(X)}{t}\,.$$

It may be possible to use an analysis similar to that here: Does a sample version of the one-sided Chebyshev inequality exist? and derive some kind of lower bound on tail probabilities for a difference in sample means.

If we can make an additional assumption, such as unimodality within the groups, we may be able to do more.

3) Indeed, since it's possible to do hypothesis testing with a single observation, at the very least one can take the difference in means and test that as a single observation. But in that case, the bounds are not likely to be useful very often.


Without some estimate of the standard deviation(s) you cannot do a statistical test. Perhaps you can find an estimate of the sd (for example, if the measure has been used elsewhere.

  • $\begingroup$ That is the main problem. it is not there. I am wondering if I can do any other test you know of . $\endgroup$ Jun 6 '14 at 21:56
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    $\begingroup$ You are going to need to get that information somewhere, or you are just stuck. Some distributions have SDs / variances that are functions of the mean, so if you knew if it were one of them (eg, Poisson), then you could figure it out. But basically you are out of luck. $\endgroup$ Jun 6 '14 at 22:16

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