I am trying to determine whether the means of two Gamma distributions are significantly different. To do this, I am trying to determine the Wald Statistic as


and thence getting the F value using


where q is the number of degrees of freedom. My first problem is determining the standard error of the mean.

Is the standard error of the mean of a gamma distribution simply the standard deviation divided by the square root of the count, as with a normal distribution? I have tried to find the answer for this using

"standard error" "gamma distribution"

and found this which initially suggested that the standard error is the same as the variance and was then corrected to say that it is the same as the standard deviation (which would not take the count into account).

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    $\begingroup$ Distributions don't have standard errors, they have standard deviations. A standard error is the standard deviation of the sampling distribution of some statistic, but you don't state any statistic. Please clarify your question. If it's the standard error of the mean you seek, the distribution only matters in so far as its variance exists. $\endgroup$
    – Glen_b
    Apr 2, 2014 at 20:38
  • $\begingroup$ Yes. It is the standard error of the mean. Thank you for pointing that out. I have obtained the SD as the square root of [the shape divided by the square of the rate]. Should I divide that by the square root of the count to get the standard error? Thanks, $\endgroup$ Apr 2, 2014 at 21:27
  • 2
    $\begingroup$ See here for a derivation of the std error of the sample mean that will apply for any distribution where variances exist. $\endgroup$
    – Glen_b
    Apr 2, 2014 at 21:30
  • 3
    $\begingroup$ If you know the parameters already, you have the whole population distribution already specified. What would you need standard errors of sampling distributions for? Can you explain the underlying problem? $\endgroup$
    – Glen_b
    Apr 2, 2014 at 21:31
  • $\begingroup$ I have expanded my question to better explain my problem. Thanks, $\endgroup$ Apr 2, 2014 at 21:54


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