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I have two matched groups consisting of 22 observations each for a Poisson distributed variable. I am interested in evaluating the power to detect a 20% decrease in the outcome for the second group. I have no idea of the correlation between matched groups and hence will have to assume independence. I am planning on evaluating the power using a simulation study and have the following ideas in mind.

1) Define rate0 and rate1 = rate0-(0.20*rate0)

2) Simulate X1~Poisson(22,rate0) and X2~Poisson(22,rate)

3) Compute pairwise differences X1-X2 and conduct a paired Wilcoxon test or a paired ttest.

4) Conduct simulation nsim times and compute the times pvalue < 0.05

I would appreciate some feedback/suggestions regarding step 3. Would anyone have any other ideas?

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1 Answer

The power analysis by simulation is ok, I think what you’re really asking for is a way to compare matched Poisson variables, other than Wilcoxon test or paired t-test.

  • A brute force approach would be: use as test statistic $\sum_i X_{1i} - X_{2i}$ ; assume H0 (same rate in two groups), estimate the common rate $\lambda$ using pooled data, and simulate N (N = big) times two groups of 22 variables $\sim \mathcal P(\lambda)$ to get an empirical distribution of you test statistic.

  • If your rates are big enough, you could also use the fact that if $X\sim\mathcal P (\lambda)$, then the distribution of $2 \sqrt X$ is approximated by a normal $\mathcal N(2\sqrt\lambda,1)$. This leads to normal based test (the distribution of $X_{1i} - X_{2i}$ is known under $H_0$, independently of $\lambda$), and can lead to a nice paper-pen computation to the power. The figure below illustrate the (in)accuracy of this approximation for various $\lambda$ (in black, the cdf of $2 \sqrt X$, in red, the normal approximation).

cdf

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+1 Good answer. Simulations suggest that a paired t-test of $\log(X+1/2)$ may be slightly more powerful still, especially for small rates $\lambda$. (The result is not very sensitive to the offset $1/2$; it's there to handle zero values of $X$.) –  whuber Jan 2 '12 at 15:33
    
An additional comment is that in all these tests, the null hypothesis is in fact $X_{1i}, X_{2i} \sim \mathcal P(\lambda_i)$: the rate is the same for the two members of the pair, but may differ from one pair to an other. –  Elvis Jan 2 '12 at 16:59
    
That seems to be a generalization of the question itself, Elvis, and is much more problematic to solve. The question refers to "22 observations each for a Poisson distributed variable." Although that could be read as allowing the parameter to vary from one observation to another, that would be an unusual interpretation. –  whuber Jan 2 '12 at 17:51
    
Thank you all. This has been extremely helpful. I am concerned about transforming mainly because of the difficulty in determining the standard errors for the mean difference estimate in its original scale. –  user1106711 Jan 2 '12 at 21:50
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@user1106711 I am glad this helped. I have been a little cryptic on the construction of the test: please consider that as each $2\sqrt{X_{1i}} - 2\sqrt{X_{2i}} \sim \mathcal N (0, 2)$, the sum $2\sum_i \sqrt{X_{1i}} - \sqrt{X_{2i}} \sim \mathcal N (0, 2n)$ and you can do a $z$-test with known variance $2n$, you don’t need to estimate the variance from the data! In my opinion, this is the main strength of this variance stabilizing transformation. –  Elvis Jan 3 '12 at 10:15
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