Donald –

 

If both patients were compared to the same group of controls in two different models each with 1 patient and N controls, then I'd do the following:

(a) convert your T-statistic to Z-score

(b) subtract the Z-scores

(c) convert to a p-value for significance testing

 

The reason that this works is that only the variance of the control group is used to determine the denominator of the T-tests since the variance of the patient is 0. Then the only difference in the contrasts is the numerator (e.g. patient1-control and patient2-control). If you subtract these two, then you get patient1-patient2. Since the denominator is the same, you can keep the denominator. Of course, all that requires more math and such, so the Z-score approach is easier.

 

Additionally, many studies that compare patients to controls using single cases report the Z-scores of the patients, so there is a history in the literature for using Z-scores.

 

Your approach may work, but I am slightly worried about the practice of reporting Z-scores in the literature that you mention. The point of the PDF I wrote (that Ben cites) is that people use Z-scores as Z-statistics, and thereby attach a p-value to a Z-score that may not be correct. A Z-score is simply a transform that subtracts the sample mean and divides by the sample standard deviation. However, this cannot be tested against a Z-distribution unless there are a large number of observations (ie in the limit of large df). The whole point of Students T is to adjust for the bias entailed in estimating the population mean and standard deviation from small samples.

 

So people who report Z-scores in papers when comparing small groups should not use Z>1.96 to imply p<.05. However, if you are talking about converting a T-statistic to a Z-statistic (not a Z-score, as you sometimes use below), that may be possible, but I am unsure whether the difference in Z-statistics for the 2 patients (vs same control group) can simply be given a p-value – but perhaps it can – hopefully someone else will comment. I guess it would correspond to constructing a T-statistic with the numerator being the difference in the 2 patient scores, and denominator as the standard error of mean of the controls alone, as you note (using an unpooled estimate of that error, in the terminology of the PDF cited above).

 

BW,R

 

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Ben and Rik,