# SPSS: 2 sample t-test: real data against fictional group with M=0 and SD=1?

We want to use SPSS to compare two groups (2 independent samples t-test). The first group contains real data, the other group is fictional. This fictional group should contain as many "subjects" as the real (1st) group. But the mean is set to 0 and the standard deviation to 1 (standard normal distribution).

There are several tools out there to compare two groups by adding the number of subjects, mean and std for every group independently. But how can we do this in SPSS? I only know how to test one variable (group1) against a test value (T=0).

EDIT: Here a more detailed description of why I wanted to use this two-sample t-test, although this kind of approach seemed strange to me in the beginning. The thing is that we investigated 20 patients and compared their data with a large database of healthy control subjects. That means we did a z-transformation using the reference database (standard procedure in this field (quantitative sensory testing,QST)).

There are other research groups which suggest to then use the 2-sample t-test I mentioned below, because it would be inappropriate to compare data of 20 patients with data of 1200 controls. So they invented a fictional group with M = 0 and SD = 1 and tested the real data against this fictional data (on a website like this --> but Two-Sample T-Test). This approach would be more conservative than using the one-sample t-test with test value=0.

To be honest, I have no idea if this is the right approach. My first idea was to do a one-sample t-test and test the patient group (i.e. their z-values) against the value 0 to see, whether the z-values differ significantly from 0 instead of using a fictional dataset.

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Katarina, what exactly would be "inappropriate" in comparing 20 patients to 1200 controls? –  whuber Aug 7 '12 at 14:40
Yes, that's the question! I don't see the point there either. And I wonder, why should I transform individual data using norm data first and then use a 2-sample t-test to test against a normal distribution? Very confusing... For those who are interested: here is the paper using and explaining that method: ncbi.nlm.nih.gov/pubmed/20965658. (But I'm not sure that you have free access to it). –  Katarina Forkmann Aug 7 '12 at 15:14

If you want to see if the first group fits a N(0,1) distribution, why not just do a goodness of fit test instead of creating an artificial N(0,1) data set to compare it with? Although what you suggest doing is unconventional it is possible to do the two sample t test to see if the means are equal (essentially both 0). Of course you might say that if the goal is to test for 0 mean why not just do the one sample test that the mean is zero rather than create the artificial second sample (which induces additional uncertainty and reduces the power of the test)? It seems that alternative approaches provide better choices.

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Don't you think this artificial two-sample t test would lose substantial power? From what I read in the question and the abstract of the referenced paper, it sounds like it compares patient data to a "control" group simulated from a statistical description of a population. That looks like an inadmissible test to me. Not only that, it will suffer from the same problem as any randomized test: it is not reproducible. –  whuber Aug 7 '12 at 15:32
@whuber I agree. I was just following along with the OPs proposal. I would not say that it is a good idea. The outcomes may tend to reproduce but the variation in the artificial sample deos mean that results from one case to another would not be identical and could possibly be contradictory. It seems to me that if the objective is to test whether the sample comes from a population with mean 0, it is best to do the one sample test. If you want to show that the sample appears to come from a N(0,1) distribution do the goodness of fit test. –  Michael Chernick Aug 7 '12 at 16:12
Thanks, Michael. I was responding to your remark that you "see nothing wrong with doing the two sample t test." It looks like it might be helpful to edit your answer at that point to reflect the (good) advice in your comment. –  whuber Aug 7 '12 at 16:14
I don't see any situation where the artificial test would be preferred. –  Michael Chernick Aug 7 '12 at 16:14
@whuber You are right. I really meant that it seems possible to do it. Maybe saying "nothing wrong" sounds too much like saying that it is all right. –  Michael Chernick Aug 7 '12 at 16:16
I would perhaps say compute a new variable in the current dataset using the random variable functions (it will that way be the same number of cases), then use varstocases to reshape the dataframe and then one can use the regular independent samples t-test command. I don't quite get the point of the exercise though. –  Andy W Aug 7 '12 at 13:28