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I'm in the process of running a repeated-measures ANOVA on two groups of participants from an experiment. In one group, there are 10 participants, and in the other group, there is only 1 participant. This is because the 1 participant is a patient who is being compared to the 10 other participants.

Now, my question is: is it fair/allowable to run an ANOVA comparing the patient with the other participants? I wasn't sure, so ran it in SPSS, and it seems to work, with a significant result. Is this naughty?

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  • $\begingroup$ One clarifying question, is the same data for the single patient being used in each comparison, or is it 10 different data from the same patient? $\endgroup$
    – probabilityislogic
    Commented Apr 7, 2011 at 15:31

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Performing the ANOVA assumes that the nature and amount of variation in the hypothetical population represented by the one patient are the same as the variation in the hypothetical population represented by the ten other patients, and that all were obtained randomly and independently from their populations.

Under these assumptions a standard approach is to use data from the ten patients to compute a prediction interval for a single additional patient. The significance test only has to check whether the lone patient falls within that interval.

The prediction interval is simple to justify and compute. It depends only on the estimated mean $\bar{m}$ from the ten patients and on their estimated variance $s^2$. Let $x$ be the value of the lone patient. Under the null hypothesis (that all patients are drawn independently from a single population), the variance of $x - \bar{m}$ equals $s^2 + s^2/10$. If you further assume all variation is Normal--that's critical here because it's hard to check with just ten patients in a group--then $\frac{x - \bar{m}}{\sqrt{s^2 + s^2/10}}$ has a Student's t distribution with 9 degrees of freedom (9 is the df used to estimate $s^2$), allowing you to erect a confidence interval in the standard way for $x$.

If this prediction interval test disagrees with SPSS's result, I would not trust the SPSS ANOVA in this case.

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  • $\begingroup$ Ah, perfect! I didn't realise that such a method was possible. Thanks :) $\endgroup$
    – user4055
    Commented Apr 6, 2011 at 17:52
  • $\begingroup$ assuming the sample of 10 has some variance, isn't the homogeneity of variance assumption of an ANOVA violated? Because error is pooled across the cells, the cell with only 1 person will pull down the MSE making a significant result more likely, no? $\endgroup$
    – Matt
    Commented Apr 7, 2011 at 12:21
  • $\begingroup$ nice answer (+1) from me. Although I do find it rather amusing that people think the normal distribution is an "agressive" assumption for variation - quite the opposite for only 1 aspect of the variation will influence the results $s^2$. Choosing any other distribution will involve more detailed statistics getting involved in the test. I'd rephrase your italicised part as unless you have a good reason to assume something else, then a normal distribution is appropriate for this problem. $\endgroup$
    – probabilityislogic
    Commented Apr 7, 2011 at 15:27
  • $\begingroup$ @probability Thank you. Those of us who frequently find outliers and large amounts of skew in good data will be less emphatic about this and say only that the assumption of normality depends on the nature of the data and the importance of the decision(s) made with the test. $\endgroup$
    – whuber ♦
    Commented Apr 7, 2011 at 16:19
  • $\begingroup$ but you see, the assumption of normality is not about "the data", it is about "the noise". If you see "outliers" or "skewness" (which only exist relative to a model in this context), this is telling you something about the model - i.e. that something is missing from it, and it gives you a clue as to how to improve it. The normal distribution is very good a picking these kind of things up. And when there is outliers, the residual variance becomes large, and so you get appropriately wide prediction/confidence intervals, as you should. $\endgroup$
    – probabilityislogic
    Commented Apr 8, 2011 at 6:57

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