I was wondering which test I should use to compare two proportions in one sample? My research is about white matter hyperintensities (WMH) in migraine compared to tension-type headache. There are three options: only deep WMH (63,6%), only periventricular WHM (6,1%) or both (30,3%). I want to know if there are more 'only deep'WMH compared to only periventricular WMH (only for migraine). I've a small sample (N=33). Can I use an one sample t-test?

Thank you in advance

Indeed pv+both and deep+both but how can I compare 33 patients with 12 pv and 31 deep to each other? Also with binominal?

  • $\begingroup$ A one-sample t-test is used to compare a sample mean with a specified mean. Since your dealing with proportions the answer is no. I'm not sure what exactly you are trying to compare but it sounds like a job for a test of independence or Fisher exact test $\endgroup$ May 17, 2017 at 9:19
  • $\begingroup$ Thank you. I want to compare deep WMH (63,6%) to periventricular WMH (6,1%) in the group with migraine. I came across this site statpac.com/statistics-calculator/percents.htm, so that's why I thought I could use an one-sample t-test $\endgroup$ May 17, 2017 at 9:32
  • $\begingroup$ I stand (partly) corrected. Apparently people use t-test although not completely appropriate. Still given that there are three possibilities instead of two, I'm unsure whether a t-test is really appropriate. Perhaps this is more what you're looking for $\endgroup$ May 17, 2017 at 13:09
  • $\begingroup$ Maybe a chi square goodness of fit is appropriate? $\endgroup$ May 17, 2017 at 13:42

1 Answer 1


You have 2 p-WMH and 21 d-WMH. Under assumption of equal probability of being p- or d-WMH, it should follow binomial distribution Bin(0.5, 23). Under this null hypothesis, the probability of getting 2 or less cases in p-WMH is 3.3 x $10^{-5}$. Same for 2 or less cases in d-WMH. So the p-value is 6.6 x $10^{-5}$.

Conclusion: Reject the null hypothesis and support the idea that the chances to be p- or d-WMH are not the same.

This method utilizes the symmetry of the binomial distribution when event probability is 0.5. So if the null hypothesis is not event probability = 0.5, this method will give you wrong answer.

  • $\begingroup$ Thank you very much. One more question because there are patients who have both. So migraine (N = 33): in total there are 12 patients with periventricular and 31 patients have deep.. can I also use binominal distribution (0.5) ? Should I use n=43? $\endgroup$ May 17, 2017 at 14:04
  • $\begingroup$ My answer is based on "if there are more 'only deep'WMH compared to only periventricular WMH (only for migraine)." For this question both-WMH dose not provide any useful information, so just exclude them. If you add both-WMH into question as group, you get the multinomial distribution. The idea that leads to n=43 is unacceptable, because sample size cannot be changed. $\endgroup$
    – user158565
    May 17, 2017 at 14:15
  • $\begingroup$ Which test should I use because 31 patients have deep and 12 have periventricular of the 33 patients? I hope you understand what I mean because my first question was the difference to only deep versus only periventricular $\endgroup$ May 17, 2017 at 14:27
  • $\begingroup$ From you data, you can work on the probabilities of p-WMH, d-WMH and both-WMH. Your first question is p-WMH = d-WMH or not. We already have the answer. It seems your second question is: If p-WMH + both-WMH = d-WMH + both-WMH. If my understanding is correct, the second question = first question, because subtracting both-WMH in both sides of the equation in second question leads to the first question. If my understanding is incorrect, please give the more explanation in terms of that three probabilities. $\endgroup$
    – user158565
    May 17, 2017 at 15:26
  • $\begingroup$ Indeed pv+both and deep+both but how can I compare 33 patients with 12 pv and 31 deep to each other? Also with binominal? $\endgroup$ May 17, 2017 at 16:10

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