Why do we use Z test for proportions and why not T test. I have found a similar question here but I am unable to get what the answer tries to convey. It would be of great help if anyone could explain the reason in comparatively easy words.


As this answer says in detail, the assumptions underlying the t-test only strictly hold when the individual data values are sampled from a normal distribution.

Proportions are limited to values between 0 and 1, while values taken from a normal distribution can be any real number. And unlike a normal distribution, where the mean and variance of a sample are independent, once you know the proportion you have some information about the variance. So proportions don't meet the assumptions needed for a t-test to be valid.

As you take more and more samples, however, the distribution of average values in most practical applications comes close to a normal distribution. The z-test is based directly on the normal distribution. So although the z-test might not be exact with very few observations it doesn't take very many observations for it to be a very good approximation.

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  • $\begingroup$ How do we know the variance once we know the proportion. I didn't get that part. $\endgroup$ – learnToCode May 2 at 3:50
  • $\begingroup$ @learnToCode that was an oversimplification; I edited a bit. If you have a probability p of success in each of n independent trials, you are working with a binomial distribution. The mean number of successes is np and the variance of the number of successes is known to be np(1-p). So if you know the true probability of success and the number of trials, you already know the variance. (continued...) $\endgroup$ – EdM May 2 at 14:43
  • $\begingroup$ @learnToCode You estimate the true probability p from the fraction of successes in your sample of trials. The t-test assumes that the mean and variance of your sample are independent. If your sample size is only 1 or 2 then examples on this page show that the sample mean exactly predicts the sample variance for a binomial distribution. For larger sample sizes the relationship isn't so strict, but the mean and variance still aren't independent (as they would be for sampling from a normal distribution) so the t-test assumptions aren't met. $\endgroup$ – EdM May 2 at 14:54

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