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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.

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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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