Can we carry out proportion test as usual and use t-values instead of Z value. For example in a college for two classes the smoking incidence data is Class A - 6/18; Class B - 7/22. Instead of using Chi-square, I will do proportion test and use t - values for significance?
Hi Ramnath, what are your sample sizes?
I agree with Babak Jamshidi , for small sample size T test is less robust but you can use it.
Z test is used ,both for means and proportions comparison. While for small samples mean test is shown to be following t test, extending the same logic, the proportion test can also be expected to follow t distribution. Hence, the question for expert to answer?
As per my understanding, you should do the "chi-square test for independence", also known as "chi-square test for association". All of the expected values will be above 5 in your example, so you can do the test without any problem. If you do this test on SPSS or Jamovi or similar programs, it will give you both the chi-square value and p-value.
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