I'm having a hard to deciding which of a two-sample t-test and a chi-squared test is more appropriate for these sorts of questions. Examples would be:

70 out of 120 americans like hotdogs. 50 out of 95 europeans like hotdogs. Do the two groups differ in their hotdog opinions?

1000 each of caucasians, african-americans, and asians were asked if they preferred red, white, or yellow onions (counts provided in a table). Do respondents from different ethnicities differ in their responses?

I realize that those questions are worded slightly differently - I think it may be the wording that differentiates them? In either case, i'm confused about which of the two tests is more appropriate.

  • $\begingroup$ If this is homework, please add the relevant tag ('homework'). $\endgroup$
    – Glen_b
    Commented Nov 1, 2012 at 20:32
  • $\begingroup$ Not exactly homework. Also, your answers so far seem as conflicted as I am! $\endgroup$
    – Seth
    Commented Nov 1, 2012 at 23:08
  • $\begingroup$ Dualinity and Dan are completely wrong, I'm sorry to say. The "size of the model" is nothing to do with it, nor the number of groups. $\endgroup$
    – Glen_b
    Commented Nov 2, 2012 at 7:56
  • $\begingroup$ I've given more detail in my answer to help you see why the other answers aren't correct. If you need more details, I can give more. ... When you say 'not exactly homework' what does that mean? How did you arrive at the the question? $\endgroup$
    – Glen_b
    Commented Nov 2, 2012 at 8:07
  • $\begingroup$ Do you wish for more information? $\endgroup$
    – Glen_b
    Commented Nov 4, 2012 at 8:36

1 Answer 1


You could use chi-squared tests to answer those questions. (Note that they're all about comparing proportions/counts - that's usually a clue.)

An alternative (particularly handy if you want a one-tailed test) would be a two-sample proportions test. In the two-tailed case they will give the same p-value (as long as they both do or both don't do the continuity correction).

(Let's say you plan to do a t-test. Once you figure out how to get a sample standard deviation, the resulting statistic doesn't have a t-distribution. The numerator isn't normal, the denominator isn't the square root of a scaled chi-squared and the numerator and denominator aren't independent.)

Google things like comparing two proportions and comparison of proportions and note how they're done. They're not often done with t-tests.

You can make a t-test work (by using 0s and 1s as the data), and it's generally going to work reasonably well, but as far as I know we have no solid basis for arguing that it's going to be better than doing the usual normal approximation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.