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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.

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  • $\begingroup$ If this is homework, please add the relevant tag ('homework'). $\endgroup$ – Glen_b -Reinstate Monica Nov 1 '12 at 20:32
  • $\begingroup$ Not exactly homework. Also, your answers so far seem as conflicted as I am! $\endgroup$ – Seth Nov 1 '12 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 -Reinstate Monica Nov 2 '12 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 -Reinstate Monica Nov 2 '12 at 8:07
  • $\begingroup$ Do you wish for more information? $\endgroup$ – Glen_b -Reinstate Monica Nov 4 '12 at 8:36
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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.

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