t-tests vs chi square test/Fisher's exact test

Hello forum friends!

So I have been using excel to compare if two groups are statistically significantly different. My null Hypothesis is that these two groups are the same (or that there is no difference between groups).

When comparing my Hispanic population vs non-Hispanic population and their knowledge of STIs (based of yes/no data converted to 1/0 data), I used two-tailed t-tests assuming equal variance (excel: t-test 2 tails, type 2).

However, I was told that this was wrong because t-tests compare means and I would want to do a proportion comparison.

Thus, I computed p-values using Fisher's exact test (and/or chi square test).

My question is: which one is the correct method of testing statistical significance?
I got the SAME p-values with both tests (t-tests vs Fisher's) and was wondering why that is so.

Much thanks on advance,


Super Moderator
Could you describe your dependent variable a bit more? Do you mean that each person simply receives a score of 0 or 1 for STI knowledge? How big are your samples?
Thanks for responding!

So for example, one question was simply: Do you know how to protect yourself from STIs? (a simple yes/no response which I then converted to 1 or 0 [yes=1, no=0] for data computation).

So I want to know if one group is more likely to know how to protect themselves against STIs than the other.

I have 2 groups: Hispanic vs non-Hispanic, N=100.
51 out of 72 Hispanics (70.8%) said yes.
26 out of 28 non-Hispanics (93%) said yes.

When I did a t-test (two-tailed, assuming equal variance), I got p=0.0186.
When I did a Fisher's exact, I got p=0.0188.

So I concluded that because the p-value was <0.05, Hispanics were less likely to know how to protect themselves from an STI (in this particular case).

However, which statistical test is the correct method of determining significance?