Is a T test the right thing to do? I have performed a survey which contains 2 logical sets of people (A and B). 
In my data set there were roughly twice as many respondents in set A as there were in set B.
The survey asks "Do you have a qualification in your field". So my answers are all Yes or No.
I want to know if the difference between the answers is statistically significant.
Does a 2 sample t-test work here?
 A: You could use a t test. But you might consider a z test instead:
$H_{0}: p_{A} - p_{B} = 0; H_{A}: p_{A} - p_{B} \ne 0$
$z = \frac{p_{A}-p_{B}}{\sigma_{p_{A}-p_{B}}}$, where
$\sigma_{p_{A}-p_{B}} = \sqrt{\hat{p}\left(1-\hat{p}\right)\left[\frac{1}{n_{A}} + \frac{1}{n_{B}}\right]}$, and where
$\hat{p}$ is the proportion from all your data, regardless of which group the data came from. ($n_{A}$ and $n_{B}$ are the sample sizes of each group.)
The p-value from this test is $p = P(|Z| \ge z)$ (where $Z$ is the standard normal distribution).
That will work pretty well if you have a large sample enough sample size in both $A$ and $B$. But a continuity correction for $z$ can help:
$z = \frac{|p_{A}-p_{B}| - \frac{1}{2}\left[\frac{1}{n_{A}} + \frac{1}{n_{B}}\right]}{\sigma_{p_{A}-p_{B}}}$.
Using this continuity correction means you will need to interpret the direction of a significant difference (i.e. which group's proportion is larger) by comparing $p_{A}$ and $p_{B}$, as the sign of the test is positive.
Finally, you have yet another option, which is to use a $\chi^{2}$ contingency table test of the same null hypothesis above. Let me know if you want the details on that.
