Significance testing questionnaire answers I apologize, there may be an answer for this but I couldn't find it.
I have a survey of 1000 people, and for a particular question they were asked which do you prefer of A, B & C. 44% answered A, 40% answered B and 16% answered C.
C is clearly significantly lower than the others, but what would be the correct way to test whether preference for A is significantly different from preference for B?
 A: I don't know about the "correct" way. But, you can ignore C altogether and have your null be that A and B have equal probabilities (one half each). Then your null hypothesis model is a binomial one and you can calculate how extreme your result is. In your case, you have 440 people voting A, 400 voting B for a total of $n = 840$ voters. It is straight forward to calculate a significance value for your data under the binomial (null) model $B(840, \frac{1}{2})$.
A: I get an answer from Bogdan Ćmiel (working at AGH University in Poland). Here it is:
What you need is a model of the distribution which than can be tested by $\chi^2$ test with corrected degree of freedom. I will use the specific case to explain, than I link to general solution. 
Since you have three possible cases and you are interested if answer A and B have the same probability you are actually asking if multinomial distribution with probabilities $p$, $p$ and $1-2p$ is correct. Since $p$ is not know the first step is to estimate $p$ by Maximum Likelihood Estimation (MLE). Let us assume we have answers $n_A$, $n_B$, and $n-n_A-n_B$ for answers A, B, and C, respectively.  Than we can show that $\hat{p}=\frac{n_A+n_B}{2n}$ is solution for MLE. 
The next step is classical $\chi^2$ test given by:
$$ \chi^2 = \frac{(n_A-n \hat{p})^2}{n \hat{p}}+\frac{(n_B-n \hat{p})^2}{n \hat{p}}+\frac{(n-n_A-n_B - n(1-2\hat{p}))^2}{n (1-2\hat{p})}$$
Normally such test has 2 degree of freedom (two independent probabilities) but we estimated one parameter already so we have to decrease degree of freedom by one, so the final result is just one degree of freedom. 
In this specific case the first two terms are the same and the last term is 0 since $1-2\hat{p}=\frac{n-n_A-n_B}{n}$, so $$ \chi^2 = \frac{(n_A-n_B)^2}{n_A+n_B}$$
Using number you mentioned $\chi^2 = \frac{(400-440)^2}{840} = \frac{40}{21}$ and the final p-value is $0.1675$.
More general explanation can be found here: http://home.agh.edu.pl/~cmiel/wmsstatystyka/W07.pdf (page 7).
If Polish is a big problem I can translate it. 
