One Sample t-test between percents I have survey data for N=200. One of the questions is a multiple choice question:
"Which Proposal will you vote for - A, B, Do not know"
Following % of respondents selected :
A = 35%
B = 45%
Do not know = 20%
How can I determine whether the difference between the response rate for choice A (35%) and response rate for choice B (45%) is statistically significant? Should I use a One Sample t-test?
I think I know how to determine the statistical significance of the response rate for one choice (E.g. A=35%. Probability that one of three options will be selected randomly = 33%. Compute Standard Error SE = SQRT (33x(1-33)/N). Compute z = (35%-33%)/SE. Find p-value that corresponds to this z value. Compute two tail-tailed p-value. Compare if two tailed p-value < alpha (e.g. 5%))
However, I am struggling to understand how to determine the statistical significance of the difference between the response rates for 2 choices of the same question
Thanks
 A: I think that to use a t-test, the data would need to come from a normal distribution.
But you can use a $\chi^2$-test to determine if the proportions are sufficiently uniform (which is to say, sufficiently similar to $40\%-40\%$).
Your null hypothesis would be that $P(A) = P(B)$ and your alternative hypothesis $P(A)\neq P(B)$.
Hopefully that helped a little. (I'm keeping it short because of the self-study tag.)
A: To me, if the question is to compare the frequencies of A and B, to see if they are not equal, I would disregard the "don't know" responses, and use a binomial test.  The following uses the same values as @DemetriPananos . You can run this code at rdrr.io if you don't want to download R. 
A=70
B=90
binom.test(A, (A+B))

   ### Exact binomial test

   ### number of successes = 70, number of trials = 160, p-value = 0.1328
   ### alternative hypothesis: true probability of success is not equal to 0.5
   ### probability of success 
   ###                 0.4375 

A: This is easily done in R with a chi-squared test
x = c(70,90,40)
p = c(80,80,40) #Assume the proportions of A and B are the same

chisq.test(x = x, p = p, rescale.p = T)

>>>
    Chi-squared test for given probabilities

data:  x
X-squared = 2.5, df = 2, p-value = 0.2865

We fail to reject the null that the null hypothesis that proportions of A and B are different.
