How to determine if % selecting Option A is statistically different from selecting Option B in a survey? I ran a survey on pricing and asked respondents to choose between option A or Option B. 61% chose option A and 39% chose Option B. The sample size is 90. How do I determine if the % selection Option A is statistically significant? What test do I run and how?
The cross tab tables show stat tests across the rows (e.g., the % of Group 1 choosing option A is statistically different from group 2 choosing option A). However, I want to compare within the column (i.e., Is the % choosing Option A statistically different from the % choosing Option B at a confidence interval of 95% and 90%)
I would like to do in Excel. Not sure how to use Binom.Dist formula. I have a feeling that would be the right test but I am not certain.
 A: 
I ran a survey on pricing and asked respondents to choose between option A or Option B. 61% chose option A and 39% chose Option B. The sample size is 90. How do I determine if the % selection Option A is statistically significant? What test do I run and how?

Note that with selection among options, the proportions selecting the options are not independent (with two options, there's perfect negative dependence).
Testing if the proportion of A and B are different is the same as testing whether the proportion of A is different from $\frac{1}{2}$.
So if we were to take the number choosing A (55 out of 90) as the number of successes in a binomial, where under the null, $p=\frac{1}{2}$, we can use binom.dist.
The required probability for a two tailed test would be the probability of 55 or more and the symmetrical probability on the other side (but since we have p=0.5 under the null, we can just double the tail area).
binom.dist with the cumulative argument set to be TRUE returns the cumulative distribution function (cdf). 
So =1-binom.dist(54,90,0.5,TRUE) is the probability of 55 or more; we need to double that for the two tailed test.
Hence =2*(1-binom.dist(54,90,0.5,TRUE)) is the required p-value (which gives $p= 0.0446$).
But with such large samples, you could easily use the normal approximation (I'd use the continuity correction in the symmetric case)
