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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.

gung - Reinstate Monica
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John
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1 Answers1

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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)

Glen_b
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  • Thank you for the quick and great response. I would like to ask a quick follow up. How do I know if the p-value proves a statistically significant or disproves the null? Also, I am not sure what is meant by cdf. – John Dec 10 '13 at 03:06
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    I've expanded the abbreviation "cdf" in the answer. You compare the p-value (see the first sentence [here](http://en.wikipedia.org/wiki/P-value)) with your significance level (if $p\leq \alpha$, you reject the null). See also [here](http://stats.stackexchange.com/questions/44769/understanding-p-value), [here](http://stats.stackexchange.com/questions/60825/is-a-p-value-of-0-04993-enough-to-reject-null-hypothesis/72594#72594) and [here](http://stats.stackexchange.com/questions/45889/confused-about-region-of-rejection-vs-p-value) – Glen_b Dec 10 '13 at 03:15
  • Very helpful - last question if I may. assuming 95% confidence, would significance be .05 or .025? I assume .05, however, I am not sure if we should divide .05 by two due to the two-tailed test. Thanks again - you are a huge help. – John Dec 10 '13 at 04:02
  • One more "last" question as well. You used 54 in the formula and not 55. Why 54? – John Dec 10 '13 at 04:20
  • Because I wanted $P(X\geq 55)=1-P(X<55) = 1-P(X\leq 54)$, where $X$ is the number selecting $A$. (Then of course, I wanted to double that.) ... The first thing follows from the definition of *p-value*, the rest is elementary probability on a variable representing a count. – Glen_b Dec 10 '13 at 04:37