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I have data from two separate experiments, which I'll call "control" and "experiment". The outcome from each trial is a "yes" or a "no".

In the control group, 87% of participants got a "yes". I am using this as the underlying probability of "yes".

I'd like to know whether my experiment group, in which 96% of participants got a "yes", has a statistically significantly higher chance of getting "yes" compared to control.

I am therefore using the binomial distribution to compute the probability that 96% of participants get a "yes" given that the underlying probability of "yes" is 87%.

Having done so, for my numbers, I get 0.1. Does that mean that there is a probability of 0.1 that 96% of people got a "yes"? Did I interpret that correctly? Is it correct to use the control group as the estimator of the probability of "yes" and then the experiment group's data in the binomial distribution?

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  • $\begingroup$ Could you show us your exact computations? How did you get those numbers? What are the sample sizes for both groups? $\endgroup$
    – Tim
    Commented May 8, 2019 at 16:55
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    $\begingroup$ Please see stats.stackexchange.com/search?q=two+sample+binomial+test for extensive discussion of this situation. $\endgroup$
    – whuber ♦
    Commented May 8, 2019 at 16:59

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No. The 87% is not a fixed true probability but rather an estimated probability, as is the 96%, but for a different group. What you need is a test comparing two proportions, for example prop.test in R (with references given on its help page); all statistical software should have this test.

It is by the way unclear what you actually computed (you don't write how you used your binomial), and the interpretation of the 0.1 is possibly wrong even if 87% were the true probability, but I can't comment on that due to lack of details.

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