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I did an experiment with 12 participants. They were asked to rate 3 stimuli, with the following 3 possible answers: "increased", "decreased", "no difference". I have as a result this table:

          Increase Decreased No_difference 

StimulusA   10      1           1
StimulusB   12      0           0
StimulusC    8      2           2

Now,if I want to understand in one shot if the "Increase" answer is chosen significantly more than the other two answers in all the Stimuli, is it correct if I simply use the binomial distribution summing the successes along all the 3 stimuli?

I mean, is it correct doing this?

> prop.test(30,36)

    1-sample proportions test with continuity correction

data:  30 out of 36, null probability 0.5 
X-squared = 14.6944, df = 1, p-value = 0.0001264
alternative hypothesis: true p is not equal to 0.5 
95 percent confidence interval:
 0.6652978 0.9303666 
sample estimates:
    p 
0.8333333 

Or it is better to use the function in this way instead?

> prop.test(c(10,12,8),c(12,12,12),c(0.5,0.5,0.5))

    3-sample test for given proportions without continuity correction

data:  c(10, 12, 8) out of c(12, 12, 12), null probabilities c(0.5, 0.5, 0.5) 
X-squared = 18.6667, df = 3, p-value = 0.0003204
alternative hypothesis: two.sided 
null values:
prop 1 prop 2 prop 3 
   0.5    0.5    0.5 
sample estimates:
   prop 1    prop 2    prop 3 
0.8333333 1.0000000 0.6666667 
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If what you mean to test is whether more people reported an increase than the combined number who reported a decrease or no difference (which is what I think you mean) then your first version is closer to the correct one. Your null hypothesis in that case is that people choose 50-50 between "increase" and "no increase, or decrease" and you are open to evidence either way (greater or less than 50% choose increase).

However, you actually are interested in testing the alternative hypothesis that >50% chose it, so you need a one sided test. You can call this explicitly in prop.test by stating that your alternative hypothesis is only for p being greater than 0.5:

prop.test(30,36, p=0.5, "greater")

It's worth pointing out though that there is nothing special about the 0.5 proportion here - why have you chosen it as the cut-off point for your alternative hypothesis? For example, why not have as a null hypothesis that a third of people choose each option? or any other set of probabilities, perhaps based on an experiment of having people fill in the survey having received a placebo. Having said that, there is strong intuitive appeal in the 0.5 cut-off point and there is no doubt your experiment shows statistically significant evidence that more than 50% do report an increase. The only question is, how does this compare to the percentage who report an increase under other circumstances? (which is not what you've asked here, so I won't worry about it).

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  • $\begingroup$ Great answer and even better questions. $\endgroup$ – Roman Luštrik Jan 23 '12 at 11:27

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