Is chi squared test appropriate

Question

I have conducted a survey in which the participants were asked to answer questions using 3 different techniques. At the end they were asked which technique the preferred, the results are as follows

A = 26
B = 8
C = 6

I want to verify that technique A is better than the other techniques. I was thinking of using the chi-squared test, with the null hypothesis that all media are equally good. So my expected value would be 13.3. I want to know if this is the right way to do it?

Did you hypothesize that A is preferred a-priori, or are you wondering that now because of the strong preference in the data? — gung - Reinstate Monica, Mar 22 '16 at 22:17
I did hypothesise (hope) that there will be strong preference for A. — salman, Mar 22 '16 at 22:26

score 3 · Answer 1 · edited Apr 13 '17 at 12:44

We want to compare different counts or proportions to a theoretical uniform distribution. Hence, we are setting up a goodness-of-fit test.
This can be calculated as an exact procedure using a multinomial test, or with an asymptotic approximation, using a chi-square goodness-of-fit. Although both tests yield a significant result, the difference between the exact multinomial test and the chi square is quite remarkable.

In R, the two calls are:

observed <- c(A = 26, B = 8, C = 6); prob <- rep(1/length(observed), 3)

For the multinomial test {EMT}:

multinomial.test(observed, prob, useChisq = FALSE, MonteCarlo = FALSE)
 Exact Multinomial Test, distance measure: p

    Events    pObs    p.value
       861       0     0.0002

and

chisq.test(observed, p = prob, correct = F)
    Chi-squared test for given probabilities
data:  observed
X-squared = 18.2, df = 2, p-value = 0.0001117

These are "omnibus" tests, and post-hoc comparisons are needed. In this regard, this post was useful, and already indirectly discussed by @Glen_b here. The idea is to compare each one of the proportions to the aggregate of the other two groups with a binomial test, and compensate with the Bonferroni correction.

In this case, since we are doing $3$ comparisons, and with a significance level of $0.05$ the correction is: $0.05/3 = 0.01666667$.

But we really won't need it, because the results are quite obvious after running the first test, which doesn't even come close to being questionable:

binom.test(26, sum(observed), 1/3, alternative="greater")

    Exact binomial test

data:  26 and sum(observed)
number of successes = 26, number of trials = 40, p-value =
0.00004174
alternative hypothesis: true probability of success is greater than 0.3333333
95 percent confidence interval:
 0.5080545 1.0000000
sample estimates:
probability of success 
                  0.65

Is chi squared test appropriate

1 Answers1

Linked