Find Statistical Significance of Binary Data I would like to calculate whether the result of my survey is significant or not. There are two designed interfaces and in the survey, I have asked one question because for example if an interface A is liked 23 over B among 26 people, then interface B is liked 3 over A among 26 people. Therefore I kept the questions just one.
I tried to perform chi-square test but whenever I try to apply the formula, I am stuck because all examples I see require at least 2*2 matrix.





New Interface




Liked
23


Disliked
3




I tried to use the chi-square formula, but I can't do it because of the reason I explained above. I have calculated my conversion rate for the Like as %88.4 and Disliked as %11.5. But I can't do any further since I can't apply an example similar to my situation. I tried to perform the formula below for the chi-square but I got stuck.

I was wondering if anyone could help me with this?
Thanks in advance!
 A: A chi-squared test is OK because $n = 26$ is large enough
for the chi-squared statistic to have approximately
the distribution $\mathsf{Chisq}(\nu=1),$ giving the
P-value $0.00009 < 0.001 = .1\%.$ (See @whuber's Comment.)
 1 - pchisq(15.3, 1)
 [1] 9.171651e-05

Here is an exact binomial test in R of $H_0: p_A = .5$ against
$H_a: p_A \ne .5,$ where $p_A$ is the population
proportion favoring A. [Unless another null value is
stated, binom.test assumes $H_0: p_A = 0.5.]$
binom.test(23, 26)

        Exact binomial test

data:  23 and 26
number of successes = 23, number of trials = 26, p-value = 8.798e-05
alternative hypothesis: 
 true probability of success is not equal to 0.5
95 percent confidence interval:
 0.6984596 0.9755419
sample estimates:
probability of success 
             0.8846154 

The exact P-value can be computed as $$P(X \le 3)+P(X \ge 23)=8.797646e-05\approx 0.000087976,$$
where $X\sim\mathsf{Binom}(26,0.5).$
sum(dbinom(c(0:3,23:26), 26, .5))
[1] 8.797646e-05

Note: If you had suspected, before seeing data, that A would be
favored over B, then you might have used a one-sided binomial test of $H_0: p_A = .5$ against
$H_a: p_A > .5,$ For that test, the P-value (half as large as above) would be computed by
looking only in the right tail.
binom.test(23, 26, p=.5, alt="greater")

        Exact binomial test

data:  23 and 26
number of successes = 23, number of trials = 26, p-value = 4.399e-05
alternative hypothesis: 
 true probability of success is greater than 0.5
95 percent confidence interval:
 0.728098 1.000000                # one-sided CI
sample estimates:
probability of success 
             0.8846154 

In effect, the chi-squared test is inherently two-sided--on account of the squaring.
A: Chi square test requires you have a prior notion of what is "expected".  Under the assumption there is no difference in interfaces, you would expect an equal proportion of people would like and dislike the interface.
Therefore, $E=13$ is the expected number of people who would like the interface (your entire sample multiplied by the expected proportion who would like it).  This is also the expected number of people who would dislike it.
The $X^2$ statistic is then
$$ \dfrac{(23-13)^2}{13} + \dfrac{(3-13)^2}{13} = \dfrac{2}{13}100 \approx 15.3  $$
This test should have one degree of freedom, so we reject the null hypothesis that equal proportions of people like and dislike the interface with a p value far below 0.001.
