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 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.
