Thanks to @StatsStudent for the well-targeted question in a Comment and for your answer.
Below are simulated data for three ads, with 1000 subjects seeing each ad.
Each subject makes a choice for 1 = Weekly, 2 = Monthly, or 3 = No.
The preference vectors (parameter p in the sample procedure
of R) show slight differences in prevalence in the population.
For example, vector c(11, 11, 10) for group B amounts to probabilities
$11/32,\, 11/32,\, 10/32$ for respective options $1, 2, 3.$
The question is whether the slightly different preferences, presumably
due to seeing different ads, result in count differences among the three
groups that rise to the level of statistical significance.
Data and contingency table. Here are simulation results of counts
in the three groups, and the resulting contingency table of counts.
set.seed(811) # for reproducibility
a = sample(1:3, 1000, rep=T, p=c(10,10,12))
A = tabulate(a); A
[1] 326 319 355
b = sample(1:3, 1000, rep=T, p=c(11,11,10))
B = tabulate(b); B
[1] 348 350 302
c = sample(1:3, 1000, rep=T, p=c(12,10,10))
C = tabulate(c); C
[1] 387 310 303
MAT = rbind(A,B,C); MAT
[,1] [,2] [,3]
A 326 319 355
B 348 350 302
C 387 310 303
Null hypothesis and expected counts. The null hypothesis is that
the choices $1,2,3$ were made independently of which ad A, B, C was seen.
The total counts in row A is, of course, 1000; the total count in column 1
is 1061; the grand total is
$n=3000.$ If the null hypothesis is true we would expect that the number subjects
seeing ad A and making choice 1 would be $P(A)P(1) = P(A\cap 1)$ estimated
as follows $\hat P(A) = 1000/3000, \hat P(1) = 1061/3000$ so that the
expected count for ad A and choice 1 is
$$E_{A1} = E_{11} = n\hat P(A)\hat P(1) = \frac{1000(1061)}{3000} = 353.6667,$$
Expected counts corresponding to the remaining eight cells of the table are
computed similarly. [For the computations below, expected counts should not be
rounded to integers.]
Chi-squared test. The test statistic for the chi-squared test of independence
is
$$ Q = \sum_{i=1}^3\sum_{j=1}^3 \frac{(X_{ij} - E_{ij})^2}{E_{ij}},$$
where $X_{ij}$ is the observed count in cell $(i,j)$ of the contingency matrix.
The first of the nine components of $Q$ is $C_{11} = \frac{(326 - 353.67)^2}{353.67}=2.165.$
Under the null hypothesis, $Q \stackrel{aprx}{\sim}\mathsf{Chisq}(\nu),$
where the 'degrees of freedom' $\nu = (r-1)(c-1) = 4,$ where $r$ and $c$ are
the number of rows and columns, respectively, of the contingency matrix.
[Notice that given the row and column totals and the four expected counts $X_{11},X_{12},X_{21}, X_{22},$ the remaining five $X_{ij}$ could be determined.]
This approximately chi-squared distribution is sufficiently accurate to give
reliable results, provided that all of the expected counts $E_{ij} > 5,$ which
is easily true for our data.
If $Q > c = 9.488,$ then we can reject the null hypothesis that Choices 1,2,3
are independent of Ads A,B,C at the 5% level of significance.
The critical value $c$ cuts probability $0.05 = 5\%$ from the upper tail
of $\mathsf{Chisq}(4).$
qchisq(.95, 4)
[1] 9.487729
Chi-squared test in R. Below we show results of this chi-squared test of independence, computed in R. $Q = 13.839 > 9.488.$ so we reject at the 5% level
of significance.
out = chisq.test(MAT); out
Pearson's Chi-squared test
data: MAT
X-squared = 13.839, df = 4, p-value = 0.007826
The P-value is $P(Q > 13.839),$ computed according to $\mathsf{Chisq}(4).$ (The
slight discrepancy from the computation in R below is that the output rounds $Q,$
labeled as X-squared.
1 - pchisq(13.839, 4)
[1] 0.007827032
Additional information on intermediate computations is available by using $-notation.
In particular, we can verify the observed counts, see the expected counts to see that
all exceed 5, and look at the 'Pearson residuals'. The residuals are the signed
square roots of the contributions $C_{ij}$ to the chi-squared statistic $Q.$
Specifically, we verify our computation of $E_{11} = 353.67.$ Also, recall that we
computed $C_{11} = 2.165,$ and $\sqrt{C_{11}} = 1.471.$ The negative sign in
the output signifies that $X_{11} = E_{11} < 0.$ Residuals with large expected
values show where agreement between observed and expected count is most important.
out$obs
[,1] [,2] [,3]
A 326 319 355
B 348 350 302
C 387 310 303
out$exp
[,1] [,2] [,3]
A 353.6667 326.3333 320
B 353.6667 326.3333 320
C 353.6667 326.3333 320
out$res
[,1] [,2] [,3]
A -1.4711595 -0.4059483 1.9565595
B -0.3013218 1.3101058 -1.0062306
C 1.7724814 -0.9041575 -0.9503289
Below is a graph of the density function of $\mathsf{Chisq}(4).$
The critical value is marked with a vertical dotted red line and
the observed value of $Q$ is marked with a solid black line.
