interpreting chi square test of independence result so I've conducted a chi square test of independence to determine whether there was a relationship between payment method (paper and no paper) vs churn (churn and no churn) and determined that there was indeed a relationship between the two as seen in the picture I attached. however, how can I know whether someone who uses no paper is more likely or less likely to churn as compared to those who uses paper? can I just look at the frequencies?

 A: Compare observed and expected counts in the cell of the table
for 'NoPaper' by 'Churn'. Alternatively, look at Pearson residuals.
Here is the chi-squared test in R for your data.
MAT = matrix(c(102, 195, 26, 172), byrow=T, nrow=2)
chisq.out = chisq.test(MAT, cor=F)
chisq.out

        Pearson's Chi-squared test

data:  MAT
X-squared = 27.882, df = 1, p-value = 1.29e-07
chisq.out$obs
     [,1] [,2]
[1,]  102  195
[2,]   26  172
chisq.out$exp
     [,1]  [,2]
[1,] 76.8 220.2
[2,] 51.2 146.8
chisq.out$resi
          [,1]      [,2]
[1,]  2.875543 -1.698212
[2,] -3.521807  2.079876

The chi-squared statistic is $Q = \sum \frac{(X_{ij} - E_{ij})^2}{E_{ij}},$
where $X_{ij}$ are observed counts, $E_{ij}$ are expected counts (obtained from row and column totals), and the sum is taken over all four cells of
the table. The null hypothesis that paperless payments and churn are independent is strongly rejected with a very small P-value.
The observed count in the cell for paperless and churn is only 26, whereas the expected count
assuming Paper and Churn are independent factors is 51.2. So, according to your data, it seems
that paperless payments are associated with reduced churn.
Pearson residuals are $\pm\sqrt{\frac{(X_{ij} - E_{ij})^2}{E_{ij}}},$ where the sign is the same as for the difference $X_{ij} - E_{ij}.$
When $Q$ is sufficiently large to reject the null hypothesis of independence,
then it is customary to look at the Pearson residuals with the larger
absolute values (especially absolute values exceeding 2 or 3) to interpret the practical consequences of the observed association. Here the key Pearson residual $-3.52$ is for the cell I
mentioned earlier.
