4 question survery - how to determine statistical significance? What is the proper approach to determine the highest response rate in a multiple question survey? I did a chi square test of the most popular answer against the 2nd most popular but it feels like that's not the right approach. But I want to be able to say that for example answer b is the most popular answer at 95% significance
 A: I believe this question boils down to the following. Consider a question that has four choices with the following results. Users ($n = 100$ users) chose


*

*A: 45

*B: 35

*C: 10

*D: 10
The first thing is to compare frequencies across all groups. You would do this with a $\chi^2$-test. The $\chi^2$-test tests the following hypothesis
$$
H_0: p_A = p_B = p_C = p_D \\
H_A: p_A \neq p_B \text{ or } p_A \neq p_C \text{ or } p_A \neq p_D \text{ or } ... \text{ or } p_C \neq p_D  
$$
In R this can be one with
> chisq.test(x = c(45, 35, 10, 10))

Chi-squared test for given probabilities

data:  c(45, 35, 10, 10)
X-squared = 38, df = 3, p-value = 2.826e-08

You can conclude from the test that there is some difference between all groups.
Now, what you're interested in is comparing if responses for A are greater than all other choices. You would thus need to test the following hypotheses (comparing proportions):
$$
H_A: p_A > p_B \\
H_A: p_A > p_C \\
H_A: p_A > p_D \\
$$
For the first hypothesis, the null would be that $H_0: p_A = p_B = 0.5$, or that the chances of choosing A or B is 50%. You can then compare proportions. You can do this manually, or in R run the following:
> prop.test(as.table(c(45, 10)), alternative = "greater")

1-sample proportions test with continuity correction

data:  as.table(c(45, 10)), null probability 0.5
X-squared = 21.018, df = 1, p-value = 2.275e-06
alternative hypothesis: true p is greater than 0.5
95 percent confidence interval:
  0.7082738 1.0000000
sample estimates:
  p 
0.8181818 

From this result, you can see that there is no statistical significance. However, because you'd be running multiple comparisons it is important to adjust the p-values. You can look up p-value corrections like Bonferroni correction.
In R you can do this by extracting the p-value from each test, then passing it into p.adjust (I'm using Bonferroni corection).
> p = c(prop.test(as.table(c(45, 35)), alternative = "greater")$p.val,
    prop.test(as.table(c(45, 10)), alternative = "greater")$p.val,
    prop.test(as.table(c(45, 10)), alternative = "greater")$p.val)


> p.adjust(p = p, method = "bonferroni")
[1] 4.714570e-01 6.824182e-06 6.824182e-06

From these final results, you'd be able to state with some statistical significance if there is a difference between choices.
You may also want to summarise your results in terms of effect sizes. See: Using Effect Size—or Why the P Value Is Not Enough
