One possibility is to consider binomial proportions of
False responses in the two groups. Is the difference in the observed proportions false (roughly $0.098$ and $0.108,$ respectively), significantly different between the two groups?
In R, this test (which uses a normal approximation) is done using the
prop.test procedure. (I have opted not to use continuity correction on account of the large
sample sizes.) The null hypothesis that proportions are equal is strongly rejected
with P-value $0.00012 < .05 = 5\%.$
prop.test(c(3648, 2205), c(37128, 20337), cor=F)
2-sample test for equality of proportions
without continuity correction
data: c(3648, 2205) out of c(37128, 20337)
X-squared = 14.851, df = 1, p-value = 0.0001163
alternative hypothesis: two.sided
95 percent confidence interval:
prop 1 prop 2
An almost-equivalent test is to consider whether a chi-squared test of homogeneity
across Groups is rejected. The appropriate $2\times 2$ table and results of the
test are shown below. Again the null hypothesis of homogeneity between the two groups is strongly rejected.
False = c(3648, 2205); True = c(33480, 18132)
TBL = rbind(False, True); TBL
False 3648 2205
True 33480 18132
Pearson's Chi-squared test
with Yates' continuity correction
X-squared = 14.74, df = 1, p-value = 0.0001234
Notes: You can look at some of the "Related" pages (right margin) found by our robots for more details.
Alternatively, you can look at this NIST page on tests of binomial proportions. Another recent related page.