Comment: Suppose that Gp 1 had 345 agreements (Likert 4 or 5)
out of 1000 subjects and Gp 2 had 551 agreements out of 1500 subjects. Then as @whuber suggests prop.test
seems appropriate. However, in R this test would not--even nearly--find a significant
difference at the 5% level between proportions
$0.345$ and $0.367$ "in favor." [Even if Gp 2 had a much larger proportion of Likert 5's.]
prop.test(c(345, 551), c(1000, 1500), cor=F)
2-sample test for equality of proportions
without continuity correction
data: c(345, 551) out of c(1000, 1500)
X-squared = 1.3014, df = 1, p-value = 0.2539
alternative hypothesis: two.sided
95 percent confidence interval:
-0.06058569 0.01591902
sample estimates:
prop 1 prop 2
0.3450000 0.3673333
If you need a discussion using a t test, then
please give an example of Likert-5 sample distributions for Groups 1 and 2. (There are
several reasons why a two-sample Wilcoxon test
might not work well.)