(1) In 2018, you have 11,000 voters out of 14,000 potentially available,
and in 2020, you have 10,000 voters out of 13,500.
In R, assuming randomness in willingness and ability to vote, you could compare the two proportions of voters
using prop.test
as follows (declining continuity correction with parameter cor=F
on account of large sample sizes):
prop.test(c(11000,10000), c(14000,13500), cor=F)
2-sample test for equality
of proportions without
continuity correction
data: c(11000, 10000) out of c(14000, 13500)
X-squared = 77.015, df = 1, p-value < 2.2e-16
alternative hypothesis: two.sided
95 percent confidence interval:
0.03493139 0.05501570
sample estimates:
prop 1 prop 2
0.7857143 0.7407407
The proportions $.786$ and $.741$ are significantly
different on account of the P-value near $0.$
Alternatively, you could do a chi-squared test on
a $2\times 2$ table:
TBL = rbind(c(11000, 10000), c(3000, 3500)); TBL
[,1] [,2]
[1,] 11000 10000
[2,] 3000 3500
chisq.test(TBL, cor=F)
Pearson's Chi-squared test
data: TBL
X-squared = 77.015, df = 1, p-value < 2.2e-16
Except for the input syntax, the two tests are
essentially equivalent.
(2) You need to use counts (as above) instead of
percentages.