A test of two binomial proportions in R, seems appropriate to test $H_0: p_1=p_2$ against $H_a: p_1 \ne p_2.$ The two estimated
proportions are $\hat p_1 = 40/300 = 0.13$ and $\hat p_2 = 200/1000 = 0.20,$ so the observed proportions are different.
Then prop.test in R gives a P-value $0.009 < 0.01 = 1\%,$
so the difference is statistically significant at the 1% level.
prop.test(c(40, 200), c(300,1000), cor=F)
2-sample test for equality of proportions
without continuity correction
data: c(40, 200) out of c(300, 1000)
X-squared = 6.8134, df = 1, p-value = 0.009048
alternative hypothesis: two.sided
95 percent confidence interval:
-0.11243026 -0.02090307
sample estimates:
prop 1 prop 2
0.1333333 0.2000000
Notes: (1) Your table is in the correct format for a chi-squared test, shown below. (The different sample sizes are not a problem.) It gives the same P-value as 'prop.test',
TAB = rbind(c(200,40), c(800, 260))
TAB
[,1] [,2]
[1,] 200 40
[2,] 800 260
chisq.test(TAB, cor=F)
Pearson's Chi-squared test
data: TAB
X-squared = 6.8134, df = 1, p-value = 0.009048
(2) I did not use the continuity correctios for the normal approximation in either test on account of the sample sizes over 100.
(3) A test similar to prop.test, which you can try with hand computation is
described on this NIST page.