# Interpreting the result of 2-sample test for equality of proportions with continuity correction

I've done 2-sample test for equality of proportions with continuity correction. (This is the second course of statistic I'm studying so I know basically nothing, at least so it feels.)

I'm using R and Rstudio for the first time in this course. I have no idea how to interpret the results of this test.

y1 <- 52
n1 <- 89
y2 <- 40
n2 <- 91
prop.test(c(y1,y2), c(n1,n2),
alternative = "greater", conf.level=0.99)

2-sample test for equality of proportions with continuity correction

data:  c(y1, y2) out of c(n1, n2)
X-squared = 3.2138, df = 1, p-value = 0.03651
alternative hypothesis: greater
99 percent confidence interval:
-0.03792798  1.00000000
sample estimates:
prop 1    prop 2
0.5842697 0.4395604


So p-value is 0.03651. What do I compare it with? And what are the conclusions? Do I reject or retain the null hypothesis? I know pretty much nothing, so please keep it simple.

• What is your $\alpha$? Ie, what is the long-run type I error rate you are willing to live with? Relatedly, why did you specify conf.level=0.99 in prop.test()? – gung - Reinstate Monica Jan 19 '14 at 1:49

As with all statistical tests, you compare the P-value with your significance level. If you use 0.05 as your significance level, you would reject the null hypothesis of equal success probabilities, since 0.03651 < 0.05. If, however, you use 0.01 as your significance level (as implied by your conf.level=0.99), you would not reject the null hypothesis, since 0.03651 ≰ 0.01.