Measuring the magnitude and the significance of the difference of two binomial samples I have two samples with binomial distribution. In one of them, the proportion of successes was 10/20 (0.5), and in the other the proportion of successes was 5/20 (0.25). I want to test whether the difference between the two is significant and what is the magnitude of the difference.
I thought of using the following code in R:
prop.test(c(10,5),c(20,20))

Which generates the following result:
2-sample test for equality of proportions with continuity correction

data:  c(10, 5) out of c(20, 20)
X-squared = 1.7067, df = 1, p-value = 0.1914
alternative hypothesis: two.sided
95 percent confidence interval:
 -0.08988258  0.58988258
sample estimates:
prop 1 prop 2 
  0.50   0.25

Question: Can I use the chi square value (1.7067) as a measure of magnitude and the p value (0.1914) as a measure of the significance of the difference between the two samples?
Is there a better way to answer these questions?

This test takes sample size into account to see if the two
proportions are 'significantly' different in a statistical
sense. Results 10 out of 20 and and 5 out of 20 could easily
occur by chance. As @Henry comments (+1), the same proportions
at larger sample sizes would be significantly different at the 5% level. Double the sample size to consider 20 out of 40 vs. 10 out of 40. Now the P-value is about $0.04 < 0.05 = 5\%.$ [Using R.]
prop.test(c(20,10),c(40,40))

        2-sample test for equality of proportions 
        with continuity correction

data:  c(20, 10) out of c(40, 40)
X-squared = 4.32, df = 1, p-value = 0.03767
alternative hypothesis: two.sided
95 percent confidence interval:
 0.02002206 0.47997794
sample estimates:
prop 1 prop 2 
  0.50   0.25 

Below is a plot of confidence intervals for $n = 16, 20, 24, \dots, 60$ corresponding to $\hat p_1 = 1/4, \hat p_2 = 1/2.$
We see that $n$ must be at least $36$ for the CI for $p_2 - p_1$ not to include $0.$ [For $n < 16$ the normal approximation is unreliable.]

plot(c(-.7,.2),c(11,65), col="white", ylab="n", 
     xlab = "Diff in Proportions") 
 abline(v = 0, col="red")
 abline(h = 36, col="green")
for (i in 4:15) 
 { CI = prop.test(c(i,2*i), c(4*i,4*i))$conf.int 
 lci = CI[1]; uci = CI[2]
 lines(c(lci,uci), c(4*i,4*i), lwd=2, col="blue")
}    

