Say I have two samples and I am measuring the amount of times a molecule appears in each. In sample 1, this particular molecule appears 200 out of the 1000 total molecules measured. For sample 2, it's 40 out of 300 total molecules.

If I want to see if this difference is statistically significant, do I use a chi-square test where the contingency table would be something like this?

200 | 800 
40  | 260

Or is a different test more appropriate? Does it matter if the two samples have very different numbers of total molecules measured?


1 Answer 1


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))
     [,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 various correctios in these two tests (arguments cor=F) 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.