What test to choose for comparing two population proportions and how to present it

Question

I have a study where mothers and fathers of patients completed a questionnaire. I want to compare the proportion of both samples that chose a response n: 348/913 mothers chose n 171/378 fathers chose n

With an online z-calculator i got a Z value of 2.37 which is significant at p=0.0178 My question is, is this test appropriate? Would you not use a t-test instead? Why/Whynot?

And how would i report my result? "There was a significant difference between mothers' and fathers' choice of the response "n" (z = 2.37, p = .0178) with 45% of fathers choosing n while 38% of mother chose "n"". Is this fine?

Thanks

Answer 1

Several approaches are possible. I think the most direct is a test of binomial proportions. There are several slight variations of this test, including differences of opinion whether make a continuity correction. Here is output from the procedure prop.test in R which implements one frequently used version. Depending on the exact formula you use and whether you make a continuity correction, you might get slightly different results. But there is strong evidence (P-value about 0.02) that the proportions choosing n are different for mothers and fathers.

prop.test(c(348,171), c(913, 378))

    2-sample test for equality of proportions 
    with continuity correction

data:  c(348, 171) out of c(913, 378)
X-squared = 5.348, df = 1, p-value = 0.02075
alternative hypothesis: two.sided
95 percent confidence interval:
 -0.13233610 -0.01010379
sample estimates:
  prop 1   prop 2 
0.381161 0.452381 

You could also make a 'contingency table' as shown below and do a chi-squared test for homogeneity of probabilities for mothers and fathers, which gives a significant result (also with P-value 0.02).

Choice     Mothers   Fathers     TOTAL
--------------------------------------
  Yes        348       171         549
  No         565       207         772
--------------------------------------
TOTAL        913       378        1291

One advantage to using the table method is that it provides an opportunity for an important 'reality check': the grand total must be the total number of subjects in the study.

TBL = rbind(c(348,171), c(565,207))
chisq.test(TBL)

TBL
     [,1] [,2]
[1,]  348  171
[2,]  565  207

      Pearson's Chi-squared test 
      with Yates' continuity correction

data:  TBL
X-squared = 5.348, df = 1, p-value = 0.02075

For counts as large as yours, some people feel that the Yates correction should not be used. If it is not used, the P-value is still very nearly 0.02.

chisq.test(TBL, cor=F)$p.val
[1] 0.01755152

Notes: (1) Please look at the link to the NIST handbook or at a basic statistics text for one of the two types of tests I illustrated above. And make sure you understand the hypotheses being tested and how to compute the test statistic.

(2) It is important for subjects in such a study to be selected at random from an appropriate population. I would want to know how random selection of subjects led to so many more mothers (913) than fathers (378). I'm not saying there isn't a valid explanation, but I would want to know it.