What is the observed threshold (p-value) of this two-tailed test?

Question

Two teams of researchers have each developed a treatment against atypical pneumonia. The effectiveness of these two treatments is compared by administering them to two separate groups of patients chosen at random from the victims of this disease. Group 1 comprises 45 individuals while group 2 comprises 35. We observed 34 cures in the first group and 28 in the second group.

What is the observed threshold (p-value) of this two-tailed test?

I found 0.34 here but I am really not sure.

Answer 1

This is a test to compare to proportions. Usually, the test uses a normal approximation for sufficiently large samples, which you have. There are several formulas for the z statistic, some using a pooled estimate for the common proportion (under $H_0)$ and some estimating proportions separately.

Here is output for the prop.test in R:

prop.test(c(34, 28), c(45,35), cor=F)


        2-sample test for equality of proportions 
        without continuity correction

data:  c(34, 28) out of c(45, 35)
X-squared = 0.22302, df = 1, p-value = 0.6367
alternative hypothesis: two.sided
95 percent confidence interval:
 -0.2270020  0.1381131
sample estimates:
   prop 1    prop 2 
0.7555556 0.8000000 

The P-value is 0.6368 for this version of the test. Thus the null hypothesis is not rejected at the 5% level of significance. [This version of the test gives a chi-squared statistic, which is the square of the relevant z-statistic.]

Here is the version of the test implemented in a recent release of Minitab.

Test and CI for Two Proportions 

Sample   X   N  Sample p
1       34  45  0.755556
2       28  35  0.800000

Difference = p (1) - p (2)
Estimate for difference:  -0.0444444
95% CI for difference:  (-0.227002, 0.138113)
Test for difference = 0 (vs ≠ 0):  Z = -0.47  P-Value = 0.637

The P-value is about the same as from R.

I suspect that your P-value 0.34 is for a one-sided test. (If I do the appropriate one-sided test in R, I get P-value 0.317.) Without knowing the formula for your z-statistic, I can't say for sure how you got your P-value.