Question 1203708
you want to test whether adding a dish of pizzazz gets more replies or whether the sample differences are due to random variations in different samples.


no pizzazz gets 65 out of 100 responses.
pizzazz gets 78 out of 98 responses.


p1 = 65/100 = .65.
p2 = 78/98 = .7959183673.
p0 = (65 + 78) / (100 + 98) = 143/198 = .7222222222.
1 - p0 = .2777777777.


z-score formula is:
z = (p1 - p2) / s
s = standard error.


s = sqrt(p0 * (1-p0) * (1/n1 + 1/n2))
n1 is sample size of sample 1.
n2 is sample size of sample 2.
s becomes equal to sqrt(.7222222222 * .2777777777 * (1/100 + 1/98)) = .0636654379.


z = (p1 - p2) / s becomes z = (.65 - .7959183673) / .0636654379 which is equal to -2.291955764.


the area to the left of that z-score of -2.291955764 is equal to .0109540673.
that's your test alpha.
it is less than your critical alpha of .05.
that makes the results of the test significant, which means that the sample results are most likely not due to random variations in sample proportions, but an actual difference.
the conclusion is that the use of pizzazz does generate greater response rates.


i used an online calculator and got the same results.
here they are:


<img src = "http://theo.x10hosting.com/2023/091902.jpg">