Answer:This is a test of 2 population proportions. Let 1 and 2 be the subscript for the subjects treated with echinacea and subjects treated with a placebo. The population proportion of subjects treated with echinacea and subjects treated with a placebo would be p1 and p2 respectively.
P1 - P2 = difference in the proportion of sample of subjects treated with echinacea the sample of subjects treated with a placebo.
The null hypothesis is
H0 : p1 = p2
p1 - p2 = 0
The alternative hypothesis is
Ha : p1 ≠ p2
p1 - p2 ≠ 0
it is a two tailed test
Sample proportion = x/n
Where
x represents number of success(number of complaints)
n represents number of samples
For women
x1 = 37
n1 = 44
P1 = 37/44 = 0.84
For men,
x2 = 91
n2 = 104
P2 = 91/104 = 0.88
The pooled proportion, pc is
pc = (x1 + x2)/(n1 + n2)
pc = (37 + 91)/(44 + 104) = 0.86
1 - pc = 1 - 0.86 = 0.14
z = (P1 - P2)/√pc(1 - pc)(1/n1 + 1/n2)
z = (0.84 - 0.88)/√(0.86)(0.14)(1/44 + 1/104) = - 0.04/0.045
z = - 0.64
Since it is a two tailed test, the curve is symmetrical. We will look at the area in both tails. Since it is showing in one tail only, we would double the area
From the normal distribution table, the area below the test z score in the left tail 0.26
We would double this area to include the area in the right tail of z = 0.64 Thus
p = 0.26 × 2 = 0.52
By using the p value,
Since 0.01 < 0.52, we would accept the null hypothesis.