Answer:
Explanation:
This is a test of 2 population proportions. Let 1 and 2 be the subscript for men and women players. The population proportions of men and women challenges for calls would be p1 and p2
P1 - P2 = difference in the proportion of men and women challenges for calls.
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 old dough
x1 = 416
n1 = 1406
P1 = 416/1406 = 0.3
For new dough,
x2 = 217
n2 = 778
P2 = 217/778 = 0.28
The pooled proportion, pc is
pc = (x1 + x2)/(n1 + n2)
pc = (416 + 217)/(1406 + 778) = 0.29
1 - pc = 1 - 0.29 = 0.71
z = (P1 - P2)/√pc(1 - pc)(1/n1 + 1/n2)
z = (0.3 - 0.28)/√(0.29)(0.71)(1/1406 + 1/778) = 0.02/√0.00553857907
z = 0.99
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 to the left of 0.99 is 1 - 0.84 = 0.16
We would double this area to include the area in the left tail of z = - 0.99 Thus
p = 0.16 × 2 = 0.32
Since 0.1 < 0.5, we would accept the null hypothesis.
p value = 0.337
Since 0.05 < 0.32, we would accept the null hypothesis