Question

Bob and Jill each conduct a survey. Bob finds that 22 out of his 100 people read more online books. Jill on the other hand finds that 77 out of her 100 people read more physical books

Calculate a 2 sample Z interval for proportions to determine the interval of the difference between the amount of readers for physical books and online books.

Answer 1

Since the confidence level is not mentioned, I am assuming it to be 95%

n1 = 100

n2 = 100

p1 = 0.77

p2 = 0.22

% = 95

Pooled Proportion, p = (n1 p1 + n2 p2)/(n1 + n2) = (100 * 0.77 + 100 * 0.22)/(100 + 100) = 0.495

q = 1 - p = 1 - 0.495 = 0.505

SE = √(pq * ((1/n1) + (1/n2))) = √(0.495 * 0.505 * ((1/100) + (1/100))) = 0.070707142

z- score = 1.959963985

Width of the confidence interval = z * SE = 1.95996398454005 * 0.0707071424963561 = 0.138583453

Lower Limit of the confidence interval = (p1 - p2) - width = 0.55 - 0.138583452742599 = 0.411416547

Upper Limit of the confidence interval = (p1 - p2) + width = 0.55 + 0.138583452742599 = 0.688583453

The 95% confidence interval is [0.4114, 0.6886]