Question

(1 point) In a study of red/green color blindness, 650 men and 2400 women are randomly selected and tested. Among the men, 58 have red/green color blindness. Among the women, 7 have red/green color blindness. Construct the 95% confidence interval for the difference between the color blindness rates of men and women. 0.865 < (PM – Pw) < 0.9133

Answer 1

The pooled proportion here is computed as:
P = (58 + 7) / (650 + 2400) = 0.0213

The standard error here is computed as:

$SE = \sqrt{P(1-P)(\frac{1}{n_1} + \frac{1}{n_2})} = \sqrt{0.0213( 1 - 0.0213)(\frac{1}{650} + \frac{1}{2400})} = 0.0064$

From standard normal tables, we have here:
P(-1.96 < Z < 1.96) = 0.95

The sample proportions here are computed as:
p1 = 58/650 = 0.0892
p2 = 7/2400 = 0.002917

Therefore the confidence interval here is obtained as:

L = p1 - p2 - z*SE = 0.0892 - 0.002917 - 1.96*0.0064 = 0.0737
U = p1 - p2 + z*SE = 0.0892 - 0.002917 + 1.96*0.0064 = 0.0988

Therefore the confidence interval here is from 0.0737 to 0.0988.