Question

A survey was taken of randomly selected​ Americans, age 65 and​ older, which found that 413 of 1005 men and 535 of 1057 women suffered from some form of arthritis.

A) Let p1 be the sample proportion of senior women suffering from some form of arthritis and let p2 be the sample proportion of senior men suffering form of arthritis . Create a 95% confidence interval for the difference in the proportions of senior men and women who have this disease p1-p2. The confidence interval is (___,___). (Round to 3 decimal places as needed).

B) There is 95% confidence, based on these samples that the proportion of American women, age 65 and older, who suffer from arthritis is between ___% and ___% more than the proportion of American men of the same age who suffer from arthritis. (round to one decimal place).

Answer 1

p1cap = X1/N1 = 535/1057 = 0.5061
p1cap = X2/N2 = 413/1005 = 0.4109

Here, , n1 = 1057 , n2 = 1005
p1cap = 0.5061 , p2cap = 0.4109

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.5061 * (1-0.5061)/1057 + 0.4109*(1-0.4109)/1005)
SE = 0.0218

For 0.95 CI, z-value = 1.96
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.5061 - 0.4109 - 1.96*0.0218, 0.5061 - 0.4109 + 1.96*0.0218)
CI = (0.052 , 0.138)

B)
There is 95% confidence, based on these samples that the proportion of American women, age 65 and older, who suffer from arthritis is between 5.2% and 13.8% more than the proportion of American men of the same age who suffer from arthritis.