Question

A survey was taken of randomly selected​ Americans, age 65 and​ older, which found that 420 of 1012 men and 531 of 1061 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 from some 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 and interpret the results.

b) Does this suggest that arthritis is more likely to afflict women than​ men? Explain.

Answer 1

a)

Here, , n1 = 1061 , n2 = 1012
p1cap = 0.5005 , p2cap = 0.415

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.5005 * (1-0.5005)/1061 + 0.415*(1-0.415)/1012)
SE = 0.0218

For 0.95 CI, z-value = 1.96
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.5005 - 0.415 - 1.96*0.0218, 0.5005 - 0.415 + 1.96*0.0218)
CI = (0.0428 , 0.1282)

b)
p1cap = X1/N1 = 531/1061 = 0.5005
p1cap = X2/N2 = 420/1012 = 0.415
pcap = (X1 + X2)/(N1 + N2) = (531+420)/(1061+1012) = 0.4588

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p1 = p2
Alternate Hypothesis, Ha: p1 > p2

Rejection Region
This is right tailed test, for α = 0.05
Critical value of z is 1.64.
Hence reject H0 if z > 1.64

Test statistic
z = (p1cap - p2cap)/sqrt(pcap * (1-pcap) * (1/N1 + 1/N2))
z = (0.5005-0.415)/sqrt(0.4588*(1-0.4588)*(1/1061 + 1/1012))
z = 3.91

P-value Approach
P-value = 0
As P-value < 0.05, reject the null hypothesis.