Question

In a random sample of 1200 Americans age 20 and over, the proportion with diabetes was found to be 0.115 (or 11.5%). Test the hypothesis that one in 10 Americans ages 20 and over has diabetes at significance level α=0.1. Find the 90% confidence interval for the proportion of all Americans age 20 and over with diabetes and compute the p-value of your test. Report and explain your conclusion.

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p = 0.1
Alternative Hypothesis, Ha: p ≠ 0.1

Test statistic,
z = (pcap - p)/sqrt(p*(1-p)/n)
z = (0.115 - 0.1)/sqrt(0.1*(1-0.1)/1200)
z = 1.73

P-value Approach
P-value = 0.0836

As P-value < 0.1, reject the null hypothesis.

Confidence Interval
sample proportion, = 0.115
sample size, n = 1200
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.115 * (1 - 0.115)/1200) = 0.0092

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.64

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.115 - 1.64 * 0.0092 , 0.115 + 1.64 * 0.0092)
CI = (0.0999 , 0.1301)