A random sample of 1500 American adults included 450 with hy

A random sample of 1500 American adults included 450 with hypertension. Construct and interpret a 95% confidence interval for the true proportion of American adults with hypertension.

Solution

A random sample of 1500 American adults included 450 with hypertension.

Proportion =450/1500=0.3

95% confidence interval for the true proportion of American adults with hypertension

p1.96

==0.01

=[0.3-1.96*0.01, 0.3+1.96*0.01]

=[0.3-0.02, 0.3+0.02]

=[0.28, 0.32]