Answer:
90% confidence interval for the proportion of all Americans age 20 and over with diabetes is [0.0998 , 0.1301].
Step-by-step explanation:
We are given that in a simple random sample of 1200 Americans age 20 and over, there were 138 people with diabetes.
Assuming the data has a normal distribution.
Firstly, the pivotal quantity for 90% confidence interval for the proportion of all Americans age 20 and over with diabetes is given by;
P.Q. = ~ N(0,1)
where, = proportion of Americans having diabetes in a sample of 1200
Americans = = 0.115
n = sample of Americans = 1200
p = population proportion
<em>Here for constructing 90% confidence interval we have used One-sample z proportion statistics.</em>
So, 90% confidence interval for the population proportion, p is ;
P(-1.6449 < N(0,1) < 1.6449) = 0.90 {As the critical value of z at 5% level of
significance are -1.6449 & 1.6449}
P(-1.6449 < < 1.6449) = 0.90
P( < < ) = 0.90
P( < p < ) = 0.90
<u>90% confidence interval for p</u>= [ , ]
= [ , ]
= [0.0998 , 0.1301]
Hence, 90% confidence interval for the proportion of all Americans age 20 and over with diabetes is [0.0998 , 0.1301].