Question

During the 7th examination of the Offspring cohort in the Framingham Heart Study there were 1219 participants being treated for hypertension and 2,313 who were not on treatment. If we call treatment a "success", find the 95% confidence interval for the proportion of people in the population being treated for hypertension.

Answer 1

95% confidence interval for the proportion of people in the population being treated for hypertension is [0.3293 , 0.3607].

Explanation:

We are given that during the 7th examination of the Offspring cohort in the Framing ham Heart Study, there were 1219 participants being treated for hypertension and 2,313 who were not on treatment.

So, the sample proportion is :
`\hat p` = x/n = 1219/3532 = 0.345

Firstly, the pivotal quantity for 95% confidence interval for the proportion of the people in the population is given by;

P.Q. =
`\frac{\hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }` ~ N(0,1)

where,
`\hat p` = sample proportion = 0.345

n = sample of participants = 1219 + 2313 = 3532

p = population proportion

Here for constructing 95% confidence interval we have used One-sample z proportion statistics.

So, 95% confidence interval for the population​ proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level of

significance are -1.96 & 1.96}

P(-1.96 <
`\frac{\hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }` < 1.96) = 0.95

P(
`-1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }` <
`{\hat p-p}` <
`1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ) = 0.95

P(
`\hat p-1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }` < p <
`\hat p+1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ) = 0.95

95% confidence interval for p= [
`\hat p-1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ,
`\hat p+1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }`]

= [
`0.345-1.96 * {\sqrt{(0.345(1-0.345))/(3532) } }` ,
`0.345+1.96 * {\sqrt{(0.345(1-0.345))/(3532) } }` ]

= [0.3293 , 0.3607]

Hence, 95% confidence interval for the proportion of people in the population being treated for hypertension is [0.3293 , 0.3607].