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", then x = 1219 and n = 3532. The sample proportion is:
p^ = x/n = 1219/3532 = 0.345
This is the point estimate, i.e our best estimate of the proportion of the population on treatment for hypertension is 34.5%.
Compute the 95% confidence interval for the proportion of the population which are on treatment. Z-value for 96% area is 1.96.

Answer 1

Answer:

95% confidence interval for the proportion of the population which are on treatment is [0.3293 , 0.3607].

Step-by-step 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.

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 population is given by;

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

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

           n = sample of participants = 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 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

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

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

95% confidence interval for p= [$\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$]

    = [ $0.345-1.96 \times {\sqrt{\frac{0.345(1-0.345)}{3532} } }$ , $0.345+1.96 \times {\sqrt{\frac{0.345(1-0.345)}{3532} } }$ ]

    = [0.3293 , 0.3607]

Hence, 95% confidence interval for the proportion of the population which are on treatment is [0.3293 , 0.3607].