Question

In a simple random sample of 1200 Americans age 20 and over, there were 138 people with diabetes.
Report the 90% confidence interval for the proportion of all Americans age 20 and over with diabetes.

Answer 1

Answer:

The 90% confidence interval for the proportion of all Americans age 20 and over with diabetes is between 0.0999 and 0.1301

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

$n = 1200, \pi = \frac{138}{1200} = 0.115$

90% confidence level

So $\alpha = 0.1$, z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$, so $Z = 1.645$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.115 - 1.645\sqrt{\frac{0.115*0.885}{1200}} = 0.0999$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.115 + 1.645\sqrt{\frac{0.115*0.885}{1200}} = 0.1301$

The 90% confidence interval for the proportion of all Americans age 20 and over with diabetes is between 0.0999 and 0.1301

Answer 2

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

where, $\hat p$ = proportion of Americans having diabetes in a sample of 1200

                  Americans = $\frac{138}{1200}$ = 0.115

           n = sample of Americans = 1200

           p = population proportion

Here for constructing 90% confidence interval we have used One-sample z proportion statistics.

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 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.6449) = 0.90

P( $-1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

90% confidence interval for p= [$\hat p-1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.6449 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$]

    = [ $0.115-1.6449 \times {\sqrt{\frac{0.115(1-0.115)}{1200} } }$ , $0.115+1.6449 \times {\sqrt{\frac{0.115(1-0.115)}{1200} } }$ ]

    = [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].