Question

A simple random sample of 2200 hospital patients admitted in a given year shows that 85.5% had an error on their medical bill. Which interval is the 95% confidence interval for the percent of all the hospital's admitted patients that year who had an error on their medical bill?
(84.0, 87.0)


(83.6, 87.4)


(84.7, 86.3)


(85.1, 85.9)

Answer 1

Answer:

95% confidence interval for the percent of all the hospital's admitted patients that year who had an error on their medical bill is [84% , 87%].

Step-by-step explanation:

We are given that a simple random sample of 2200 hospital patients admitted in a given year shows that 85.5% had an error on their medical bill.

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

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

where, $\hat p$ = sample % of patients who  had an error on their medical bill = 85.5%

n = sample of hospital patients = 2200

p = population percentage of all the hospital's admitted patients that year who had an error on their medical bill

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

= [ $0.855-1.96 \times {\sqrt{\frac{0.855(1-0.855)}{2200} } }$ , $0.855+1.96 \times {\sqrt{\frac{0.855(1-0.855)}{2200} } }$ ]

 = [0.84 , 0.87]

= [84% , 87%]

Therefore, 95% confidence interval for the percent of all the hospital's admitted patients that year who had an error on their medical bill is [84% , 87%].