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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)

User Joon Hong
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1 Answer

2 votes

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%].

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{(\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{(\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) } }]

= [
0.855-1.96 * {\sqrt{(0.855(1-0.855))/(2200) } } ,
0.855+1.96 * {\sqrt{(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%].

User ColinTea
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