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. = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = 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 < [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < 1.96) = 0.95
P( [tex]-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < [tex]{\hat p-p}[/tex] < [tex]1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.95
P( [tex]\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < p < [tex]\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.95
95% confidence interval for p = [[tex]\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] , [tex]\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex]]
= [ [tex]0.855-1.96 \times {\sqrt{\frac{0.855(1-0.855)}{2200} } }[/tex] , [tex]0.855+1.96 \times {\sqrt{\frac{0.855(1-0.855)}{2200} } }[/tex] ]
= [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%].
