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swat32
3 years ago
11

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

hich 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)
Mathematics
1 answer:
Andreyy893 years ago
8 0

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

<em>Here for constructing 95% confidence interval we have used One-sample z proportion statistics.</em>

<u>So, 95% confidence interval for the population proportion, p is ;</u>

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

<u>95% confidence interval for p</u> = [\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%].

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