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)

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. =    ~ N(0,1)

where, = 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 < < 1.96) = 0.95

P( < < ) = 0.95

P( < p < ) = 0.95

95% confidence interval for p = [ , ]

= [ , ]

 = [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%].

maybe i can

x = -4

step-by-step explanation:

- 20 - 4(x + 3) = 4x

=> -20 - 4x - 12 = 4x

=> - 32 -4x = 4x

=> -32 = 4x + 4x

=> - 32 = 8x

=> -32/8 = x

=> -4 = x