Respuesta :
We are given a sample size of 1800 hospital patients. The data for a year shows that 82.5% of the patients had an error on their medical bill, meaning:
1800 * 0.825 = 1485 patients have errors on their medical bills.
Given the confidence interval of 95% with an equivalent z value of 1.96, the number of patients who have medical errors on their medical bills is:
1485 patients +- (considering the confidence interval, z value 1.96) use the z-test to arrive at the final answer.
1800 * 0.825 = 1485 patients have errors on their medical bills.
Given the confidence interval of 95% with an equivalent z value of 1.96, the number of patients who have medical errors on their medical bills is:
1485 patients +- (considering the confidence interval, z value 1.96) use the z-test to arrive at the final answer.
Answer:
The answer is: 74 patients.
Step-by-step explanation:
If we have 1800 patients of which 82.5% have had an error in their medical bill
They are 1800 x 0.825 = 1485 patients.
If of those 1485 patients we obtain the 95% confidence interval, it would be equal to: 1485 x 0.95 = 1410.75, which would be rounded in 1411 patients.
The confidence interval between 1485 who had an error in their bill over 95%, would be equal to 1485 - 1411 = 74 patients.
The answer is: 74 patients.