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stich3 [128]
3 years ago
9

A simple random sample of 1800 hospital patients admitted in a given year shows that 82.5% 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 admitted that year who had an error on their medical bill?
Mathematics
2 answers:
Fantom [35]3 years ago
6 0
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. 
Talja [164]3 years ago
5 0

Answer:

<em><u>The answer is</u></em>: <u>74 patients.</u>

<u />

Step-by-step explanation:

If we have <u>1800</u> patients of which <u>82.5%</u> have had an error in their medical bill

They are 1800 x 0.825 = 1485 patients.

If of those <u>1485</u> patients we obtain the <u>95%</u> 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 <u>95%</u>, would be equal to 1485 - 1411 = 74 patients.

<em><u>The answer is</u></em>: <u>74 patients.</u>

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