Health insurers and the federal government are both putting pressure on hospitals to shorten the average length of stay (LOS)of their patients.In 1996,the average LOS for non-heart patient was 4.6 days.A random sample of 20 hospitals in one state had a mean LOS for non-heart patients in 2000 of 3.8 days and a standard deviation of 1.2 days.How large a sample of hospitals would we need to be 99% confident that the sample mean is within 0.5 days of the population mean?

Question

Quizplus · Accepted Answer

We need to use the formula for the margin of error:

Margin of error = z*(standard deviation/sqrt(sample size))

Where z is the z-score for the confidence level, which is 2.58 for 99% confidence.

We want the margin of error to be 0.5 days, so we can rearrange the formula to solve for the sample size:

sample size = [(z*standard deviation)/margin of error]^2

Plugging in the values given,

sample size = [(2.58*1.2)/0.5]^2 = 48.38

We always round up for sample size, so we need a sample size of at least 49 hospitals.

Therefore, the best choice is D) 48.

Health Insurers and the Federal Government Are Both Putting Pressure