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 had a mean LOS for non-heart patients in 2000 of 3.8 days and a standard deviation of 1.2 days.Assume LOS is normally distributed.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?
A)3
B)7
C)32
D)48
E)96

Question

Quizplus · Accepted Answer

To calculate the required sample size, we can use the formula:

n = (zα/2 * σ / E)^2

where:

zα/2 = 2.576 (from the standard normal distribution table for 99% confidence level)
σ = 1.2 (the known standard deviation of the population)
E = 0.5 (the margin of error)

Plugging in the values, we get:

n = (2.576 * 1.2 / 0.5)^2
n = 48.98

We round up to the nearest whole number, so the required sample size is 49 hospitals. Therefore, the best choice is D (48).

Health Insurers and the Federal Government Are Both Putting Pressure