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 patients 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 percent 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 find the sample size needed for a 99 percent confidence interval with a margin of error of 0.5 days, we can use the formula:

n = (Z*(sigma / E))^2

where:
Z = the z-score for a 99% confidence interval, which is 2.576
sigma = the population standard deviation, which is given as 1.2 days
E = the margin of error, which is 0.5 days

Plugging in these values, we get:

n = (2.576*(1.2/0.5))^2
n = 48.7

Since we can't have a fraction of a hospital, we must round up to the nearest whole number. Therefore, we need a sample size of 49 hospitals to be 99 percent confident that the sample mean is within 0.5 days of the population mean. The best choice is D, 48.

Health Insurers and the Federal Government Are Both Putting Pressure