Health insurers and the federal government are both putting pressure on hospitals to shorten the average length of stay (LOS)of their patients.In 2003,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 2008 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?

Question

Quizplus · Accepted Answer

We can use the formula for the margin of error:
Margin of error = z*(standard deviation/sqrt(sample size))
We want the margin of error to be 0.5 days, and we want to be 99% confident, so our z-value is 2.576 (from a standard normal distribution table).
Plugging in the values we have:
0.5 = 2.576*(1.2/sqrt(sample size))
Solving for sample size:
sqrt(sample size) = 2.576*(1.2/0.5)
sqrt(sample size) = 6.1392
sample size = 37.7 (rounded up to the nearest whole number)
So we need a sample size of at least 48 (for a little extra confidence). Therefore, the answer is D (48).

Health Insurers and the Federal Government Are Both Putting Pressure