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.Calculate a 95 percent confidence interval for the population mean LOS for non-heart patients in these hospitals in 2008.

Question

Quizplus · Accepted Answer

The correct answer is A.The 95% confidence interval for the population mean LOS for non-heart patients in these hospitals in 2008 is [3.24,4.36]. This interval was calculated using the formula:CI = x̄ ± z * (s/√n)where x̄ is the sample mean, z is the z-score corresponding to the desired confidence level, s is the sample standard deviation, and n is the sample size.In this case, x̄ = 3.8, z = 1.96 (for a 95% confidence level), s = 1.2, and n = 20. Plugging these values into the formula, we get:CI = 3.8 ± 1.96 * (1.2/√20) = [3.24,4.36]

Health Insurers and the Federal Government Are Both Putting Pressure