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.A 95% confidence interval for the population mean LOS for non-heart patients in hospitals for 2000 would be _____.
A)[3.24,4.36]
B)[3.67,3.93]
C)[3.34,4.26]
D)[3.38,4.22]
E)[3.27,4.33]

Question

Quizplus · Accepted Answer

The 95% confidence interval for the population mean can be calculated using the formula: $\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $s$ is the sample standard deviation, $n$ is the sample size, and $t_{\alpha/2}$ is the t-score for a 95% confidence level with $n-1$ degrees of freedom. For a sample size of 20, the degrees of freedom is 19. Looking up the t-score for 19 degrees of freedom for a 95% confidence interval gives us approximately 2.093. Plugging in the values: $3.8 \pm 2.093 \cdot \frac{1.2}{\sqrt{20}}$ gives us the interval $[3.24, 4.36]$.

Health Insurers and the Federal Government Are Both Putting Pressure