Question

In: Statistics and Probability

Suppose you monitor quality assurance for a local hospital and want to estimate the average length...

Suppose you monitor quality assurance for a local hospital and want to estimate the average length of stay (LOS) at your hospital. You take a random sample of 30 patients and find that the average LOS is 3.4 days with a sample standard deviation of 1.2 days. What is the 99% confidence interval for the population average length of stay?

Expert Solution

Solution :

Given that,

Point estimate = sample mean = = 3.4

sample standard deviation = s = 1.2

sample size = n = 30

Degrees of freedom = df = n - 1 = 30 - 1 = 29

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t_/2,df = t_0.005,29 = 2.756

Margin of error = E = t_/2,df * (s / $\sqrt$ n)

= 2.756 * (1.2 / $\sqrt$ 30)

E = 0.6

The 99% confidence interval estimate of the population mean is,

- E < < + E

3.4 - 0.6 < < 3.4 + 0.6

2.8 < < 4.0

( 2.8,4.0 )

The 99% confidence interval is (2.8 , 4.0)

orchestra answered 1 year ago

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