Question

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 3.8 days with a sample standard deviation of 1.2 days. What is the 95% confidence interval for the population average length of stay?

Select one:

(3.43, 4.17)

(3.35, 4.25)

(3.20, 4.40)

(3.61, 3.82)

none of these are correct

Answer 1

Solution :

Given that,

t$\alpha$ /2,df = 2.045

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

= 2.045 * (1.2 / $\sqrt$ 30)

Margin of error = E = 0.45

The 95% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

3.8 - 0.45 < $\mu$ < 3.8 + 0.45

3.35 < $\mu$ < 4.25

(3.35 , 4.25)