Question

a hospital administrator wants to estimate the mean length of stay for all inpatients in the hospital. Based on a random sample of 676 patients from the previous year, she finds that the sample mean is 5.3 days with a standard deviation of 1.2 days. Construct and interpret a 95% and a 99% confidence interval for the mean.

Answer 1

a)

df = n - 1 = 676 - 1 = 675

t critical value at 0.05 significance level with 675 df = 1.963

95% confidence interval for is

- t * S / sqrt(n) < < + t * S / sqrt(n)

5.3 - 1.963 * 1.2 / sqrt(676) < < 5.3 + 1.963 * 1.2 / sqrt(676)

5.21 < < 5.39

95% CI is ( 5.21 , 5.39 )

Interpretation - We are 95% confident that the mean length of stay for all inpatients in the hospital is between

5.21 day and 5.39 day.

b)

df = n - 1 = 676 - 1 = 675

t critical value at 0.01 significance level with 675 df = 2.583

99% confidence interval for is

- t * S / sqrt(n) < < + t * S / sqrt(n)

5.3 - 2.583 * 1.2 / sqrt(676) < < 5.3 + 2.583 * 1.2 / sqrt(676)

5.18 < < 5.42

99% CI is ( 5.18 , 5.42 )

Interpretation - We are 99% confident that the mean length of stay for all inpatients in the hospital is between

5.18 day and 5.42 day.