Question

Length of Stay ~ Health insurers and federal government are both putting pressure on hospitals to shorten the average length of stay (LOS) of their patients. The average LOS for women is 4.55 days according to Statistical Abstract of the United States: 2005.

A first study conducted on a random sample of 25 hospitals in Michigan had a mean LOS for women of 3.8 days and a standard deviation of 1.17 days.

A medical researcher wants to determine the sample size required to estimate the mean LOS for women in Michigan at 95% confidence with a margin of error of no more than 0.25 using the following formula,

n = (t*s/ME)^2.

A. What is the degrees of freedom used to calculate the t* multiplier for this scenario?

B. What is the value of the t* multiplier you would substitute in the formula above? Give your answer to 4 decimal places.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 3.8

sample standard deviation = s = 1.17

sample size = n = 25

A) Degrees of freedom = df = n - 1 = 25 - 1 = 24

B) At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t_0.025,24 = 2.0639

margin of error = E = 0.25

sample size = n = [t/2,df* s / E]²

n = [2.0639 * 1.17 / 0.25 ]²

n = 93.29

Sample size = n = 94