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 4.6425 days according to Statistical Abstract of the United States: 2005.

A first study conducted on a random sample of 23 hospitals in Michigan had a mean LOS for women of 3.8372 days and a standard deviation of 1.1503 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 0.3022 using the following formula,

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

Note: Numbers are randomized for each instance of this question. Use the numbers given above.

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,

sample standard deviation = s = 1.1503

sample size = n = 23

Degrees of freedom = df = n - 1 = 23 - 1 = 22

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t0.025,22 = 2.074

margin of error = E = 0.3022

sample size = n = [t/2,df* s / E]2

n = [2.074 * 1.1503 / 0.3022 ]2

n = 62.32

Sample size = n = 63