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 4.6471 days according to Statistical Abstract of the United States: 2005.
A first study conducted on a random sample of 20 hospitals in Michigan had a mean LOS for women of 3.8917 days and a standard deviation of 1.1767 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 0.3885 using the following formula,
n = (t*sME)2.n = t*sME2.
Note: Numbers are randomized for each instance of this question. Use the numbers given above.
What is the required sample size? Give your answer as a whole number.
Your Answer:
Question 16 options:
Answer |
Solution :
Given that,
sample standard deviation = s = 1.1767
sample size = n = 20
Degrees of freedom = df = n - 1 = 20 - 1 = 19
At 95% confidence level
= 1 - 95%
=1 - 0.95 =0.05
/2
= 0.025
t/2,df
= t0.025,19 = 2.093
margin of error = E = 0.3885
sample size = n = [t/2,df* s / E]2
n = [2.093 * 1.1767 / 0.3885 ]2
n = 40.18
Sample size = n = 41
Get Answers For Free
Most questions answered within 1 hours.