Question

Length of Stay ~ Health insurers and federal government are both putting pressure on hospitals to...

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 1.15 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*sME)2. What is the value of the t* multiplier you would substitute in the formula above? Give your answer to 4 decimal places.

Please explain with detail how to do this. I am really lost.

Homework Answers

Answer #1

Solution :

Given that,

Point estimate = sample mean = = 3.8

sample standard deviation = s = 1.15

sample size = n = 25

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

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t0.025,24 = 2.0639

Margin of error = E = 0.25

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

n = [2.0639 * 1.15 / 0.25 ]2

n = 90.13

Sample size = n = 91

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Length of Stay ~ Health insurers and federal government are both putting pressure on hospitals to...
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...
Length of Stay ~ Health insurers and federal government are both putting pressure on hospitals to...
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...
Length of Stay ~ Health insurers and federal government are both putting pressure on hospitals to...
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...
Health insurers and the federal government are both putting pressure on hospitals to shorten the average...
Health insurers and the federal government are both putting pressure on hospitals to shorten the average length of stay of their patients from the current average of 3.8 days. The correct alternate hypothesis is?
The length of maternity stay in U.S. hospitals is normally distributed, with mean of 2.5 days...
The length of maternity stay in U.S. hospitals is normally distributed, with mean of 2.5 days and a standard deviation of 1.1 days. We randomly survey 92 women who recently bore children in U.S. hospitals. Let X= maternity stay in U.S. hospital, and X¯= average maternity stay for a sample of size 92 .   X∼ (pick one)PBEN ( , ) .   X¯∼ (pick one)PBEN ( , ) .    What is the probability that the average stay is more than...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT