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Hospital length of stay. Health insurers and the federal
Chapter 6, Problem 32E(choose chapter or problem)
Problem 32E
Hospital length of stay. Health insurers and the federal government are both putting pressure on hospitals to shorten the average length of stay (LOS) of their patients. The average LOS for men in the United States is 5.4 days, and the average for women is 4.6 days (Statistical Abstract of the United States: 2012). A random sample of 20 hospitals in one state had a mean LOS for women of 3.8 days and a standard deviation of 1.2 days.
a. Use a 90% confidence interval to estimate the population mean LOS for women for the state’s hospitals.
b. Interpret the interval in terms of this application.
c. What is meant by the phrase “90% confidence interval”?
Questions & Answers
QUESTION:
Problem 32E
Hospital length of stay. Health insurers and the federal government are both putting pressure on hospitals to shorten the average length of stay (LOS) of their patients. The average LOS for men in the United States is 5.4 days, and the average for women is 4.6 days (Statistical Abstract of the United States: 2012). A random sample of 20 hospitals in one state had a mean LOS for women of 3.8 days and a standard deviation of 1.2 days.
a. Use a 90% confidence interval to estimate the population mean LOS for women for the state’s hospitals.
b. Interpret the interval in terms of this application.
c. What is meant by the phrase “90% confidence interval”?
ANSWER:
Answer
Step 1 of 3
(a)
The average length of stay for men in the United States is days, and the average for women is
days.
A random sample of 20 hospitals in one state had a mean LOS for women of 3.8 days and a standard deviation of 1.2 days.
We are asked to find a confidence interval to estimate the population mean LOS for women for the state’s hospitals.
For a small random sample, we use to compute the confidence interval.
Assume the measurements come from a normal population.
Let be a small random sample from a normal population with mean
. then a level
confidence interval for
is
……….(1)
Where is a Student’s
distribution with
degrees of freedom,
is sample mean,
is a sample standard deviation and
is a number of samples.
Hence ,
and
.
For a 90% confidence interval, the value of is,
and
Putting the value into equation (1)
The value of = 1.729 from t table III, Appendix D, hence our equation become,
Hence a confidence interval to estimate the population mean LOS for women for the state’s hospitals is
.