Question

An administrator for a hospital wished to estimate the average number of days required for inpatient treatment of patients between the ages of 25 and 34. A random sample of 500 hospital patients between these ages produced a mean and standard deviation equal to 5.4 and 3.1 days, respectively. Assume that ages of patients are normally distributed.

(a) Construct a 95% confidence interval for the mean length of stay for the population of patients from which the sample was drawn.

(b) Choose a correct interpretation to your result in part (a):

i. We can be 95% certain that the hospital stay for any individual patient is in the interval calculated in part (a).

ii. We can be 95% certain that average stay at the hospital for 500 surveyed patients is in the interval calculated in part (a).

iii. We can be 95% certain that true average stay at the hospital for all patients is in the interval calculated in part (a).

(c) Assume that the population standard deviation of hospital stays, σ, is 3.1 days. You wish to maintain the margin of error obtained in part (a) and construct a 99% confidence interval for the true average number of days required for inpatient treatment of patients between the ages of 25 and 34. Determine the required sample size.

Answer 1

sample size is n = 500

mean = 5.4 and standard deviation (sigma) = 3.1

(A) z score for 95% confidence interval is 1.96 (using z distribution table)

confidence interval =

(B) Correct interpretation of 95% confidence interval is that one can be 95% confident that the true mean value is between (5.13, 5.67)

So, option (iii) is correct

(C) Margin of error in part (A) is 0.2717

z score for 99% confidence interval is 2.58 (using z distribution table)

standard deviation (sd)= 3.1

sample size n = ((z*sd)/ME)^2

= ((2.58*3.1)/0.2717)^2

= 29.4369^2

= 866.53

or sample size n = 867 (rounded to nearest integer)