Question

You are interested in finding a 95% confidence interval for the mean number of visits for physical therapy patients. The data below show the number of visits for 10 randomly selected physical therapy patients.

17 24 24 19 7 25 9 9 11 15

a. To compute the confidence interval use a distribution.

b. With 95% confidence the population mean number of visits per physical therapy patient is answer ___ between and answer ____ visits.

c. If many groups of 10 randomly selected physical therapy patients are studied, then a different confidence interval would be produced from each group. About answer ____ percent of these confidence intervals will contain the true population mean number of visits per patient and about answer ___ percent will not contain the true population mean number of visits per patient.

Answer 1

a)
Student's t distribution

b)
sample mean, xbar = 16
sample standard deviation, s = 6.86
sample size, n = 10
degrees of freedom, df = n - 1 = 9

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.262

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (16 - 2.262 * 6.86/sqrt(10) , 16 + 2.262 * 6.86/sqrt(10))
CI = (11.09 , 20.91)

c)
About 95 percent of these confidence intervals will contain the true population mean number of visits per patient and about 5 percent will not contain the true population mean number of visits per patient.