Question

You are interested in finding a 98% confidence interval for the mean number of visits for physical therapy patients. The data below show the number of visits for 12 randomly selected physical therapy patients. Round answers to 3 decimal places where possible.

27

18

25

26

25

11

23

16

18

28

15

26

a. To compute the confidence interval use a distribution.

b. With 98% confidence the population mean number of visits per physical therapy patient is between answer ____ and answer _______ visits.

c. If many groups of 12 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)
t-distribution

b)
sample mean, xbar = 21.5
sample standard deviation, s = 5.6165
sample size, n = 12
degrees of freedom, df = n - 1 = 11

Given CI level is 98%, hence α = 1 - 0.98 = 0.02
α/2 = 0.02/2 = 0.01, tc = t(α/2, df) = 2.718

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (21.5 - 2.718 * 5.6165/sqrt(12) , 21.5 + 2.718 * 5.6165/sqrt(12))
CI = (17.093 , 25.907)

With 98% confidence the population mean number of visits per physical therapy patient is between 17.093 and 25.907 visits.

c)
About 98 percent of these confidence intervals will contain the true population mean number of visits per patient and about 2 percent will not contain the true population mean number of visits per patient.