Question

You are interested in finding a 90% confidence interval for the mean number of visits for physical therapy patients. The data below show the number of visits for 15 randomly selected physical therapy patients. Round answers to 3 decimal places where possible.

13

26

16

13

6

27

26

6

13

25

6

11

20

14

16

a. To compute the confidence interval use a ? z t  distribution.

b. With 90% confidence the population mean number of visits per physical therapy patient is between  and   visits.

c. If many groups of 15 randomly selected physical therapy patients are studied, then a different confidence interval would be produced from each group. About  percent of these confidence intervals will contain the true population mean number of visits per patient and about  percent will not contain the true population mean number of visits per patient.

Answer 1

a)
use t distribution

b)

sample mean, xbar = 15.867
sample standard deviation, s = 7.434
sample size, n = 15
degrees of freedom, df = n - 1 = 14

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.761

ME = tc * s/sqrt(n)
ME = 1.761 * 7.434/sqrt(15)
ME = 3.4

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (15.867 - 1.761 * 7.434/sqrt(15) , 15.867 + 1.761 * 7.434/sqrt(15))
CI = (12.487 , 19.247)

With 90% confidence the population mean number of visits per physical therapy patient is between 12.487 and 19.247 visits.

c)
. If many groups of 15 randomly selected physical therapy patients are studied, then a different confidence interval would be produced from each group. About 90 percent of these confidence intervals will contain the true population mean number of visits per patient and about 10 percent will not contain the true population mean number of visits per patient.