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 13 randomly selected physical therapy patients. Round answers to 3 decimal places where possible.

25 18 18 18 17 11 27 28 9 18 25 23 12

a. To compute the confidence interval use a distribution.

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

c. If many groups of 13 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)

use a t distribution

b)

sample mean, xbar = 19.1538
sample standard deviation, s = 6.1488
sample size, n = 13
degrees of freedom, df = n - 1 = 12

Given CI level is 98%, hence α = 1 - 0.98 = 0.02
α/2 = 0.02/2 = 0.01, tc = t(α/2, df) = 2.681

ME = tc * s/sqrt(n)
ME = 2.681 * 6.1488/sqrt(13)
ME = 4.572

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (19.1538 - 2.681 * 6.1488/sqrt(13) , 19.1538 + 2.681 * 6.1488/sqrt(13))
CI = (14.582 , 23.726)

With 98% confidence the population mean number of visits per physical therapy patient is 14.582 between and 23.726 visits

c)

If many groups of 13 randomly selected physical therapy patients are studied, then a different confidence interval would be produced from each group. 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.