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

19

25

10

7

19

24

11

27

16

19

24

25

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 12 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

Answer:

Given

19      25      10      7        19      24      11      27      16      19      24      25

a)

t distribution

b)

Sample mean, xbar = 18.833

Sample standard deviation, s = 6.631

Sample size, n = 12

Degrees of freedom, df = n - 1 = 11

Given CI level is 90%, hence α = 1 - 0.9 = 0.1

α/2 = 0.1/2 = 0.05, tc = t (α/2, df) = 1.796

ME = tc * s/sqrt (n)

ME = 1.796 * 6.631/sqrt (12)

ME = 3.4379

CI = (xbar - tc * s/sqrt (n), xbar + tc * s/sqrt (n))

CI = (18.833 - 1.796 * 6.631/sqrt (12), 18.833 + 1.796 * 6.631/sqrt (12))

CI = (15.395, 22.271)

Between 15.395 and 22.271

c)

About 90% will contain the population mean and about 10% will not contain