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

5

24

21

21

15

25

23

12

7

6

9

20

19

19

a. To compute the confidence interval use a Z or T distribution?

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

c. If many groups of 14 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

Solution:

x	x2
5	25
24	576
21	441
21	441
15	225
25	625
23	529
12	144
7	49
6	36
9	81
20	400
19	361
19	361
∑x=226	∑x2=4294

Mean ˉx=∑xn

=5+24+21+21+15+25+23+12+7+6+9+20+19+19//14

=226/14

=16.1429

Sample Standard deviation S=√∑x2-(∑x)2nn-1

=√4294-(226)214/13

=√4294-3648.2857/13

=√645.7143/13

=√49.6703

=7.0477

Degrees of freedom = df = n - 1 = 25 - 1 = 24

At 98% confidence level the t is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

t /2,df = t0.01,13 =2.650

Margin of error = E = t/2,df * (s /n)

= 2.650 * (7.042 / 14)

= 4.988

Margin of error = 4.988

The 98% confidence interval estimate of the population mean is,

- E < < + E

16.143 - 4.988 < < 16.143 + 4.988

11.155 < < 21.131

c. If many groups of 14 randomly selected physical therapy patients are studied, then a different confidence interval would be produced from each group. About 70% percent of these confidence intervals will contain the true population mean number of visits per patient and about 30%percent will not contain the true population mean number of visits per patient