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 13 randomly selected physical therapy patients. Round answers to 3 decimal places where possible. 19 19 14 13 26 19 5 14 17 27 13 20 28 a. To compute the confidence interval use a distribution. b. With 90% confidence the population mean number of visits per physical therapy patient is between and 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 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
19	361
19	361
14	196
13	169
26	676
19	361
5	25
14	196
17	289
27	729
13	169
20	400
28	784
∑x=234	∑x2=4716

Mean ˉx=∑xn

=19+19+14+13+26+19+5+14+17+27+13+20+28/13

=23413

=18

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

=√4716-(234)213/12

=√4716-4212/12

=√504/12

=√42

=6.4807

Degrees of freedom = df = n - 1 = 13 - 1 = 12

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t /2,df = t0.05,12 =1.782

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

= 1.782 * (6.481 / 13)

= 3.204

Margin of error = 3.204

The 90% confidence interval estimate of the population mean is,

- E < < + E

18 - 3.204 < < 18 + 3.204

14.796 < < 21.204

(14.796, 21.204 )