Question

You are interested in finding a 95% 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.

17

6

7

9

7

15

15

11

6

15

25

14

26

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

b. With 95% 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

a) t -distribution is used because  population std dev is unknown

b)

sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   6.6506
Sample Size ,   n =    13
Sample Mean,    x̅ = ΣX/n =    13.3077

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   12          
't value='   tα/2=   2.1788   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   6.651   / √   13   =   1.8446
margin of error , E=t*SE =   2.1788   *   1.845   =   4.019
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    13.31   -   4.019   =   9.2888
Interval Upper Limit = x̅ + E =    13.31   -   4.019   =   17.3266
95%   confidence interval is (   9.29   < µ <   17.33   )

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 ___95__ percent of these confidence intervals will contain the true population mean number of visits per patient and about __5__percent will not contain the true population mean number of visits per patient.