Question

5. An investigator predicts that dog owners in the country spend more time walking their dogs than dog owners in the city. Two groups of individuals were studied, and the minutes of walking dogs in a day was recorded. Using the following data in the table below, compute the appropriate analysis to test the hypothesis. Set alpha = .05. Country City 85 80 S2 10 9 n 20 20

a. State the null and alternative hypotheses in symbols. (2 points)

b. Set up the criteria for making a decision. That is, find the critical value(s). (1 point)

c. Compute the appropriate test statistic. Show your work. (3 points) Complete by NOT Pooling the Variance!

d. Based on your answers above, evaluate the null hypothesis. (1 point) Reject Fail to reject (circle one)

e. State your conclusion in words. (1 point) e. Compute the 95% confidence interval. (2 points)

f. Interpret the confidence interval you calculated above. (1 point)

Answer 1

a)

Ho :   µ1 - µ2 =   0
Ha :   µ1-µ2 >   0
      

b)

Degree of freedom, =   37  
          
          
t-critical value , t* =        1.6871   (excel function: =t.inv(α,df)

c)

Sample #1   ---->   1          
mean of sample 1,    x̅1=   85.00          
standard deviation of sample 1,   s1 =    10          
size of sample 1,    n1=   20          
                  
Sample #2   ---->   2          
mean of sample 2,    x̅2=   80.000          
standard deviation of sample 2,   s2 =    9.00          
size of sample 2,    n2=   20          
                  
difference in sample means = x̅1-x̅2 =    85.000   -   80.0000   =   5.0000
                  
std error , SE =    √(s1²/n1+s2²/n2) =    3.0083          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   5.0000   /   3.0083   ) =   1.6621

d)

test stat < critical value, Fail to reject Null

e)

Degree of freedom, DF=       37          
t-critical value =    t α/2 =    2.026   (excel formula =t.inv(α/2,df)      
                  
                  
                  
std error , SE =    √(s1²/n1+s2²/n2) =    3.008          
margin of error, E = t*SE =    2.026   *   3.008   =   6.095439
                  
difference of means = x̅1-x̅2 =    85.0000   -   80.000   =   5.0000
confidence interval is                   
Interval Lower Limit = (x̅1-x̅2) - E =    5.0000   -   6.095   =   -1.095
Interval Upper Limit = (x̅1-x̅2) + E =    5.0000   -   6.095   =   11.095

f)

There is 95% confidence that true mean difference lies within CI