Question

Answer ALL questions please!

1. Records of 40 used passenger cars and 40 used pickup trucks (none used commercially) were randomly selected to investigate whether there was any difference in the mean time in years that they were kept by the original owner before being sold. For cars the mean was 5.3 years with standard deviation 2.2 years. For pickup trucks the mean was 7.1 years with standard deviation 3.0 years. a. Construct the 95% confidence interval for the difference in the means based on these data. b. Test the hypothesis that there is a difference in the means against the null hypothesis that there is no difference. Use the 1% level of significance.

Answer 1

1)

a)α=0.05

Degree of freedom, DF=   n1+n2-2 =    78              
t-critical value =    t α/2 =    1.9908   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    2.6306              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    0.5882              
margin of error, E = t*SE =    1.9908   *   0.59   =   1.17  
                      
difference of means =    x̅1-x̅2 =    5.3000   -   7.100   =   -1.8000
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    -1.8000   -   1.1711   =   -2.9711
Interval Upper Limit=   (x̅1-x̅2) + E =    -1.8000   +   1.1711   =   -0.6289

b)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.01                  
                          
Sample #1   ---->   1                  
mean of sample 1,    x̅1=   5.300                  
standard deviation of sample 1,   s1 =    2.200                  
size of sample 1,    n1=   40                  
                          
Sample #2   ---->   2                  
mean of sample 2,    x̅2=   7.100                  
standard deviation of sample 2,   s2 =    3.000                  
size of sample 2,    n2=   40                  
                          
difference in sample means =    x̅1-x̅2 =    5.3000   -   7.1   =   -1.800  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    2.6306                  
std error , SE =    Sp*√(1/n1+1/n2) =    0.5882                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -1.8000   -   0   ) /    0.59   =   -3.0601
                          
Degree of freedom, DF=   n1+n2-2 =    78                  
p-value =        0.003033   (excel function: =T.DIST.2T(t stat,df) )              
Conclusion:     p-value <α , Reject null hypothesis                      

There is enough evidence that there is a difference between means