Question

A technician compares repair costs for two types of microwave ovens (type I and type II). He believes that the repair cost for type I ovens is greater than the repair cost for type II ovens. A sample of 5959 type I ovens has a mean repair cost of $73.27$⁢73.27, with a standard deviation of $10.76$⁢10.76. A sample of 4747 type II ovens has a mean repair cost of $69.22$⁢69.22, with a standard deviation of $23.21$⁢23.21. Conduct a hypothesis test of the technician's claim at the 0.050.05 level of significance. Let μ1μ1 be the true mean repair cost for type I ovens and μ2μ2 be the true mean repair cost for type II ovens.

Step 2 of 4 :

Compute the value of the test statistic. Round your answer to two decimal places.

Step 3 of 4: Determine the decision rule for rejecting the null hypothesis H0. Round the numerical portion of your answer to two decimal places

Step 4 of 4: Make the decision for the hypothesis test. Fail or reject to fail.

Answer 1

2)

Ho :   µ1 - µ2 =   0          
Ha :   µ1-µ2 >   0          
                  
Level of Significance ,    α =    0.05          
                  
Sample #1   ---->   1          
mean of sample 1,    x̅1=   73.27          
standard deviation of sample 1,   s1 =    10.76          
size of sample 1,    n1=   59          
                  
Sample #2   ---->   2          
mean of sample 2,    x̅2=   69.220          
standard deviation of sample 2,   s2 =    23.21          
size of sample 2,    n2=   47          
                  
difference in sample means = x̅1-x̅2 =    73.270   -   69.2200   =   4.0500
                  
std error , SE =    √(s1²/n1+s2²/n2) =    3.6639          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   4.0500   /   3.6639   ) =   1.11

3) at α=0.05,df=61

t-critical value , t* =        1.67 (excel function: =t.inv(α,df)

rejection region: reject Ho, if test stat >1.67

4)

Decision:     t-stat < critical value , so,Fail to Reject Ho