Question

14. A researcher compares two compounds (1 and 2) used in the manufacture of car tires that are designed to reduce braking distances for SUVs equipped with the tires. SUVs equipped with tires using compound 1 have a mean braking distance of 44 feet and a standard deviation of 8.1 feet. SUVs equipped with tires using compound 2 have a mean braking distance of 47 feet and a standard deviation of 11.7 feet. Suppose that a sample of 51 braking tests are performed for each compound. Using these results, test the claim that the braking distance for SUVs equipped with tires using compound 1 is shorter than the braking distance when compound 2 is used. Let μ1 be the true mean braking distance corresponding to compound 1 and μ2 be the true mean braking distance corresponding to compound 2. Use the 0.05 level of significance.

Step 1 of 4: State the null and alternative hypotheses for the test.

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 H0H0. Round the numerical portion of your answer to three decimal places.

Step 4 of 4: Make the decision for the hypothesis test.

Answer 1

1)

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

2)

Level of Significance ,    α =    0.05          
                  
Sample #1   ---->   compund 1          
mean of sample 1,    x̅1=   44.000          
standard deviation of sample 1,   s1 =    8.10          
size of sample 1,    n1=   51          
                  
Sample #2   ---->   compound 2          
mean of sample 2,    x̅2=   47.00          
standard deviation of sample 2,   s2 =    11.70          
size of sample 2,    n2=   51          
                  
difference in sample means = x̅1-x̅2 =    44.000   -   47.0000   =   -3.0000
                  
std error , SE =    √(s1²/n1+s2²/n2) =    1.9926          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -3.0000   /   1.9926   ) =   -1.51

3)

DF = min(n1-1 , n2-1 )=   50  
      
      
t-critical value , t* =    -1.676 (excel function: =t.inv(α,df)

Decision rule : reject Ho , if t-stat <  -1.676 ,otherwise not

4)

since, test stat < -1.676, do not reject Ho