Question

Suppose that a sample of 36 brand X tires has a sample mean life of 54000 miles and a sample standard deviation of 6000 miles, while a sample of 36 brand Y tires produces a sample mean of 60000 miles and a sample standard deviation of 9000 miles.

3. The company manufacturing the brand Y tires claims that their tires last longer than the brand X tires, on average. Is there enough evidence, at significance level ? = 0.05, to support the company’s claim? Also, give a range for the p-value of the test.

4. The company manufacturing the brand X tires claims that their manufacturing process yields more consistently performing tires (lower variance of the lifespan) than the brand Y tires. Is there enough evidence, at significance level ? = 0.05, to support the company’s claim?

Answer 1

3)

Ho :   µ1 - µ2 =   0          
Ha :   µ1-µ2 <   0          
                  
Level of Significance ,    α =    0.05          
                  
Sample #1   ---->   1          
mean of sample 1,    x̅1=   54000.00          
standard deviation of sample 1,   s1 =    6000          
size of sample 1,    n1=   36          
                  
Sample #2   ---->   2          
mean of sample 2,    x̅2=   60000.000          
standard deviation of sample 2,   s2 =    9000.00          
size of sample 2,    n2=   36          
                  
difference in sample means = x̅1-x̅2 =    54000.000   -   60000.0000   =   -6000.0000
                  
std error , SE =    √(s1²/n1+s2²/n2) =    1802.7756          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -6000.0000   /   1802.7756   ) =   -3.3282
                  
                  
                  
p-value =        0.0007   [ excel function: =T.DIST(t stat,df) ]       
Conclusion:     p-value<α , Reject null hypothesis              
                  
There is enough evidence to support the brand Y tires claims that their tires last longer than the brand X tires, on average

4)

   Sample 1:   Y
Sample Standard deviation,    s₁ =    9000.000
Sample size,    n₁ =    36

   Sample 2:   X
Sample Standard deviation,    s₂ =    6000
Sample size,    n₂ =    36
   α =    0.05

Null and alternative hypothesis:  
Hₒ : σ₁ = σ₂  
H₁ : σ₁ > σ₂  
Test statistic:  
F = s₁² / s₂² = 9000² / 6000² =    2.2500
Degree of freedom:  
df₁ = n₁-1 =    35
df₂ = n₂-1 =    35
  
P-value :  
P-value = F.DIST.RT(2.25, 35, 35) =    0.0094
Conclusion:  
As p-value < α, we reject the null hypothesis.
there is enough evidence to support the brand X tires claims that their manufacturing process yields more consistently performing tires (lower variance of the lifespan) than the brand Y tires.