Two methods of filling cereal boxes are being compared. Both methods fill the bo

Question

Two methods of filling cereal boxes are being compared. Both methods fill the box

with the same amount of cereal on average so the company wants to select the

method with the lower variance. In a sample of 31 boxes filled using method 1, the

standard deviation of fill weights is .75 ounces and in a sample of 41 boxes filled

using method 2, the standard deviation is .65 ounces. At the 5% level of significance,

does it appear that the variance of method 1 is significantly larger than the variance of

method 2? Please show work.

Explanation / Answer

Formulating the null and alternative hypotheses,              
              
Ho:   sigma1^2 / sigma2^2   <=   1  
Ha:    sigma1^2 / sigma2^2   >   1  
              
As we can see, this is a    right   tailed test.      
              
Thus, getting the critical F, as alpha =    0.05   ,      
alpha =    0.05          
df1 = n1 - 1 =    30         
df2 = n2 - 1 =    40          
F (crit) =    1.744431964         
              
Getting the test statistic, as              
s1 =    0.75          
s2 =    0.65          
              
Thus, F = s1^2/s2^2 =    1.331360947          
                      
              
Comparing F < Fcrit, we   FAIL TO REJECT THE NULL HYPOTHESIS.

Thus, there is no significant evidence that the variance of method 1 is greater than that of method 2. [CONCLUSION]  
              

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