Question Part Points Submissions Used A sample of 25 pieces

Question Part Points Submissions Used A sample of 25 pieces of laminate used in the manufacture of circuit boards was selected, and the amount of warpage (in.) under particular conditions was determined for each piece, resulting in a sample mean warpage of 0.0638 and a sample standard deviation of 0.0067.

(b) Calculate an interval for which you can have a high degree of confidence that at least 95% of all pieces of laminate result in amounts of warpage that are between the two limits of the interval. (Use a 99% tolerance interval. Round your answers to four decimal places.

Please don\'t copy and paste solutions. please show equations only

Solution

b)

As the middle area is          
          
Middle Area = P(x1<x<x2) =    0.95      
          
Then the left tailed area of the left endpoint is          
          
P(x<x1) = (1-P(x1<x<x2))/2 =    0.025      
          
Thus, the z score corresponding to the left endpoint, by table/technology, is          
          
z1 =    -1.959963985      
By symmetry,          
z2 =    1.959963985      
          
As          
          
u = mean =    0.0638      
s = standard deviation =    0.0067      
          
Then          
          
x1 = u + z1*s =    0.050668241      
x2 = u + z2*s =    0.076931759      

Thus, the interval is (0.050668241, 0.076931759). [ANSWER]