Statistics 419 6 An airplane manufacturer intends to establi

Statistics 419 6. An airplane manufacturer intends to establish a component acceptance criterion that is based upon sound statistical methods. Preliminary tests on 61 acceptable components have determined that the mean load to produce component failure is 500 psi with a standard deviation of 25 psi. Based upon this information, provide (a) an estimate, with 99 % confidence, of the value of the next (the 62nd) measured load to produce failure, (b) an estimate, with 99 % confidence, of the true mean load to produce failure, and (c) an estimate, with 98 % confidence, of the true variance. Finally, the manufacturer wants to be 99 % confident that if the batch sample meets the acceptance criterion. (d) Determine the range of sample standard deviation values (in psi) that the batch sample can have and still meet the test criterion. ighte ocaule 0 42 N and the samnle

Solution

6

a)

As the middle area is          
          
Middle Area = P(x1<x<x2) =    0.99      
          
Then the left tailed area of the left endpoint is          
          
P(x<x1) = (1-P(x1<x<x2))/2 =    0.005      
          
Thus, the z score corresponding to the left endpoint, by table/technology, is          
          
z1 =    -2.575829304      
By symmetry,          
z2 =    2.575829304      
          
As          
          
u = mean =    500      
s = standard deviation =    25      
          
Then          
          
x1 = u + z1*s =    435.6042674      
x2 = u + z2*s =    564.3957326  

[ANSWER, BETWEEN 435.604 AND 564.396]

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b)

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.005          
X = sample mean =    500          
z(alpha/2) = critical z for the confidence interval =    2.575829304          
s = sample standard deviation =    25          
n = sample size =    61          
              
Thus,              
Margin of Error E =    8.245028682          
Lower bound =    491.7549713          
Upper bound =    508.2450287          
              
Thus, the confidence interval is              
              
(   491.7549713   ,   508.2450287   ) [ANSWER]

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c)

As              
              
df = n - 1 =    60          
alpha = (1 - confidence level)/2 =    0.01          
              
Then the critical values for chi^2 are              
              
chi^2(alpha/2) =    88.3794189          
chi^2(alpha/2) =    37.48485153          
              
Thus, as              
              
lower bound = (n - 1) s^2 / chi^2(alpha/2) =    424.306931          
upper bound = (n - 1) s^2 / chi^2(1 - alpha/2) =    1000.404122          
              
Thus, the confidence interval for the variance is              
              
(   424.306931   ,   1000.404122   ) [ANSWER]

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d)

As              
              
df = n - 1 =    60          
alpha = (1 - confidence level)/2 =    0.005          
              
Then the critical values for chi^2 are              
              
chi^2(alpha/2) =    91.95169816          
chi^2(alpha/2) =    35.53449108          
              
Thus, as              
              
lower bound = (n - 1) s^2 / chi^2(alpha/2) =    407.8228108          
upper bound = (n - 1) s^2 / chi^2(1 - alpha/2) =    1055.312708          
              
Thus, the confidence interval for the variance is              
              
(   407.8228108   ,   1055.312708   )
              
Also, for the standard deviation, getting the square root of the bounds,              
              
(   20.19462331   ,   32.48557692   ) [ANSWER]