Question

A production line operation is tested for filling weight accuracy using the following hypotheses.

Hypothesis       Conclusion and Action
H ₀:  = 16         Filling okay, keep running
H _a:     16       Filling off standard; stop and adjust machine

The sample size is 37 and the population standard deviation is  = 0.8. Use  = .05. Do not round intermediate calculations.

1. What would a Type II error mean in this situation?
   SelectConcluding that the mean filling weight is not 16 ounces when it actually isConcluding that the mean filling weight is 16 ounces when it actually isn'tItem 1
2. What is the probability of making a Type II error when the machine is overfilling by .5 ounces (to 4 decimals)?
   
3. What is the power of the statistical test when the machine is overfilling by .5 ounces (to 4 decimals)?

Answer 1

a)

Concluding that the mean filling weight is 16 ounces when it actually isn't

b)

true mean ,    µ =    16.5              
                      
hypothesis mean,   µo =    16              
significance level,   α =    0.05              
sample size,   n =   37              
std dev,   σ =    0.8              
                      
δ=   µ - µo =    0.5              
                      
std error of mean,   σx = σ/√n =    0.8000   / √    37   =   0.13152

Zα/2   = ±   1.960   (two tailed test)                      
We will fail to reject the null (commit a Type II error) if we get a Z statistic between                           -1.960   and   1.960
these Z-critical value corresponds to some X critical values ( X critical), such that                                  
                                  
-1.960   ≤(x̄ - µo)/σx≤   1.960                          
15.742   ≤ x̄ ≤   16.258                          
                                  
now, type II error is ,ß =        P (   15.742   ≤ x̄ ≤   16.258   )          
       Z =    (x̄-true mean)/σx                      
       Z1 = (   15.742   -   16.5   ) /   0.13152   =   -5.762
       Z2 = (   16.258   -   16.5   ) /   0.13152   =   -1.842
                                  
   so, P(   -5.762   ≤ Z ≤   -1.842   ) = P ( Z ≤   -1.842   ) - P ( Z ≤   -5.762   )
                                  
       =   0.033   -   0.000   =   0.0328 [ Excel function: =NORMSDIST(z) ]  

c)

                                  
power =    1 - ß =   0.9672         

              

Thanks in advance!

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