Question

The mean oull-off force of an adhesive used in manufacturing a connector for an automotive engine application should be at least 75 pounds. This adhesive will be used unless there is strong evidence that the pull-off force does not meet this requirement. A test of an appropiate device is to be conducted with a sample size n = 30 and a = 0.01. Assume the standard deviation is known.

a) If the true standard deviation s = 1, what is the risk that the adhesive will be judged acceptable when the true mean pull-off force is only 73pounds?

b)What sample size is required to give a 90% chance of detecting that the true mean pull-off force is only 72 pounds when s =1?

Answer 1

a)

true mean ,    µ =    73
      
hypothesis mean,   µo =    75
significance level,   α =    0.01
sample size,   n =   30
std dev,   σ =    1
      
δ=   µ - µo =    -2
      
std error of mean,   σx = σ/√n =    0.1826

Zα =       -2.3263   (left tail test)  
              
We will fail to reject the null (commit a Type II error) if we get a Z statistic >               -2.3263
this Z-critical value corresponds to X critical value( X critical), such that              
              
       (x̄ - µo)/σx ≥ Zα      
       x̄ ≥ Zα*σx + µo      
       x̄ ≥    74.575   (acceptance region)
              
now, type II error is ,ß =    P( x̄ ≥    74.575   given that µ =   73
              
   = P ( Z > (x̄-true mean)/σx )           
   = P ( Z >    8.628   )      =   0.00

b)

True mean,   µ =    72  
hypothesis mean,   µo =    75  
          
Level of Significance ,    α =    0.01  
std dev =    σ =    1  
power =    1-ß =    0.9  
ß=       0.1  
δ=   µ - µo =    -3  
          
Z ( α ) =       2.3263   [excel function: =normsinv(α)
          
Z (ß) =        1.2816   [excel function: =normsinv(ß)
          
sample size needed =    n = ( σ [ Z(ß)+Z(α) ] / δ )² =    1.4463  
          
so, sample size =        2.000