Question

In a certain chemical process, it is very important that a particular solution that is to be used as a reactant have a pH of exactly 8.20. A method for determining pH that is available for solutions of this type is known to give measurements that are actual pH and with a standard deviation of 0.03. Suppose 36 independent measurements yielded sample mean pH of 8.19 1) Design you specify Ho, H1, explain parameter u, and then calculate test statistic and give the rejection region, at last state final conclusion in the context.) 2) If the true pH is 8.18, please compute the power of the above test. Note that power 1-B where B is the probability of Type I error normally distributed with equal to the mean a a test to determine if the pH is less than 8.20 at significance level 0.05. (Make sure

Answer 1

1)

Ho :   µ =   8.2                  
Ha :   µ <   8.2       (Left tail test)          
                          
Level of Significance ,    α =    0.01                  
population std dev ,    σ =    0.0300                  
Sample Size ,   n =    36                  
Sample Mean,    x̅ =   8.1900                  
                          
'   '   '                  
                          
Standard Error , SE = σ/√n =   0.0300   / √    36   =   0.0050      
Z-test statistic= (x̅ - µ )/SE = (   8.190   -   8.2   ) /    0.005   =   -2.0000
                          
  
p-Value   =   0.0228    [ Excel formula =NORMSDIST(z) ]              
Conclusion: There is enough evidence that pH is less than 8.2

b)

true mean ,    µ =    8.18              
                      
hypothesis mean,   µo =    8.2              
significance level,   α =    0.05              
sample size,   n =   36              
std dev,   σ =    0.03              
                      
δ=   µ - µo =    -0.02              
                      
std error of mean,   σx = σ/√n =    0.03   / √    36   =   0.0050

Zα =       -1.6449   (left tail test)          
                      
We will fail to reject the null (commit a Type II error) if we get a Z statistic >               -1.645      
this Z-critical value corresponds to X critical value( X critical), such that                      
                      
(x̄ - µo)/σx ≥ Zα                      
x̄ ≥ Zα*σx + µo                      
x̄ ≥    -1.645   *   0.0050   +   8.2  
       x̄ ≥    8.192   (acceptance region)      
                      
now, type II error is ,ß =    P( x̄ ≥    8.192   given that µ =   8.18   )  
                      
   = P ( Z > (x̄-true mean)/σx )                   
=P(Z > (   8.192   -   8.18   ) /   0.0050   )
                      
   = P ( Z >    2.355   ) =   0.00926   [ Excel function: =1-normsdist(z) ]  
                      
                      
power =    1 - ß =   0.9907