:8. Solution.  Under the null hypothesis, the expected value of
the percentage of defectives in the sample is 20%.  The observed
percentage of defectives in the sample is 23%.  To figure out the
SE of the sample percentage, we consider a 0-1 box with 10,000
tickets, 20% of them marked with 1.  The number of defectives in
the sample is like the sum of 100 draws from this box, and the SE
of this sum is

 	
 10 x (.20x.80)^(1/2) = 10 x .16^(1/2) 

                         = 10 x .4

 	
                      = 4 

This is the SE of the number of defectives in the sample.  The SE
of the percentage of defectives is 

 	
               4/100 x 100% = 4% 
So

 	
  (observed - expected)/SE  =  (23-20)/4 = 3/4 = .75 . 

The observed level of significance is the area under the normal
curve to the right of .75.  From the table, we compute this area
to be about 22.5% percent. Of the choices, 25% is the closest.