1. For some population, the proportion of subjects in a particular class 
   is 20%. Considering the sampling distribution of p, what percentage of 
   samples of size n=400 would you expect to have a sample proportion 
   between 18% and 22%.

	Since S(p)=sqrt(0.2*0.8/400)=sqrt(0.0004)=0.02 then
	z1=(0.18-0.20)/0.02=-1 and z2=(0.22-0.20)/0.02=1. Hence 
        68.27% of samples of size 400 would result in sample 
        proportions between 18% and 22%.

2. You are interested in estimating the proportion of modules in a software
   portfolio that are Y2K compliant. You examine a sample of 120 modules and
   discover that 60% are compliant.
 
   a) Estimate the proportion of modules in the portfolio that are Y2K
      compliant

           Since the sample proportion is 60% then assuming the 
           sample is representative of the population (i.e. the 
           portfolio) then 60% is a good estimate of the portfolio
           proportion.
         
   b) Determine the "error" in your estimate for a) above.

           The error in the estimate is given by S(p) which is:
                      S(p)=sqrt(0.6*0.4)/120)=0.0447
           Hence the "error" is 4.47%.

   c) What is the "margin of error". Interpret your answer.
 
           Margin of error is z(95)*S(p)=1.95*S(p)=1.95*4.47=8.72%.
           Since p +/- "margin of error" is a 95% confidence interval
           then we may say that we are 95% confident that the 
           portfolio proportion is between 51.28% and 68.72%.