1. Of 600 disk drives sampled 150 are defective. 
 
a) Estimate the population proportion that are defective:

The true proportion of defective drives is estimated by the sample
proportion p=150/600=0.25
 
b) Compute the error in your estimate:

The standard deviation of this estimate denoted s(p) may be 
computed by sqrt{p*(1-p)/n} where n is the sample size. 
Hence sqrt{0.25*0.75/600}=0.0177
 
c) Find a 90% CI for the proportion of defective drives in the population: 

A 90% CI of defective drives is p plus/minus 1.65(s(p)) which is 
0.25 plus/minus 1.65(0.0177)=[0.22, 0.28]

 
2. Of 1000 DePaul students selected in a sample 380 approved
of eliminating charges for internet access. 
 
a) Estimate the proportion of all DePaul students who approve:

The true proportion of DePaul students who approve of free internet
is estimated by 380/1000=0.38
 
b) Estimate the total number who approve if there are 9500 students:

Since the actual DePaul student population is 9500 students 
the total number who approve of free internet is estimated 
by 9500*0.38=3610
 
c) Construct a 90% CI for the total number of students who
approve of free internet access. (Hint: Base your answer
on the 90% CI for the population proportion):

To construct a 90% confidence interval for the total number 
of students first construct a 90% confidence interval for pi.
Since we need s(p) for the 90% confidence interval for p, 
we need to compute s(p)=sqrt{0.38*0.62/1000}=0.0154. 
The 90% confidence interval for p is 0.38 plus/minus 
1.65*0.0154=[0.355,0.405]. Now simply multiply the boundaries
of this confidence interval by 9500 to get the required population 
boundaries as a count rather than a proportion. Therefore
the boundaries are 9500*0.355 and 9500*0.405 which is 
[3369, 3851]