Here is the problem and I honestly am not sure  what to do. Is Sample size all
voters in U.S or there is a computation to be done? 

A New York Times article about Poll results states, "In theory, in 19 cases out
of 20, the results from such a poll should differ by no more than one
percentage point in either direction from what would have been obtained by
interviewing all voters in the United States. Find the sample size suggested by
this statement.


We do not need the standard deviation s as the previous 
tutor stated. We do not even need an estimate of the
proportion who will say "yes".  The previous tutor 
also gave the wrong value for z.  He gave
za/2 = z0.01/2 = z0.005 = 2.5758.  That's wrong. He should 
have given
za/2 = z0.05/2 = z0.025 = 1.96

We use the formula. He also gave the wrong formula. 
This is a proportion problem.

       ^^
     z2pq
n = -----
      E2

We will assume this is a poll where people
polled answer "yes, I will vote for the candidate"
or "no, I will not vote for the candidate".

The confidence level is 19 out of 20, that is,
19/20, which equals 0.95.  From the standard
normal tables, we know that  

P(-1.96 < z < 1.96) = 0.95

so we use z = 1.96

(this is often written za/2 = z0.05/2 = z0.025 = 1.96)

Now E represents the maximum allowable error,
which, the words "no more than one percentage point
in either direction" tells us that E = 1% = 0.01

                   ^    ^
Now we do not know p or q, the estimated proportions
of people who will say "yes" and "no" respectively.
However since they will be multiplied together in
the formula, the most their product can possibly be
is when they are both 0.5. Why this is true can be 
shown from this table:

   ^           ^          ^^
If p is   then q is   and pq =
   0.0      1.0          0.0  
   0.1      0.9          0.09
   0.2      0.8          0.16
   0.3      0.7          0.21
   0.4      0.6          0.24
   0.5      0.5          0.25 <--(largest value in list)
   0.6      0.4          0.24
   0.7      0.3          0.21
   0.8      0.2          0.16
   0.9      0.1          0.09
   1.0      0.0          0.00 

          ^           ^ 
so we use p = 0.5 and q = 0.5,
and    

       ^^
     z2pq
n = ------
       E2

becomes
     (1.96)2(0.5)(0.5) 
n = ------------------
         (0.01)2

n = 9604

So from the information given, we'd need to
interview 9604 people at random to ensure
that we can get the proportion within 1
percentage point with a probability of 95%
of being right.

Edwin