Question 59559: 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.
Found 2 solutions by stanbon, Edwin McCravy:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 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.
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The formula for sample size is n=[z^*sigma/E]^2
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You want n.
You have the following information:
"19 out of 20" indicates you want a 95% confidence result which means
z^*=2.5758...
"differ no more than one percentage point" means E=0.01
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Missing information:
You are missing the value of sigma.
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If you have sigma you can calculate n.
Cheers,
Stan H.
Answer by Edwin McCravy(20056)  (Show Source):
You can put this solution on YOUR website! 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
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