We now revisit the polling problem that we
have started earlier.
When we first looked at that problem, we used the Chebyshev
inequality to obtain certain bounds and numerical results.
What we want to do now is instead to use a central limit
theorem-type approximation, which we hope that it will be
more accurate and more informative.
Let us remind ourselves of the setting.
We want to estimate a certain number, p, which is the
fraction of the population that will vote yes in a
certain referendum.
And we estimate p by picking a sample out of the population.
We pick n people.
We pick them randomly, uniformly over the population
and independently.
For each one of the people in the sample, we ask them if
they will vote to yes or no, and then we record their
answers in Bernoulli random variables, Xi.
So by the assumptions that we have made, these Xi's are
independent Bernoulli random variables, and their mean is
equal to p.
We count how many X's were equal to 1.
That's the number of yeses.
We divide by n, and that gives us the fraction in the
population that have responded yes.
This is the sample mean of the X's.
And we use this sample mean to estimate the
unknown fraction p.
We would like the error in our estimation to be small, that
is the difference between the sample mean and the true value
p to be small, less, let's say, than
one percentage point.
Now there's no way of guaranteeing that this spec
will be met with certainty, unless we sample almost
everyone in the population.
But what we can do instead is to ask that these
specifications are violated with only a small probability.
So we look at the probability that our estimation error is
larger than what we want.
This is the case that we do not meet the specs, and we
would like this probability to be small.
One possible question is what the value of n should be in
order to meet the specs.
But in order to do any calculations, we first need a
way of approximating this probability.
We will do that using the central limit theorem.
The central limit theorem involves this standardized
version of the random variable Sn, where Sn stands for the
sum of the X's.
We know that this random variable is
approximately normal.
And what we want to do now is to take this event and rewrite
it in an equivalent way but which involves this random
variable Zn.
Let us start.
First, we note that here we a mu and a sigma, so we should
know what these are.
For a Bernoulli random variable, the mean is what we
already wrote down, and sigma is the square root of p
times 1 minus p.
Now let's look at this event.
Mn is the same as Sn/n, by definition.
And we can write p in this form, minus n times
p divided by n.
And we want this quantity to be larger
than or equal to 0.01.
So this event here is identical to
that event up there.
This starts to look like this expression.
p is the same as Mu.
But there is a little bit of a difference in
the denominator terms.
So let's see what we can do.
Let's take this same event but multiply both sides of the
inequality by a square root of n.
This causes this denominator term to become just square
root of n, and we get a square root of n term in the
numerator on the other side.
This is an equivalent description of the event.
Now we can multiply both sides of this inequality by sigma--
actually the denominators on both sides by sigma--
and we obtain this equivalent representation.
But now we notice that here we do have the random variable Zn
that we wanted.
And so we managed to express this event in terms of the
random variable Zn.
In particular what we have is that this probability is the
same as the probability that the absolute value of Zn is
larger than or equal to 0.01 square root of
n divided by sigma.
Then we can use the central limit theorem approximation to
approximate this probability by the corresponding
probability where we now use a standard normal random
variable instead of the Zn random variable.
So here, Z stands for a standard normal random
variable with mean 0 and variance equal to 1.
Let us now continue on a new slide so that we have some
working space.
And here is the result that we have derived so far.
If somebody gives us the value of n, we would like to be able
to calculate this probability using this approximation.
However, there's a slight difficulty because sigma is a
function that depends on p, and it is not known.
However, as we discussed when we first started the polling
problem, we do know that sigma is always less
than or equal to 1/2.
And this suggests that we could use here the worst-case
value of the standard deviation, replace sigma by
1/2 and instead look at this probability here.
How are these two probabilities related?
Which direction does the inequality go?
A sketch will be useful here.
Z is a standard normal, and it's centered at 0.
Somewhere here, we have a value of 0.02 square root n.
And somewhere further out, we have the value of 0.01 square
root n divided by sigma.
Why are these two values ordered this way?
Since sigma is less than 1/2, 1 over sigma is bigger than 2.
So this expression here is bigger than
this expression there.
Since the inequality goes this way, now we can compare these
two events.
This event, that Z is larger in absolute value than this
number, is the probability of this tail of the distribution.
And we will have a similar probability from the other end
of the tail of the distribution.
Here we're talking about the probability of being larger
than or equal to this number, which would correspond only to
this part of the tail and, similarly, a small part of the
tail from the other side.
The blue event is smaller than the red event.
This is the probability of the blue event, so it's going to
be no larger than the probability of the red event.
Now if somebody gives us a value of n, we should be able
to calculate this probability.
How do we calculate it?
The probability that the absolute value is above a
certain number is equal to the probability of this tail plus
the probability of that tail.
But because of the symmetry of the normal distribution, this
is twice the probability of each one of the tails.
What is the probability of this tail?
It's 1 minus the probability of whatever is below that.
So it's 1 minus.
And the probability of being below that, this is the
standard normal CDF evaluated at 0.02 square root n.
So we do have now an expression for the desired
probability, or at least a bound for it, which is
expressed in terms of the standard normal CDF.
If somebody gives you a value of n, you can plug in here.
If n is 10,000, then square root of n is 100.
And this number becomes equal to 2.
And so in this case, what we obtain is that the probability
of interest is less than or equal to 2 times 1
minus Phi of 2.
Now we invoke the standard normal table.
From the normal table, we obtain that this quantity is
equal to twice 1 minus 0.9772, which evaluates to 0.046.
So if we use 10,000 people in our sample, then we will get
an accuracy of one percentage point with very high
probability.
The probability that we do not meet the specification so that
the accuracy that we get is worse than one percentage
point, that probability is quite small.
It's 0.046.
That is 4 and something percent.
This is pretty good.
And suppose that your boss now tells you, I only want the
probability of not meeting the specs to be 5%.
You look at this result, and you say, with 10,000, I
achieved a probability of a large error
that's less than 5%.
This means that I probably have some leeway and that I
can reduce the size of my sample.
What could the size of the sample be and
still meet those specs?
What we're trying to do here is that we have this
approximation for the probability of interest, and
we want to set this probability
to a value of 0.05.
Then we want to ask, what is the value of n that will
result in this particular probability of
not meeting the specs?
Now we can do the algebra.
And we find that this corresponds to requiring that
phi of 0.02 square root n to be equal to 0.975.
What's the interpretation of this?
We want to choose n so that the probability of the two
tails is 5%.
This means that we want this probability here to be 2 and
1/2 percent.
This means that the probability of whatever is to
the left of this number should be 0.975, including the tail.
This means, again, that we have to look at the standard
normal table and ask, what's the value for which the CDF is
equal to 0.975?
So we look around, and we find 0.975 to be here, and it
corresponds to 1.96.
This tells us that 0.02 square root n
should be equal to 1.96.
Then we solve for n, and we find that the value of n is
9,604, which is indeed some reduction from the 10,000 that
we had originally.
How does this relate to the real world?
When you read newspapers about polls, you will never see
sample sizes that are about 10,000.
You will usually see sample sizes of the order of 1,000,
sometimes even smaller.
How can they do that?
Well, they can do that because the specs that they impose are
not as tight as the specs that we have here.
Usually, they tell you that the results are accurate
within three percentage points, let's say, instead of
one percentage point.
And by moving from 0.01 to 0.03, and if you repeat those
calculations, you will find that the sample size of about
1,000 will actually do.