WEBVTT
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We now revisit the polling
problem that we
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have started earlier.
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When we first looked at that
problem, we used the Chebyshev
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inequality to obtain certain
bounds and numerical results.
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What we want to do now is
instead to use a central limit
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theorem-type approximation,
which we hope that it will be
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more accurate and more
informative.
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Let us remind ourselves
of the setting.
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We want to estimate a certain
number, p, which is the
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fraction of the population
that will vote yes in a
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certain referendum.
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And we estimate p by picking a
sample out of the population.
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We pick n people.
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We pick them randomly, uniformly
over the population
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and independently.
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For each one of the people in
the sample, we ask them if
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they will vote to yes or no,
and then we record their
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answers in Bernoulli random
variables, Xi.
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So by the assumptions that we
have made, these Xi's are
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independent Bernoulli random
variables, and their mean is
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equal to p.
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We count how many X's
were equal to 1.
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That's the number of yeses.
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We divide by n, and that gives
us the fraction in the
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population that have
responded yes.
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This is the sample
mean of the X's.
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And we use this sample
mean to estimate the
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unknown fraction p.
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We would like the error in our
estimation to be small, that
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is the difference between the
sample mean and the true value
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p to be small, less,
let's say, than
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one percentage point.
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Now there's no way of
guaranteeing that this spec
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will be met with certainty,
unless we sample almost
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everyone in the population.
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But what we can do instead
is to ask that these
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specifications are violated with
only a small probability.
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So we look at the probability
that our estimation error is
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larger than what we want.
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This is the case that we do
not meet the specs, and we
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would like this probability
to be small.
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One possible question is what
the value of n should be in
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order to meet the specs.
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But in order to do any
calculations, we first need a
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way of approximating
this probability.
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We will do that using the
central limit theorem.
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The central limit theorem
involves this standardized
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version of the random variable
Sn, where Sn stands for the
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sum of the X's.
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We know that this random
variable is
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approximately normal.
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And what we want to do now is to
take this event and rewrite
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it in an equivalent way but
which involves this random
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variable Zn.
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Let us start.
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First, we note that here we a
mu and a sigma, so we should
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know what these are.
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For a Bernoulli random variable,
the mean is what we
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already wrote down, and sigma
is the square root of p
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times 1 minus p.
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Now let's look at this event.
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Mn is the same as Sn/n,
by definition.
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And we can write p in this
form, minus n times
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p divided by n.
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And we want this quantity
to be larger
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than or equal to 0.01.
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So this event here
is identical to
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that event up there.
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This starts to look like
this expression.
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p is the same as Mu.
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But there is a little bit
of a difference in
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the denominator terms.
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So let's see what we can do.
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Let's take this same event but
multiply both sides of the
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inequality by a square
root of n.
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This causes this denominator
term to become just square
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root of n, and we get a square
root of n term in the
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numerator on the other side.
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This is an equivalent
description of the event.
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Now we can multiply both sides
of this inequality by sigma--
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actually the denominators
on both sides by sigma--
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and we obtain this equivalent
representation.
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But now we notice that here we
do have the random variable Zn
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that we wanted.
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And so we managed to express
this event in terms of the
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random variable Zn.
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In particular what we have is
that this probability is the
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same as the probability that
the absolute value of Zn is
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larger than or equal to
0.01 square root of
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n divided by sigma.
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Then we can use the central
limit theorem approximation to
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approximate this probability
by the corresponding
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probability where we now use
a standard normal random
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variable instead of the
Zn random variable.
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So here, Z stands for a
standard normal random
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variable with mean 0 and
variance equal to 1.
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Let us now continue on a new
slide so that we have some
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working space.
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And here is the result that
we have derived so far.
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If somebody gives us the value
of n, we would like to be able
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to calculate this probability
using this approximation.
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However, there's a slight
difficulty because sigma is a
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function that depends on
p, and it is not known.
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However, as we discussed when
we first started the polling
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problem, we do know that
sigma is always less
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than or equal to 1/2.
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And this suggests that we could
use here the worst-case
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value of the standard deviation,
replace sigma by
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1/2 and instead look at
this probability here.
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How are these two probabilities
related?
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Which direction does
the inequality go?
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A sketch will be useful here.
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Z is a standard normal, and
it's centered at 0.
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Somewhere here, we have a value
of 0.02 square root n.
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And somewhere further out, we
have the value of 0.01 square
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root n divided by sigma.
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Why are these two values
ordered this way?
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Since sigma is less than 1/2, 1
over sigma is bigger than 2.
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So this expression here
is bigger than
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this expression there.
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Since the inequality goes this
way, now we can compare these
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two events.
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This event, that Z is larger
in absolute value than this
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number, is the probability of
this tail of the distribution.
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And we will have a similar
probability from the other end
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of the tail of the
distribution.
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Here we're talking about the
probability of being larger
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than or equal to this number,
which would correspond only to
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this part of the tail and,
similarly, a small part of the
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tail from the other side.
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The blue event is smaller
than the red event.
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This is the probability of the
blue event, so it's going to
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be no larger than the
probability of the red event.
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Now if somebody gives us a value
of n, we should be able
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to calculate this probability.
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How do we calculate it?
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The probability that the
absolute value is above a
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certain number is equal to the
probability of this tail plus
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the probability of that tail.
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But because of the symmetry of
the normal distribution, this
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is twice the probability of
each one of the tails.
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What is the probability
of this tail?
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It's 1 minus the probability
of whatever is below that.
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So it's 1 minus.
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And the probability of being
below that, this is the
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standard normal CDF evaluated
at 0.02 square root n.
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So we do have now an expression
for the desired
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probability, or at least a
bound for it, which is
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expressed in terms of the
standard normal CDF.
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If somebody gives you a value
of n, you can plug in here.
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If n is 10,000, then square
root of n is 100.
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And this number becomes
equal to 2.
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And so in this case, what we
obtain is that the probability
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of interest is less than
or equal to 2 times 1
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minus Phi of 2.
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Now we invoke the standard
normal table.
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From the normal table, we obtain
that this quantity is
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equal to twice 1 minus 0.9772,
which evaluates to 0.046.
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So if we use 10,000 people in
our sample, then we will get
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an accuracy of one percentage
point with very high
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probability.
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The probability that we do not
meet the specification so that
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the accuracy that we get is
worse than one percentage
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point, that probability
is quite small.
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It's 0.046.
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That is 4 and something
percent.
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This is pretty good.
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And suppose that your boss now
tells you, I only want the
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probability of not meeting
the specs to be 5%.
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You look at this result, and
you say, with 10,000, I
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achieved a probability
of a large error
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that's less than 5%.
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This means that I probably have
some leeway and that I
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can reduce the size
of my sample.
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What could the size of
the sample be and
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still meet those specs?
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What we're trying to do here
is that we have this
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approximation for the
probability of interest, and
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we want to set this probability
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to a value of 0.05.
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Then we want to ask, what is
the value of n that will
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result in this particular
probability of
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not meeting the specs?
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Now we can do the algebra.
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And we find that this
corresponds to requiring that
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phi of 0.02 square root n
to be equal to 0.975.
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What's the interpretation
of this?
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We want to choose n so that
the probability of the two
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tails is 5%.
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This means that we want this
probability here to be 2 and
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1/2 percent.
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This means that the probability
of whatever is to
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the left of this number should
be 0.975, including the tail.
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This means, again, that we have
to look at the standard
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normal table and ask, what's the
value for which the CDF is
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equal to 0.975?
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So we look around, and we find
0.975 to be here, and it
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corresponds to 1.96.
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This tells us that 0.02
square root n
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should be equal to 1.96.
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Then we solve for n, and we find
that the value of n is
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9,604, which is indeed some
reduction from the 10,000 that
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we had originally.
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How does this relate
to the real world?
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When you read newspapers about
polls, you will never see
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sample sizes that are
about 10,000.
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You will usually see sample
sizes of the order of 1,000,
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sometimes even smaller.
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How can they do that?
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Well, they can do that because
the specs that they impose are
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not as tight as the specs
that we have here.
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Usually, they tell you that
the results are accurate
00:13:35.100 --> 00:13:38.850
within three percentage points,
let's say, instead of
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one percentage point.
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And by moving from 0.01 to 0.03,
and if you repeat those
00:13:46.420 --> 00:13:50.090
calculations, you will find that
the sample size of about
00:13:50.090 --> 00:13:53.690
1,000 will actually do.