1
00:00:02,050 --> 00:00:04,500
We now revisit the polling
problem that we

2
00:00:04,500 --> 00:00:06,390
have started earlier.

3
00:00:06,390 --> 00:00:09,210
When we first looked at that
problem, we used the Chebyshev

4
00:00:09,210 --> 00:00:13,820
inequality to obtain certain
bounds and numerical results.

5
00:00:13,820 --> 00:00:18,000
What we want to do now is
instead to use a central limit

6
00:00:18,000 --> 00:00:22,110
theorem-type approximation,
which we hope that it will be

7
00:00:22,110 --> 00:00:24,920
more accurate and more
informative.

8
00:00:24,920 --> 00:00:27,350
Let us remind ourselves
of the setting.

9
00:00:27,350 --> 00:00:30,940
We want to estimate a certain
number, p, which is the

10
00:00:30,940 --> 00:00:33,510
fraction of the population
that will vote yes in a

11
00:00:33,510 --> 00:00:34,960
certain referendum.

12
00:00:34,960 --> 00:00:39,010
And we estimate p by picking a
sample out of the population.

13
00:00:39,010 --> 00:00:40,650
We pick n people.

14
00:00:40,650 --> 00:00:44,360
We pick them randomly, uniformly
over the population

15
00:00:44,360 --> 00:00:46,090
and independently.

16
00:00:46,090 --> 00:00:49,360
For each one of the people in
the sample, we ask them if

17
00:00:49,360 --> 00:00:52,520
they will vote to yes or no,
and then we record their

18
00:00:52,520 --> 00:00:56,360
answers in Bernoulli random
variables, Xi.

19
00:00:56,360 --> 00:00:59,590
So by the assumptions that we
have made, these Xi's are

20
00:00:59,590 --> 00:01:03,950
independent Bernoulli random
variables, and their mean is

21
00:01:03,950 --> 00:01:06,580
equal to p.

22
00:01:06,580 --> 00:01:09,760
We count how many X's
were equal to 1.

23
00:01:09,760 --> 00:01:11,000
That's the number of yeses.

24
00:01:11,000 --> 00:01:13,700
We divide by n, and that gives
us the fraction in the

25
00:01:13,700 --> 00:01:16,070
population that have
responded yes.

26
00:01:16,070 --> 00:01:18,610
This is the sample
mean of the X's.

27
00:01:18,610 --> 00:01:21,210
And we use this sample
mean to estimate the

28
00:01:21,210 --> 00:01:24,200
unknown fraction p.

29
00:01:24,200 --> 00:01:28,960
We would like the error in our
estimation to be small, that

30
00:01:28,960 --> 00:01:31,990
is the difference between the
sample mean and the true value

31
00:01:31,990 --> 00:01:35,280
p to be small, less,
let's say, than

32
00:01:35,280 --> 00:01:37,030
one percentage point.

33
00:01:37,030 --> 00:01:40,300
Now there's no way of
guaranteeing that this spec

34
00:01:40,300 --> 00:01:44,070
will be met with certainty,
unless we sample almost

35
00:01:44,070 --> 00:01:45,780
everyone in the population.

36
00:01:45,780 --> 00:01:49,740
But what we can do instead
is to ask that these

37
00:01:49,740 --> 00:01:54,520
specifications are violated with
only a small probability.

38
00:01:54,520 --> 00:01:58,150
So we look at the probability
that our estimation error is

39
00:01:58,150 --> 00:02:01,290
larger than what we want.

40
00:02:01,290 --> 00:02:04,310
This is the case that we do
not meet the specs, and we

41
00:02:04,310 --> 00:02:07,410
would like this probability
to be small.

42
00:02:07,410 --> 00:02:11,880
One possible question is what
the value of n should be in

43
00:02:11,880 --> 00:02:13,590
order to meet the specs.

44
00:02:13,590 --> 00:02:16,310
But in order to do any
calculations, we first need a

45
00:02:16,310 --> 00:02:19,520
way of approximating
this probability.

46
00:02:19,520 --> 00:02:22,700
We will do that using the
central limit theorem.

47
00:02:22,700 --> 00:02:26,070
The central limit theorem
involves this standardized

48
00:02:26,070 --> 00:02:30,730
version of the random variable
Sn, where Sn stands for the

49
00:02:30,730 --> 00:02:33,390
sum of the X's.

50
00:02:33,390 --> 00:02:35,090
We know that this random
variable is

51
00:02:35,090 --> 00:02:36,620
approximately normal.

52
00:02:36,620 --> 00:02:41,050
And what we want to do now is to
take this event and rewrite

53
00:02:41,050 --> 00:02:44,490
it in an equivalent way but
which involves this random

54
00:02:44,490 --> 00:02:46,900
variable Zn.

55
00:02:46,900 --> 00:02:48,030
Let us start.

56
00:02:48,030 --> 00:02:52,590
First, we note that here we a
mu and a sigma, so we should

57
00:02:52,590 --> 00:02:54,280
know what these are.

58
00:02:54,280 --> 00:02:58,480
For a Bernoulli random variable,
the mean is what we

59
00:02:58,480 --> 00:03:03,340
already wrote down, and sigma
is the square root of p

60
00:03:03,340 --> 00:03:04,930
times 1 minus p.

61
00:03:07,850 --> 00:03:09,930
Now let's look at this event.

62
00:03:09,930 --> 00:03:17,050
Mn is the same as Sn/n,
by definition.

63
00:03:17,050 --> 00:03:22,980
And we can write p in this
form, minus n times

64
00:03:22,980 --> 00:03:26,790
p divided by n.

65
00:03:26,790 --> 00:03:30,220
And we want this quantity
to be larger

66
00:03:30,220 --> 00:03:33,150
than or equal to 0.01.

67
00:03:33,150 --> 00:03:35,550
So this event here
is identical to

68
00:03:35,550 --> 00:03:37,490
that event up there.

69
00:03:37,490 --> 00:03:40,940
This starts to look like
this expression.

70
00:03:40,940 --> 00:03:42,900
p is the same as Mu.

71
00:03:42,900 --> 00:03:45,340
But there is a little bit
of a difference in

72
00:03:45,340 --> 00:03:47,190
the denominator terms.

73
00:03:47,190 --> 00:03:49,820
So let's see what we can do.

74
00:03:49,820 --> 00:03:57,860
Let's take this same event but
multiply both sides of the

75
00:03:57,860 --> 00:04:00,380
inequality by a square
root of n.

76
00:04:00,380 --> 00:04:04,490
This causes this denominator
term to become just square

77
00:04:04,490 --> 00:04:09,470
root of n, and we get a square
root of n term in the

78
00:04:09,470 --> 00:04:12,260
numerator on the other side.

79
00:04:12,260 --> 00:04:15,030
This is an equivalent
description of the event.

80
00:04:15,030 --> 00:04:20,720
Now we can multiply both sides
of this inequality by sigma--

81
00:04:20,720 --> 00:04:25,580
actually the denominators
on both sides by sigma--

82
00:04:25,580 --> 00:04:28,930
and we obtain this equivalent
representation.

83
00:04:28,930 --> 00:04:36,110
But now we notice that here we
do have the random variable Zn

84
00:04:36,110 --> 00:04:37,860
that we wanted.

85
00:04:37,860 --> 00:04:42,050
And so we managed to express
this event in terms of the

86
00:04:42,050 --> 00:04:44,000
random variable Zn.

87
00:04:44,000 --> 00:04:47,960
In particular what we have is
that this probability is the

88
00:04:47,960 --> 00:04:53,380
same as the probability that
the absolute value of Zn is

89
00:04:53,380 --> 00:04:58,030
larger than or equal to
0.01 square root of

90
00:04:58,030 --> 00:05:02,140
n divided by sigma.

91
00:05:02,140 --> 00:05:06,080
Then we can use the central
limit theorem approximation to

92
00:05:06,080 --> 00:05:09,360
approximate this probability
by the corresponding

93
00:05:09,360 --> 00:05:13,630
probability where we now use
a standard normal random

94
00:05:13,630 --> 00:05:20,100
variable instead of the
Zn random variable.

95
00:05:20,100 --> 00:05:23,860
So here, Z stands for a
standard normal random

96
00:05:23,860 --> 00:05:28,780
variable with mean 0 and
variance equal to 1.

97
00:05:28,780 --> 00:05:31,720
Let us now continue on a new
slide so that we have some

98
00:05:31,720 --> 00:05:33,010
working space.

99
00:05:33,010 --> 00:05:37,520
And here is the result that
we have derived so far.

100
00:05:37,520 --> 00:05:40,630
If somebody gives us the value
of n, we would like to be able

101
00:05:40,630 --> 00:05:44,100
to calculate this probability
using this approximation.

102
00:05:44,100 --> 00:05:49,040
However, there's a slight
difficulty because sigma is a

103
00:05:49,040 --> 00:05:55,210
function that depends on
p, and it is not known.

104
00:05:55,210 --> 00:05:58,670
However, as we discussed when
we first started the polling

105
00:05:58,670 --> 00:06:02,460
problem, we do know that
sigma is always less

106
00:06:02,460 --> 00:06:04,520
than or equal to 1/2.

107
00:06:04,520 --> 00:06:09,340
And this suggests that we could
use here the worst-case

108
00:06:09,340 --> 00:06:13,000
value of the standard deviation,
replace sigma by

109
00:06:13,000 --> 00:06:17,290
1/2 and instead look at
this probability here.

110
00:06:17,290 --> 00:06:19,990
How are these two probabilities
related?

111
00:06:19,990 --> 00:06:22,800
Which direction does
the inequality go?

112
00:06:22,800 --> 00:06:25,500
A sketch will be useful here.

113
00:06:25,500 --> 00:06:33,200
Z is a standard normal, and
it's centered at 0.

114
00:06:33,200 --> 00:06:39,650
Somewhere here, we have a value
of 0.02 square root n.

115
00:06:39,650 --> 00:06:45,210
And somewhere further out, we
have the value of 0.01 square

116
00:06:45,210 --> 00:06:49,040
root n divided by sigma.

117
00:06:49,040 --> 00:06:51,860
Why are these two values
ordered this way?

118
00:06:51,860 --> 00:06:58,450
Since sigma is less than 1/2, 1
over sigma is bigger than 2.

119
00:06:58,450 --> 00:07:01,170
So this expression here
is bigger than

120
00:07:01,170 --> 00:07:04,180
this expression there.

121
00:07:04,180 --> 00:07:09,250
Since the inequality goes this
way, now we can compare these

122
00:07:09,250 --> 00:07:12,230
two events.

123
00:07:12,230 --> 00:07:16,570
This event, that Z is larger
in absolute value than this

124
00:07:16,570 --> 00:07:22,170
number, is the probability of
this tail of the distribution.

125
00:07:22,170 --> 00:07:26,540
And we will have a similar
probability from the other end

126
00:07:26,540 --> 00:07:29,280
of the tail of the
distribution.

127
00:07:29,280 --> 00:07:31,880
Here we're talking about the
probability of being larger

128
00:07:31,880 --> 00:07:36,159
than or equal to this number,
which would correspond only to

129
00:07:36,159 --> 00:07:39,830
this part of the tail and,
similarly, a small part of the

130
00:07:39,830 --> 00:07:42,200
tail from the other side.

131
00:07:42,200 --> 00:07:45,590
The blue event is smaller
than the red event.

132
00:07:45,590 --> 00:07:49,159
This is the probability of the
blue event, so it's going to

133
00:07:49,159 --> 00:07:54,970
be no larger than the
probability of the red event.

134
00:07:54,970 --> 00:07:58,690
Now if somebody gives us a value
of n, we should be able

135
00:07:58,690 --> 00:08:00,780
to calculate this probability.

136
00:08:00,780 --> 00:08:03,740
How do we calculate it?

137
00:08:03,740 --> 00:08:07,480
The probability that the
absolute value is above a

138
00:08:07,480 --> 00:08:13,600
certain number is equal to the
probability of this tail plus

139
00:08:13,600 --> 00:08:15,570
the probability of that tail.

140
00:08:15,570 --> 00:08:18,960
But because of the symmetry of
the normal distribution, this

141
00:08:18,960 --> 00:08:24,450
is twice the probability of
each one of the tails.

142
00:08:24,450 --> 00:08:26,390
What is the probability
of this tail?

143
00:08:26,390 --> 00:08:30,740
It's 1 minus the probability
of whatever is below that.

144
00:08:30,740 --> 00:08:32,669
So it's 1 minus.

145
00:08:32,669 --> 00:08:36,270
And the probability of being
below that, this is the

146
00:08:36,270 --> 00:08:44,850
standard normal CDF evaluated
at 0.02 square root n.

147
00:08:44,850 --> 00:08:48,870
So we do have now an expression
for the desired

148
00:08:48,870 --> 00:08:52,890
probability, or at least a
bound for it, which is

149
00:08:52,890 --> 00:08:57,640
expressed in terms of the
standard normal CDF.

150
00:08:57,640 --> 00:09:01,970
If somebody gives you a value
of n, you can plug in here.

151
00:09:01,970 --> 00:09:06,410
If n is 10,000, then square
root of n is 100.

152
00:09:06,410 --> 00:09:10,640
And this number becomes
equal to 2.

153
00:09:10,640 --> 00:09:14,730
And so in this case, what we
obtain is that the probability

154
00:09:14,730 --> 00:09:19,680
of interest is less than
or equal to 2 times 1

155
00:09:19,680 --> 00:09:23,230
minus Phi of 2.

156
00:09:23,230 --> 00:09:27,960
Now we invoke the standard
normal table.

157
00:09:27,960 --> 00:09:32,730
From the normal table, we obtain
that this quantity is

158
00:09:32,730 --> 00:09:45,060
equal to twice 1 minus 0.9772,
which evaluates to 0.046.

159
00:09:45,060 --> 00:09:53,300
So if we use 10,000 people in
our sample, then we will get

160
00:09:53,300 --> 00:09:57,280
an accuracy of one percentage
point with very high

161
00:09:57,280 --> 00:09:58,460
probability.

162
00:09:58,460 --> 00:10:02,620
The probability that we do not
meet the specification so that

163
00:10:02,620 --> 00:10:05,620
the accuracy that we get is
worse than one percentage

164
00:10:05,620 --> 00:10:08,660
point, that probability
is quite small.

165
00:10:08,660 --> 00:10:12,450
It's 0.046.

166
00:10:12,450 --> 00:10:18,080
That is 4 and something
percent.

167
00:10:18,080 --> 00:10:19,880
This is pretty good.

168
00:10:19,880 --> 00:10:25,070
And suppose that your boss now
tells you, I only want the

169
00:10:25,070 --> 00:10:30,610
probability of not meeting
the specs to be 5%.

170
00:10:30,610 --> 00:10:34,670
You look at this result, and
you say, with 10,000, I

171
00:10:34,670 --> 00:10:38,810
achieved a probability
of a large error

172
00:10:38,810 --> 00:10:42,210
that's less than 5%.

173
00:10:42,210 --> 00:10:45,990
This means that I probably have
some leeway and that I

174
00:10:45,990 --> 00:10:49,820
can reduce the size
of my sample.

175
00:10:49,820 --> 00:10:52,690
What could the size of
the sample be and

176
00:10:52,690 --> 00:10:56,060
still meet those specs?

177
00:10:56,060 --> 00:10:59,630
What we're trying to do here
is that we have this

178
00:10:59,630 --> 00:11:04,880
approximation for the
probability of interest, and

179
00:11:04,880 --> 00:11:07,500
we want to set this probability

180
00:11:07,500 --> 00:11:15,250
to a value of 0.05.

181
00:11:15,250 --> 00:11:18,860
Then we want to ask, what is
the value of n that will

182
00:11:18,860 --> 00:11:22,900
result in this particular
probability of

183
00:11:22,900 --> 00:11:26,060
not meeting the specs?

184
00:11:26,060 --> 00:11:28,090
Now we can do the algebra.

185
00:11:28,090 --> 00:11:33,080
And we find that this
corresponds to requiring that

186
00:11:33,080 --> 00:11:42,320
phi of 0.02 square root n
to be equal to 0.975.

187
00:11:42,320 --> 00:11:45,380
What's the interpretation
of this?

188
00:11:45,380 --> 00:11:49,500
We want to choose n so that
the probability of the two

189
00:11:49,500 --> 00:11:53,780
tails is 5%.

190
00:11:53,780 --> 00:11:58,620
This means that we want this
probability here to be 2 and

191
00:11:58,620 --> 00:12:00,510
1/2 percent.

192
00:12:00,510 --> 00:12:03,240
This means that the probability
of whatever is to

193
00:12:03,240 --> 00:12:13,550
the left of this number should
be 0.975, including the tail.

194
00:12:13,550 --> 00:12:18,540
This means, again, that we have
to look at the standard

195
00:12:18,540 --> 00:12:25,150
normal table and ask, what's the
value for which the CDF is

196
00:12:25,150 --> 00:12:28,190
equal to 0.975?

197
00:12:28,190 --> 00:12:34,230
So we look around, and we find
0.975 to be here, and it

198
00:12:34,230 --> 00:12:38,130
corresponds to 1.96.

199
00:12:38,130 --> 00:12:42,750
This tells us that 0.02
square root n

200
00:12:42,750 --> 00:12:48,840
should be equal to 1.96.

201
00:12:48,840 --> 00:12:54,240
Then we solve for n, and we find
that the value of n is

202
00:12:54,240 --> 00:13:01,110
9,604, which is indeed some
reduction from the 10,000 that

203
00:13:01,110 --> 00:13:02,360
we had originally.

204
00:13:04,450 --> 00:13:08,020
How does this relate
to the real world?

205
00:13:08,020 --> 00:13:12,330
When you read newspapers about
polls, you will never see

206
00:13:12,330 --> 00:13:16,100
sample sizes that are
about 10,000.

207
00:13:16,100 --> 00:13:20,290
You will usually see sample
sizes of the order of 1,000,

208
00:13:20,290 --> 00:13:22,530
sometimes even smaller.

209
00:13:22,530 --> 00:13:24,400
How can they do that?

210
00:13:24,400 --> 00:13:28,100
Well, they can do that because
the specs that they impose are

211
00:13:28,100 --> 00:13:31,140
not as tight as the specs
that we have here.

212
00:13:31,140 --> 00:13:35,100
Usually, they tell you that
the results are accurate

213
00:13:35,100 --> 00:13:38,850
within three percentage points,
let's say, instead of

214
00:13:38,850 --> 00:13:40,600
one percentage point.

215
00:13:40,600 --> 00:13:46,420
And by moving from 0.01 to 0.03,
and if you repeat those

216
00:13:46,420 --> 00:13:50,090
calculations, you will find that
the sample size of about

217
00:13:50,090 --> 00:13:53,690
1,000 will actually do.