Download lecture 4

Introduction to Data
Analysis
Probability
Confidence Intervals
Today’s lecture

Some stuff on probability

Confidence intervals (A&F 5).



Standard error (part 2) and efficiency.
Confidence intervals for means.
Confidence intervals for proportions.
2
Last week’s aide memoire

Population distribution


Sample distribution


We don’t know this, but we want to know about it. In particular we
normally want to know the mean.
We know this, and calculate the statistics such as the sample mean
and the sample standard deviation from it.
Sampling distribution


This describes the variability in value of the sample means amongst
all the possible samples of a certain size.
We can work this out from information about the sample
distribution and the fact that the sampling distribution is normal if
sample size is large (ish).
3
Random Variables
Flip a coin. Will it be heads or tails?
 The outcome of a single event is random, or
unpredictable
 What if we flip a coin 10 times? How many
will be heads and how many will be tails?
 Over time, patterns emerge from seemingly
random events. These allow us to make
probability statements.

4
Heads or Tails?
A coin toss is a random event [H or T]
unpredictable on each toss but a stable pattern
emerges of 50:50 after many repetitions.
 The French naturalist, Buffon (1707-1788)
tossed a coin 4040 times; resulting in 2048
heads for a relative frequency of 2048
/4040 = .5069
5
Heads or Tails?
The English mathematician John Kerrich,
while imprisoned by Germans in WWII, tossed
a coin 5,000 times, with result 2534 heads .
What is the Relative Frequency?
 2,534 / 5,000 = .5068

6
7
Heads or tails?
A computer simulation of 10,000 coin flips
yields 5040 heads. What is the relative
frequency of heads?
 5040 / 10,000 = .5040

8
Each of the tests is the result of a sample of fair
coin tosses.
Sample outcomes vary.
• Different samples produce different results.
True, but the law of large numbers tells us
that the greater the number of repetitions the
closer the outcomes come to the true
probability, here .5.
A single event may be unpredictable but the
relative frequency of these events is lawful
over an infinite number of trials\repetitions.
9
Random Variables




"X" denotes a random variable. It is the outcome of a
sample of trials.
“X,” some event, is unpredictable in the short run but
lawful over the long run.
This “Randomness” is not necessarily unpredictable.
Over the long run X becomes probabilistically
predictable.
We can never observe the "real" probability, since the
"true" probability is a concept based on an infinite
number of repetitions/trials. It is an "idealized" version
of events
10
To figure the odds of some event
occurring you need 2 pieces of
information:
1. A list of all the possibilities – all the
possible outcomes (sample space)
2. The number of ways to get the
outcome of interest (relative to the
number of possible outcomes).
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Take a single Dice Roll
Assuming an evenly-weighted 6-sided dice,
what are the odds of rolling a 3?
 How do you know?

6
possible outcomes (equally likely)
 1 way to get a 3
 p(Roll=3) = 1 / 6
12

What are the chances of rolling numbers
that add up to “4” when rolling two sixsided dice?

What do we need to know?
 All
Possible Outcomes from rolling two dice
 Outcomes that would add up to 4
13
How Many Ways can the Two Dice Fall?
Let’s say the dice are different colors (helps us
keep track.
The White Dice could come out as:
We know how to figure out probabilities here, but
What about the other dice?
14

When the white die shows
possible outcomes.
, there are six

When the white die shows
more possible outcomes.
, there are six

We then just do that for all six possible
outcomes on the white die
15
16

Remember the Question: What is the
probability of Rolling numbers that sum to 4?

What do we need to know?
 All

Possible Outcomes from rolling two dice
(36--Check Previous Slide)
 How
many outcomes would add up to 4?
Our Probability is 3/36 = .08333
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Probability = Frequency of Occurrence
Total # outcomes
Frequency of occurrence = # of ways this
one event could happen
Total # outcomes = # ways all the
possible events could happen
Probability of a 7 is 6 ways out of 36
possibilities
p=.166
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Frequency of Sum of 2 Dice
F
R
E
Q
U
E
N
C
Y
6 5 4 3 2 1 -
*
p = .139 *
p = .167
*
p =.111 *
p=.083
p = .139
*
*
p = .111
* p=.083
* p=.056
p=.056
*
* p=.027
p=.027
*
--+----+----+----+----+----+----+----+----+----+----+
2.0
4.0
6.0
8.0
SUM OF 2 DICE
10.0
12.0
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Review of Set Notation
Capital Letters sets of points
 Lower case letters represent elements of the set
 For example:

A = {a1, a2, a3}
20
Subsets
Let S denote the full sample space (the set of
all possible elements)
 For two sets A and B, if every element of A is
also an element in B, we say that A is a subset
of B
A B

S
B
A
21
Union

The union of two arbitrary sets of points is the
set of all points that are in at least one of the
sets
A B
S
A
B
22
Intersection

The intersection of two arbitrary sets of points
is the set of all points that are in both of the
sets
A B
23
Mutual Exclusivity

Two events are said to be disjoint or mutually
exclusive if none of the elements in set A
appear in set B.
24
Independence
We will give a more rigorous definition later,
but…
 Two events are independent if the occurrence of A
is unaffected by the occurrence or nonoccurrence
of B.
 Example: You flip a coin—what is the probability
of heads?
 You flip it 10 times, getting heads each time.
What is the probability of getting heads again?

25
Axioms for Probabilities

The conventional rules for probabilities are
named the Kolmogorov Axioms. They are:
1. P( A)  0
2. P( S )  1
3. If A1, A2, A3, … are pairwise mutually exclusive
events in S, then:
P( A1  A2  A3  )   P( Ai )
26
Rules for Calculating Probabilities

Simple Additive rule for disjoint events
 a.k.a.
the “or” rule
P( A  B)  P( A)  P( B)
S
A
B
27
Example:
One community is 75% white (non-hispanic),
10% black (non-hispanic), and 15% hispanic.
They choose their mayor at random to
maximize equality.
 What is the probability that the next mayor
will be non-white?
P( Black  Hispanic)  P( Black )  P( Hispanic)
P( Black  Hispanic)  .1  .15
P( Black  Hispanic)  .25

28
Rules for Calculating Probabilities

Simple Multiplicative rule for independent
events
 a.k.a.
the “and” rule
P( A  B)  P( A)* P( B)
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Example:

Suppose in that same mythical community
(75% white, 10% black, 15% Hispanic) there
was an even division of males and females.
What is the probability of a white male mayor?
P(White  Male)  P(White) * P( Male)
P(White  Male)  (.75) *(.5)
P(White  Male)  .375
30
Rules for Calculating Probabilities

The Complement Rule
P( A )  1  P( A)
C
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Rules for Calculating Probabilities

Additive rule for events that are not mutually
exclusive events
P( A  B)  P( A)  P( B)  P(A  B)
32
Rules for Calculating Probabilities

Multiplicative rule for conditional events
P( A  B)  P( A) P( B | A)
33
But wait…

Now we know the rules for manipulating
probabilities mathematically, but to get them,
we need to calculate the sample space and the
number of ways to get the outcome
34
Tools for Counting Sample Spaces
Listing
 The mn rule
 Permutations (Pnr)
 Combinations (Cnr)

35
Listing

You are flipping 2 coins (this is one trial). List
the possible outcomes:
HH
HT

TH
TT
S=4
You are flipping 3 coins:
THH
HTT
HTH
THT
HHH
HHT
TTH
S=8
TTT
36
Listing

Advantages
 Intuitive
 Easy

(with a small sample space)
Disadvantages
 Hard
to do with a large sample space
37
The mn rule
Think of the coin example—there are 2
possible outcomes for each coin, and 2 coins.
Thus, there are 2 * 2 = 4 possible outcomes
 Think of the dice example—there are 6
possible outcomes for each dice, and 2 dice.
Thus, there are 6 * 6 = 36 possible outcomes

38
The mnp rule ?
It works for more than 2 dice/coins or
whatever too. Consider the 3 coin flips:
2 * 2 * 2 = 23 = 8
 This can be thought of as successive
applications of the mn rule. First, you get the
combinations of 2 coin flips

mn = 2 * 2 = 4

Then you get the combination of that sample
space with the 3rd coin
(mn)*p = 4 * 2 = 8
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Complex Example
Assume that leap years don’t exist and there
are thus just 365 possible birthdays.
 We want to know the sample space for
possible birthdays for 20 randomly drawn
people.
 How do we get it?
36520 = 1.76*1051

40
Complex Example

Assuming that each birthday is equally likely,
what is the probability of everyone having a
January 1 Birthday?
1
36520

What is the probability that everyone has the
same birthday?
365
1

20
365
36519
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Complex Example



What is the probability that everyone has a different
birthday?
For the first drawn person, there are 365 possible bdays because no one else has been drawn
For the second drawn person, there are 364 possible
b-days because only 1 birthday has been drawn; for
the third person it’s 363; fourth is 362…
365*364*363*
36520
*346
 .589
42
Permutations
P 
n
r
In some situations, we will be concerned with
the order in which sequences occur.
 An ordered arrangement of r objects is called a
permutation
 The number of possible permutations is denoted
n
as
Pr
 Our proof is based on our extension of the mn
rule in the last problem…

43
Permutations

P 
n
r
We are interested in filling r positions with n
distinct objects—no repeated values (selecting
20 people with 365 possible birthdays)
n!
P  n(n  1)(n  2)(n  3)...(n  r  1) 
 n  r !
n
r
44
Permutation Example

Think of a chain lock with a combination that
requires 4 digits (0-9)
4

6
2
0
How many possible combinations are there?
10!
10! 10*9*8*7*6*5*4*3*2*1
P 


6*5*4*3*2*1
10  4! 6!
10
4
10*9*8*7  5040
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Combinations
In some situations, the ordering of the symbols
in a set is unimportant—we only care what is
included (not its position)
 There will be fewer combinations than
permutations—permutation counts HHT,
HTH, and THH as being separate;
combinations don’t consider the order, just the
contents.
n
n


P
n!
n
r
Cr    

 r  r ! r ! n  r  !

46
Example

A company selects five applicants for a “short
list” for a job. They will actually interview
just 2 candidates. How combinations of two
applicants can be selected from the pool?
5
5!
5!
C  

 2  2! 5  2 ! 2! 3!
5*4*3*2*1
5*4

 5*2  10
 2*1 *  3*2*1 2
5
2
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Example 2




A company receives 10 applications for a single position.
Being short on time, but good at combinatorics, the boss
considers drawing two applications at random for
interviews.
The boss wants to know the probability of getting
exactly 1 of the two best applicants in his sample by
random chance.
The probability of choosing one of the two best is given
 2
by
 
1
The probability of choosing one of the three worst is
given by  13 
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 
Example 2

We can then use the mn rule to figure out the
number of ways we can get both
 2   3  2! 3!
3  2 1

 2
 23  6
    
2 1
 1   1  1!1! 1!2!1!

In the last example we computed the number
of possible combinations to be
5
5!
5!
5
C2    

 10
 2  2! 5  2 ! 2! 3!

Thus, the probability is 6 / 10 = .6
49
Conditional Probability
Under some circumstances the probability of
an event depends on another event.
 An unconditional probability asks what the
chances are of rain tomorrow (event A).
P( A)


A conditional probability says, “Given that
rained today (event B), what are the chances of
rain tomorrow? (event A)”
 P(A|B)
50
Computing Conditional Probabilities
P( B  A)
P( B | A) 
P( A)
51
Independence

Two events are said to be independent if
P( A | B)  P( A)

Otherwise, the events are dependent
52
Bayes’ Rule

Suppose we knew P(B|A) but wanted to know
P(A|B)?
P( B j | A) 
P( B j ) P( A | B j )
k
 P( B ) P( A | B )
i 1
i
i
53
Example
Suppose you have been tested positive for a disease;
what is the probability that you actually have the
disease? Suppose the probability of having the
disease is .01. The test is 95% accurate, and you
tested positive. Do you have the disease?
 We know:

 The
probability of anyone having the disease (.01)
 The probability of testing positive for the disease
conditional on having the disease (.95)

We want to know the probability of having the
disease if you tested positive for it…
54
Bayes’ Rule
P( HaveIt | TestPos) 
P( HaveIt ) P(TestPos | HaveIt )
P( HaveIt ) P(TestPos | HaveIt )  P( NoHaveIt ) P(TestPos | NoHaveIt )
.01 .95
P( HaveIt | TestPos) 
.01 .95  .99  .05
.0095
.0095
P( HaveIt | TestPos ) 

 .161
.0095  .0495 .059
55
What? .161? Why so low?
Out of 100 people who take this test, we
expect only 1 would have the disease.
 However, 5 people would test positive even if
they didn’t have the disease.
 Out of those 6 people, only 1 actually has the
disease…

56
Political Application

In a certain population of voters, 40% are
Republican and 60% are Democrats. It is
reported that 30% of Republicans and 70% of
Democrats support a particular issue. A
randomly selected person is found to favor the
issue—what is the conditional probability that
they are a Democrat?
57
Work it out

We want to know P(Dem | F_issue)
P( Dem) P( F _ issue | Dem)
P( Dem | F _ issue) 
P( Dem) P( F _ issue | Dem)  P(Rep) P( F _ issue | Rep)
.6  .7
P( Dem | F _ issue) 
.6  .7  .4  .3
.42
P( Dem | F _ issue) 
 .778
.42  .12
58
Standard errors continued…

Last week we managed to work out some info
about the distribution of sample means. If we
have lots of samples then:
Mean of all the sample means = population mean.
 Shape of the distribution of sample means = normal.
 Standard error tells us how dispersed the distribution of
sample means is.


Put all these together, and we can finally work
out how likely our sample mean is to be near
the population mean.
59
Churchgoing example

Let’s take a less contrived example than my
car managing to break any speed limits.
This week I’m interested in the average number of times
a year people go to church (or synagogue, or whatever).
 I take a sample of 100 people and record the number of
times they went to church last year. Some went a lot but
most went infrequently.
 So, from that sample I get a sample mean and a sample
standard deviation.

60
Church-going sample
60
Sample mean = 8.5
50
Frequency
40
Sample s.d. = 20
30
20
10
0
0
10
20
30
40
50
60
Number of times per year
61
Standard error

Since we know the s.d. and mean of the sample, we can
calculate the standard error.
 
Standard error X 


s
20

2
n
100
Because we know the standard error and the fact that the
sampling distribution is normal, we are able to calculate how
‘likely’ it is that a specific range around our estimate contains
the population mean.
We can calculate what’s called a confidence interval.
62
Confidence intervals (1)

A confidence interval for an estimate is a range
of numbers within which the parameter is
‘likely’ to fall.


Remember a parameter is something about the
population, like the population mean.
We can use the standard error (and the fact that
the sampling distribution is normal) to produce
such a range.
63
Confidence intervals (2)
estimate  (k  standard error )
k is chosen to determine what is meant by
‘likely’ to contain the actual value of the
population mean.
 We want to pick a k that is meaningful to us.

… and is not so large that the interval is useless.
 … but not so small that the interval is very unlikely to
contain the true population value.

64
Confidence intervals (3)
To return to our example, we have:
 An estimate of the population mean (the
mean number of days of church attendance)
which is the sample mean (8 ½ days).
 A standard error (2 days) which allows us to
put a range around our estimate.

8.5  ( k  2)
65
How do we pick k?
Imagine sampling repeatedly, and therefore
getting lots of sample means.
 Given what we know about the sampling
distribution (i.e. it’s normal if n is large
enough), we know what proportion of these
sample means will fall within a certain number
of standard errors of the actual population
mean.

66
Confidence coefficient

Call this proportion the confidence coefficient.



If we picked .95 then we know that if we repeatedly sampled the
population, that interval around each of the sample means would
include the true value 95% of the time.
The confidence coefficient is a number that is chosen by the
researcher which is close to 1, like 0.95 or 0.99.
Since the sampling distribution is normal, we know
the values of k that correspond to the probability of
any proportion.

So we know that approximately 95% of confidence intervals that
are 2 standard errors either side of the sample mean will include the
population mean.
67
Back to church

For our particular sample we calculate a 95%
confidence interval (i.e. k=2).
k
8.5  (2  2)
8.5  4

SE
Imagine the population mean for churchgoing in
Britain is actually 6 days a year.


So our particular sample is off by 2 ½ days a year.
The mean of all the possible sample means is equal to the
population mean so the centre of the sampling distribution is 6.
68
Sampling distribution (1)
Population mean = 6
0
1
2
3
4
5
6
7
Church-going (days a year)
8
9
10
11
12
Sample mean=8.5
69
More samples
But that’s just our one sample, let’s imagine we
took many samples, and then calculated 95%
confidence intervals for all the sample means.
70
Sampling distribution (2)
Population mean = 6
0
1
2
3
4
5
6
7
8
9
10
11
12
Church-going (days a year)
71
Sampling distribution (3)


Of my 7 samples, all the confidence intervals around
the sample mean enclosed the actual true population
mean apart from one.
If we repeated this lots of times, we would expect
95% of the confidence intervals to enclose the actual
population mean.


95% because that’s the confidence coefficient that we picked.
If we had picked a confidence coefficient of 0.99, then it would be
99% and the confidence intervals would be larger.
72
Confidence coefficients

To make things easier I’ve been rounding the
numbers off, the exact figures for k at each
confidence coefficient are slightly different.
Confidence coefficient
k
68%
95%
99%
99.9%
1.00
1.96
2.58
3.29
73
Smaller confidence intervals?

We could just be less certain.
If I was willing to pick a 90% confidence interval then
the range of the interval would be narrower as k would
be lower.
 We would be wrong more often though…


We could increase the sample size.
The bigger the sample size, the lower the standard error
and therefore the smaller the confidence interval for a
given probability.
 This isn’t always practical of course…

74
Why 95%?




Generally speaking in social science we pick 95% as
an appropriate confidence coefficient.
A 1 in 20 chance of being wrong is generally felt to
be okay.
Path dependency—Fisher integrated the normal PDF
for the 95% level, which is REALLY hard to do
without a computer.
This isn’t always the case…


Sometimes we want to be really sure we’ve enclosed the mean.
Other times we want a narrower range and are willing to accept that
we are wrong more often.
75
CIs for proportions

Since calculating the standard error is similar for
proportions, so are producing confidence intervals.


Remember we need binary data coded a 0 and 1 (yes/no,
men/women etc.)
Remember the standard error for proportions depends on the n
and the sample proportion. So the CI is as below:
P(1  P)
  P  1.96
n
76
Proportions example

Let’s take a political
opinion poll in the US,
we’re interested in the
population proportion
voting Democrat (call
this π).
 Sample
is 1000 people.
 Sample proportion is
0.45 (or 45%).
  P  1.96
P (1  P )
n
0.45(1  0.45)
1000
  0.45  1.96  0.0157
  0.45  0.03
  0.45  1.96
i.e. Actual proportion is 45%  3%,
so the interval is 42%, 48%.
77
Opinion polls

A lot of opinion polls use a sample size of
about 1000 people, and aim to estimate
proportions that are between 30% and 50%.


This is why you often see ‘margins of error’ that are 3%
either side of an estimate quoted in newspapers; these
are 95% confidence intervals.
Note that all these confidence intervals ignore
non-sampling error.

What we call sampling bias, this could be due to nonresponse, badly worded questions and so forth.
78
Exercise

If you were taking a political opinion poll,
roughly how big a sample would you need to
achieve a ‘margin of error’ (i.e. a 95%
confidence interval) of:


2% either side of the estimate?
1% either side of the estimate?
(Assume that the sample proportion of interest
is around 40%).
79
Exercise answer
Margin of error of 0.02
0.02  1.96
0.02  2
0.02

2
P( 1  P)
n
P( 1  P)
n
0.40( 1  0.40 )
n
0.24
n
0.24
0.0001 
n
n  2400
0.01 
Margin of error of 0.01
0.01  1.96
0.01  2
0.01

2
P( 1  P)
n
P( 1  P)
n
0.40( 1  0.40 )
n
0.24
n
0.24
0.000025 
n
n  9600
0.005 
80