Transcript Psychology 210 Psychometric Methods

Chapter 6
Probability
Inferential Statistics
Samples - so far we have been
concerned about describing and
summarizing samples or subsets of a
population
 Inferential stats allows us to “go
beyond” our sample and make
educated guesses about the
population

Inferential Statistics

But, we need some help which comes
from Probability theory
Inferential = Descriptive + Probability
Statistics
Statistics
Theory
What is probability theory
and what is its role?
Probability theory, or better “probability
theories” are found in mathematics
and are interested in questions about
unpredictable events
 Although there is no universal
agreement about what probability is,
probability helps us with the possible
outcomes of random sampling from
populations

Examples
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“The Chance of Rain”
The odds of rolling a five at the craps table (or
winning at black jack)
The chance that a radioactive mass will emit a
particle
The probability that a coin will come up heads
upon flipping it*
The probability of getting exactly 2 heads out of 3
flips of a coin*
The probability of getting 2 or more heads out of
3 flips*
Set Theory
(a brief digression)
Experiment - an act which leads to an
unpredictable, but measurable
outcome
 Set - a collection of outcomes
 Event - one possible outcome; a value
of a variable being measured
 Simple probability – the likelihood that
an event occurs in a single random
observation

Simple Probabilities

To compute a simple probability (read
the probability of some event),
p(event):
p(event) =
Number of equally likely observations of event in the population
Number of equally likely observations in the population
Probability Theory
Most of us understand probability in
terms of a relative frequency measure
(remember this, f/n), the frequency of
occurrences (f) divided by the total
number of trials or observations (n)
 However, probability theories are about
the properties of probability not
whether they are true or not (the

determination of a probability can come from a variety of
sources)
Example
100 marbles are
Placed
in a jar
Relative Frequency
(a reminder)
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What if we counted all
of the marbles in the
jar and constructed a
frequency distribution?
We find 50 black
marbles, 25 red
marbles, and 5 white
marbles
Relative frequency
(proportion) seems like
probability
Color
Black
Red
White
Total
f
50
25
25
100
rf
.50
.25
.25
1.00
1
0.8
0.6
0.4
0.2
Red
0
White
A graphical
representation of a
relative frequency
distribution is also
similar to a
“Probability
Distribution”
Black

Relative Frequency/
Probability
Relative Frequency and
Probability Distributions
What does this mean?
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What will happen if we choose a single marble
out of the jar?
If we chose 100 marbles from the jar, tallied
the color, and replaced them, will we get 50,
25, and 25? If so, what if we selected only
99?
If .5, .25, and .25 are the “real” probabilities,
then “in the long run” will should get relative
proportions that are close to .5, .25, and .25
Bernoulli’s Theorem
The notion of “in the long run” is
attributed to Bernoulli
 It is also known as the “law of large
numbers”
 as the number of times an experiment
is performed approaches infinity
(becomes large), the “true” probability
of any outcome equals the relative
proportion

Venn Diagrams
A
A
S
Venn Diagrams
A
“Red”
A
“not red”
S
“all the
marbles”
Mutually Exclusive Events
A
B
S
Mutually Exclusive Events
Axiom’s of Probability
1. The probability of any event A, denoted p(A),
is 0 < p(A) < 1
2. The probability of S, or of an event in sample
space S is 1
3. If there is a sequence of mutually exclusive
events (B1, B2, B3, etc.) and C represents the
event “at least one of the Bi’s occurs, then the
probability of C is the sum of the probabilities
of the Bis (p(C) = Σ p(Bi)
1. 0 < p(A) < 1 (in Venn diagrams)
A
S
S
A
The probability of event
A is between 0 and 1
2. p(S) = 1
A
S
The probability of AN
event, in S, occurring is 1
3. p(C) = Σp(Bi)
B1
B5
B4
If the events B1, B2, B3,
etc. are mutually
exclusive, the probability
of one of the Bs occurring
is C, the sum of the Bs
C
B2
B3
S
Mutually Exclusive Events

If A and B are mutually exclusive,
meaning that an event of type A
precludes event B from occurring, by
the 3rd axiom of probability
Mutually Exclusive Events

If A and B are mutually exclusive, and
set A and set B are not null sets,
Joint Events

If the events are independent, (not
mutually exclusive), meaning that the
occurrence of one does not affect the
occurrence of the other, the
intersection
Joint Events - Example
What is the probability of selecting a
black marble and white marble in two
successive selections?
 Since each selection is independent,
then
p(Black, White) = .5 • .25 = .125
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Generalization from
Joint Events
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If A, B, C, and D are independent
events, then:
What is the probability of selecting a
white marble, then red, then white,
then black?
What if the events are not
independent?
Conditional probability - the occurrence
of one event is influenced by another
event
 “Conditional Probability” refers to the
probability of one event under the
condition that the other event is known
to have occurred
 p(A | B) - read “the probability of A
given that B has occurred”

Probability Theory and
Hypothesis Testing
A man comes up to you on the street
and says that he has a “special”
quarter that, when flipped, comes up
heads more often than tails
 He says you can buy it from him for $1
 You say that you want to test the coin
before you buy it
 He says “OK”, but you can only flip it 5
times
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Probability Theory Example
How many heads would convince you
that it was a “special” coin?
 3?, 4?, 5?
 How “sure” do you want to be that it is
a “special” coin? What is the chance
that he is fooling you and selling a
“regular” quarter?
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2 Hypotheses
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The coin is not biased, it’s a normal
quarter that you can get at any bank
– The likelihood of getting a heads on a
single flip is 1/2, or .5

The coin is a special
– The probability of getting a heads on a
single flip is greater than .5
Hypothesis Testing

Let’s assume that it is a regular, old
quarter
p = .5 (the probability of getting a heads on
a SINGLE toss is .5)
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We flip the coin and get 4 heads.
What is the probability of this result,
assuming the coin is fair?
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Note that this is a problem involving conditional
probability : p(4/5 heads|coin is fair)
How do I solve this problem?
Any Ideas?
 You might think that, using the rule of
Joint events, that:

p(4 / 5) = p(H1 )· p(H 2 )· p(H 3 )· p(H 4 )
p(4 / 5) = p(H )
p(4 / 5) = .0625

NO!
4
Why not?
You have just calculated the probability
of getting exactly H, H, H, H on four
flips of our coin.
 What is the probability of getting H, T,
H, T on 4 flips?
 Exactly the same as H, H, H, H…any
single combination of 4 H and T are
equally likely in this scenario.
 Here they are:

All Possibilities: 4 flips of a coin
HHTT
HTHT
TTTT
0 Heads
f=
1
TTTH
HTTH
HHHT
TTHT
TTHH
HHTH
THTT
THTH
HTHH
HTTT
THHT
THHH
1 heads
4
2 heads
6
3 heads
4
HHHH
4 heads
1
Relative Frequency Dist
Heads
f
p
0
1
.0625
1
4
.25
2
6
.375
3
4
.25
4
1
.0625
Total
16
1.00
YES!
Relative frequency and Probability are
related by Bernoulli’s theorem
 If I did this test again, would I get the
same result? (probably not)
 If I did it over and over again, what
results would we expect given a nonbiased coin?
 How many combinations?
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What if
I figured out the total number of
possible outcomes of this experiment,
 and I figured out the total number of
outcomes that had 4/5 heads?
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Prob of 4/5 =
Freq of 4/5
Total N of outcomes
How many outcomes?
Lots
H, H, T, T, T
 T, H, T, H, T
 H, T, H, T, H
 ETC. ETC. ETC.
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How many 4 out of 5?
5 flips
 (exactly) 4 heads
 1 possibility – H, H, H, H, T
 Another – H, H, H, T, H
 More – H, H, T, H, H
 And – H, T, H, H, H
 Lastly – T, H, H, H, H

5 Flips: All possibilities
TTTTT
HHTTT
TTHHH
HTTHT
THHTH
HTHTT
THTHH
THHTT
HTTHH
THTHT
HTHTH
HTTTT
TTHHT
HHTTH THHHH
THTTT
HTTTH
THHHT HTHHH
TTHTT
THTTH
HTHHT HHTHH
TTTHT
TTHTH
HHTHT HHHTH
TTTTH
TTTHH
HHHTT HHHHT
HHHHH
0 heads 1 head 2 heads 3 heads 4 heads 5 heads
p = .03125
.15625
.3125
.3125
.15625
.03125
At least 4 heads out of 5
Given a Fair Coin:
 Getting at least 4 heads out of 5 flips is
p(4) + p(5)

B1
B5
.15625+.03125 = .1875
B4
C
B3
There is a 18.75% chance that, upon
flipping a FAIR coin 5 times, you will get
at least 4 heads.
B2
You gonna buy that quarter?
What if this guy let you flip this quarter
100 times?
 How many times do you want to flip it?
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(the more the better, yes? In the long
run???)