Practice
• You bought a ticket for a fire department
lottery and your brother has bought two
tickets. You just read that 1000 tickets
were sold.
– a) What is the probability you will win the
grand prize?
– b) What is the probability that your brother will
win?
– c) What is the probably that you or your
bother will win?
5.2
• A) 1/1000 = .001
• B)2/1000 = .002
• C) .001 + .002 = .003
Practice
• Assume the same situation at before
except only a total of 10 tickets were sold
and there are two prizes.
– a) Given that you didn’t win the first prize, what is the
probability you will win the second prize?
– b) What is the probability that your borther will win the
first prize and you will win the second prize?
– c) What is the probability that you will win the first
prize and your brother will win the second prize?
– d) What is the probability that the two of you will win
the first and second prizes?
5.3
•
•
•
•
A) 1/9 = .111
B) 2/10 * 1/9 = (.20)*(.111) = .022
C) 1/10 * 2/9 = (.10)*(.22) = .022
D) .022 + .022 = .044
Practice
• In some homes a mother’s behavior
seems to be independent of her baby's,
and vice versa. If the mother looks at her
child a total of 2 hours each day, and the
baby looks at the mother a total of 3 hours
each day, and if they really do behave
independently, what is the probability that
they will look at each other at the same
time?
5.8
• 2/24 = .083
• 3/24 = .125
• .083*.125 = .01
Practice
• Abe ice-cream shot has six different
flavors of ice cream, and you can order
any combination of any number of them
(but only one scoop of each flavor). How
many different ice-cream cone
combinations could they truthfully
advertise (note, we don’t care about the
order of the scoops and an empty cone
doesn’t count).
5.29
6!
6
1!(1  6)!
6!
 15
2!(2  6)!
6!
 20
3!(3  6)!
6!
 15
4!(6  4)!
6!
6
5!(6  5)!
6!
1
6!(6  6)!
6 + 15 + 20 +15 + 6 + 1 = 63
• Suppose you live in a place that has a
constant chance of being struck by
lightning at any time through the year.
Suppose that the strikes are random:
every day the chance of a strike is the
same, and the rate works out to one strike
a month. Your house is hit by lightning
today, Monday. What is the most likely day
for the next bolt to strike your house?
• Tuesday!
• Say the prob of lighting hitting is .025
• On Tues prob of hitting is .025
• For Wed to be next hit this would have to
be true: no hit on Tues AND hit on Wed
975*.025=.0243
• For Thursday
.975*.975*.025=.0237
Remember
• Playing perfect black jack – the probability
of winning a hand is .498
• What is the probability that you will win 8
of the next 10 games of blackjack?
Binomial Distribution
N!
X (NX )
p( X ) 
p q
X !( N  X )!
Ingredients:
N = total number of events
p = the probability of a success on any one trial
q = (1 – p) = the probability of a failure on any one trial
X = number of successful events
Binomial Distribution
10!
8
(108 )
p( X ) 
.498 .502
8!(10  8)!
Ingredients:
N = total number of events
p = the probability of a success on any one trial
q = (1 – p) = the probability of a failure on any one trial
X = number of successful events
Binomial Distribution
10!
8
(108 )
p( X ) 
.498 .502
8!(10  8)!
p = .0429
Ingredients:
N = total number of events
p = the probability of a success on any one trial
q = (1 – p) = the probability of a failure on any one trial
X = number of successful events
Binomial Distribution
• What if you are interested in the probability
of winning at least 8 games of black jack?
• To do this you need to know the
distribution of these probabilities
Probability of Winning Blackjack
Number of Wins
• p = .498, N = 10
0
1
2
3
4
5
6
7
8
9
10
p
Probability of Winning Blackjack
• p = .498, N = 10
Number of Wins
p
0
.001
1
2
3
4
5
6
7
8
9
10
Probability of Winning Blackjack
• p = .498, N = 10
Number of Wins
p
0
.001
1
.010
2
.045
3
.119
4
.207
5
.246
6
.203
7
.115
8
.044
9
.009
10
.001
Probability of Winning Blackjack
• p = .498, N = 10
Number of Wins
p
0
.001
1
.010
2
.045
3
.119
4
.207
5
.246
6
.203
7
.115
8
.044
9
.009
10
.001
1.00
Binomial Distribution
0.3
0.25
0.2
p
0.15
0.1
0.05
0
0
1
2
3
4
5
6
Games Won
7
8
9
10
Hypothesis Testing
• You wonder if winning at least 7 games of
blackjack is significantly (.05) better than
what would be expected due to chance.
• H1= Games won > 6
• H0= Games won < or equal to 6
• What is the probability of winning 7 or
more games?
Binomial Distribution
0.3
0.25
0.2
p
0.15
0.1
0.05
0
0
1
2
3
4
5
6
Games Won
7
8
9
10
Binomial Distribution
0.3
0.25
0.2
p
0.15
0.1
0.05
0
0
1
2
3
4
5
6
Games Won
7
8
9
10
Probability of Winning Blackjack
• p = .498, N = 10
Number of Wins
p
0
.001
1
.010
2
.045
3
.119
4
.207
5
.246
6
.203
7
.115
8
.044
9
.009
10
.001
1.00
Probability of Winning Blackjack
• p = .498, N = 10
• p of winning 7 or more
games
• .115+.044+.009+.001 = .169
• p > .05
• Not better than chance
Number of Wins
p
0
.001
1
.010
2
.045
3
.119
4
.207
5
.246
6
.203
7
.115
8
.044
9
.009
10
.001
1.00
Practice
• The probability at winning the “Statistical
Slot Machine” is .08.
• Create a distribution of probabilities when
N = 10
• Determine if winning at least 4 games of
slots is significantly (.05) better than what
would be expected due to chance.
Probability of Winning Slot
Number of Wins
p
0
.434
1
.378
2
.148
3
.034
4
.005
5
.001
6
.000
7
.000
8
.000
9
.000
10
.000
1.00
Binomial Distribution
p
0.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
6
Games Won
7
8
9
10
Probability of Winning Slot
Number of Wins
p
0
.434
1
.378
2
.148
3
.034
4
.005
5
.001
6
.000
7
.000
8
.000
9
.000
10
.000
• p of winning at least 4
games
• .005+.001+.000 . . . .000
= .006
• p< .05
• Winning at least 4 games
is significantly better than
chance
1.00
Binomial Distribution
• These distributions
can be described with
means and SD.
• Mean = Np
• SD = Npq
Binomial Distribution
• Black Jack; p = .498, N =10
• M = 4.98
• SD = 1.59
Binomial Distribution
0.3
0.25
0.2
p
0.15
0.1
0.05
0
0
1
2
3
4
5
6
Games Won
7
8
9
10
Binomial Distribution
• Statistical Slot Machine; p = .08, N = 10
• M = .8
• SD = .86
Binomial Distribution
Note: as N gets bigger, distributions
will approach normal
p
0.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
6
Games Won
7
8
9
10
Next Step
• You think someone is cheating at BLINGOO!
• p = .30 of winning
• You watch a person play 89 games of blingoo
and wins 39 times (i.e., 44%).
• Is this significantly bigger than .30 to assume
that he is cheating?
Hypothesis
• H1= .44 > .30
• H0= .44 < or equal to .30
• Or
• H1= 39 wins > 26.7 wins
• H0= 39 wins < or equal to 26.7 wins
Distribution
• Mean = 26.7
• SD = 4.32
• X = 39
Z-score
Results
• (39 – 26.7) / 4.32 = 2.85
• p = .0021
• p < .05
• .44 is significantly bigger than .30. There is
reason to believe the person is cheating!
• Or – 39 wins is significantly more than 26.7 wins
(which are what is expected due to chance)
BLINGOO Competition
• You and your friend enter at competition with 2,642 other
players
• p = .30
• You win 57 of the 150 games and your friend won 39.
• Afterward you wonder how many people
– A) did better than you?
– B) did worse than you?
– C) won between 39 and 57 games
• You also wonder how many games you needed to win in
order to be in the top 10%
Blingoo
• M = 45
• SD = 5.61
• A) did better than you?
• (57 – 45) / 5.61 = 2.14
• p = .0162
• 2,642 * .0162 = 42.8 or 43 people
Blingoo
• M = 45
• SD = 5.61
• A) did worse than you?
• (57 – 45) / 5.61 = 2.14
• p = .9838
• 2,642 * .9838 = 2,599.2 or 2,599 people
Blingoo
• M = 45
• SD = 5.61
• A) won between 39 and 57 games?
•
•
•
•
(57 – 45) / 5.61 = 2.14 ; p = .4838
(39 – 45) / 5.61 = -1.07 ; p = .3577
.4838 + .3577 = .8415
2,642 * .8415 = 2,223.2 or 2, 223 people
Blingoo
• M = 45
• SD = 5.61
• You also wonder how many games you needed
to win in order to be in the top 10%
• Z = 1.28
• 45 + 5.61 (1.28) = 52.18 games or 52 games
Is a persons’ size related to if
they were bullied
• You gathered data from 209 children at
Springfield Elementary School.
• Assessed:
• Height (short vs. not short)
• Bullied (yes vs. no)
Results
Ever Bullied
Height
Yes
No
Short
42
50
Not short
30
87
Results
Ever Bullied
Height
Yes
No
Total
Short
42
50
92
Not short
30
87
117
Total
72
137
209
Results
Ever Bullied
Height
Yes
No
Total
Short
42
50
92
Not short
30
87
117
34%
66%
209
Total
Results
Ever Bullied
Height
Yes
No
Total
Short
42
50
44%
Not short
30
87
56%
Total
72
137
209
Results
Ever Bullied
Height
Yes
No
Total
Short
42
50
92
Not short
30
87
117
Total
72
137
209
Results
Ever Bullied
Height
Yes
No
Total
Short
45%
55%
92
Not short
26%
74%
117
72
137
209
Total
Is this difference in proportion
due to chance?
• To test this you use a Chi-Square (2)
• Notice you are using nominal data
Hypothesis
• H1: There is a relationship between the two
variables
– i.e., a persons size is related to if they were
bullied
• H0:The two variables are independent of
each other
– i.e., there is no relationship between a
persons size and if they were bullied
Logic
• 1) Calculate an observed Chi-square
• 2) Find a critical value
• 3) See if the the observed Chi-square falls
in the critical area
Chi-Square
O = observed frequency
E = expected frequency
Results
Ever Bullied
Height
Yes
No
Total
Short
42
50
92
Not short
30
87
117
Total
72
137
209
Observed Frequencies
Ever Bullied
Height
Yes
No
Total
Short
42
50
92
Not short
30
87
117
Total
72
137
209
Expected frequencies
• Are how many observations you would
expect in each cell if the null hypothesis
was true
– i.e., there there was no relationship between a
persons size and if they were bullied
Expected frequencies
• To calculate a cells expected frequency:
• For each cell you do this formula
Expected Frequencies
Height
Ever Bullied
Yes
No
Short
42
50
92
Not short
30
87
117
Total
72
137
209
Total
Expected Frequencies
Height
Ever Bullied
Yes
No
Short
42
50
92
Not short
30
87
117
Total
72
137
209
Total
Expected Frequencies
Row total = 92
Height
Ever Bullied
Yes
No
Short
42
50
92
Not short
30
87
117
Total
72
137
209
Total
Expected Frequencies
Row total = 92
Column total = 72
Height
Ever Bullied
Yes
No
Short
42
50
92
Not short
30
87
117
Total
72
137
209
Total
Expected Frequencies
Row total = 92
N = 209
Column total = 72
Height
Ever Bullied
Yes
No
Short
42
50
92
Not short
30
87
117
Total
72
137
209
Total
Expected Frequencies
E = (92 * 72) /209 = 31.69
Height
Ever Bullied
Yes
No
Short
42
50
92
Not short
30
87
117
Total
72
137
209
Total
Expected Frequencies
Height
Short
Not short
Total
Ever Bullied
Yes
No
Total
42
(31.69)
30
50
92
87
117
72
137
209
Expected Frequencies
Height
Short
Not short
Total
Ever Bullied
Yes
No
Total
42
(31.69)
30
50
92
87
117
72
137
209
Expected Frequencies
E = (92 * 137) /209 = 60.30
Height
Short
Not short
Total
Ever Bullied
Yes
No
Total
42
(31.69)
30
50
(60.30)
87
92
117
72
137
209
Expected Frequencies
E = (117 * 72) / 209 = 40.30
E = (117 * 137) / 209 = 76.69
Height
Short
Not short
Total
Ever Bullied
Yes
No
42
(31.69)
30
(40.30)
72
50
(60.30)
87
(76.69)
137
Total
92
117
209
Expected Frequencies
The expected frequencies are what you would expect if
there was no relationship between the two variables!
Height
Short
Not short
Total
Ever Bullied
Yes
No
42
(31.69)
30
(40.30)
72
50
(60.30)
87
(76.69)
137
Total
92
117
209
How do the expected
frequencies work?
Looking only at:
Ever Bullied
Height
Yes
No
Total
Short
42
(31.69)
30
(40.30)
72
50
(60.30)
87
(76.69)
137
92
Not short
Total
117
209
How do the expected
frequencies work?
If you randomly selected a person from these 209
people what is the probability you would select a
person who is short?
Ever Bullied
Height
Yes
No
Total
Short
42
(31.69)
30
(40.30)
72
50
(60.30)
87
(76.69)
137
92
Not short
Total
117
209
How do the expected
frequencies work?
If you randomly selected a person from these 209
people what is the probability you would select a
person who is short? 92 / 209 = .44
Ever Bullied
Height
Yes
No
Total
Short
42
(31.69)
30
(40.30)
72
50
(60.30)
87
(76.69)
137
92
Not short
Total
117
209
How do the expected
frequencies work?
If you randomly selected a person from these 209
people what is the probability you would select a
person who was bullied?
Ever Bullied
Height
Yes
No
Total
Short
42
(31.69)
30
(40.30)
72
50
(60.30)
87
(76.69)
137
92
Not short
Total
117
209
How do the expected
frequencies work?
If you randomly selected a person from these 209
people what is the probability you would select a
person who was bullied? 72 / 209 = .34
Ever Bullied
Height
Yes
No
Total
Short
42
(31.69)
30
(40.30)
72
50
(60.30)
87
(76.69)
137
92
Not short
Total
117
209
How do the expected
frequencies work?
If you randomly selected a person from these 209
people what is the probability you would select a
person who was bullied and is short?
Ever Bullied
Height
Yes
No
Total
Short
42
(31.69)
30
(40.30)
72
50
(60.30)
87
(76.69)
137
92
Not short
Total
117
209
How do the expected
frequencies work?
If you randomly selected a person from these 209
people what is the probability you would select a
person who was bullied and is short? (.44) (.34) = .15
Ever Bullied
Height
Yes
No
Total
Short
42
(31.69)
30
(40.30)
72
50
(60.30)
87
(76.69)
137
92
Not short
Total
117
209
How do the expected
frequencies work?
How many people do you expect to have been bullied
and short?
Ever Bullied
Height
Yes
No
Total
Short
42
(31.69)
30
(40.30)
72
50
(60.30)
87
(76.69)
137
92
Not short
Total
117
209
How do the expected
frequencies work?
How many people would you expect to have been
bullied and short? (.15 * 209) = 31.35 (difference due
to rounding)
Ever Bullied
Height
Yes
No
Total
Short
42
(31.69)
30
(40.30)
72
50
(60.30)
87
(76.69)
137
92
Not short
Total
117
209
Back to Chi-Square
O = observed frequency
E = expected frequency
2
O
E
O-E
2
(O - E) (O - E)
E
2
2
O
42
50
30
87
E
O-E
2
(O - E) (O - E)
E
2
2
O
E
42
31.69
50
60.30
30
40.30
87
76.69
O-E
2
(O - E) (O - E)
E
2
2
O
E
O-E
42
31.69
10.31
50
60.30
-10.30
30
40.30
-10.30
87
76.69
10.31
2
(O - E) (O - E)
E
2
2
2
O
E
O-E
42
31.69
10.31
(O - E) (O - E)
E
106.30
50
60.30
-10.30
106.09
30
40.30
-10.30
106.09
87
76.69
10.31
106.30
2
2
2
O
E
O-E
42
31.69
10.31
(O - E) (O - E)
E
106.30
3.35
50
60.30
-10.30
106.09
1.76
30
40.30
-10.30
106.09
2.63
87
76.69
10.31
106.30
1.39
2
2
2
O
E
O-E
42
31.69
10.31
(O - E) (O - E)
E
106.30
3.35
50
60.30
-10.30
106.09
1.76
30
40.30
-10.30
106.09
2.63
87
76.69
10.31
106.30
1.39
2 =
9.13
2
Significance
• Is a 2 of 9.13 significant at the .05 level?
• To find out you need to know df
Degrees of Freedom
• To determine the degrees of freedom you
use the number of rows (R) and the
number of columns (C)
• DF = (R - 1)(C - 1)
Degrees of Freedom
Rows = 2
Ever Bullied
Height
Yes
No
Total
Short
42
(31.69)
30
(40.30)
72
50
(60.30)
87
(76.69)
137
92
Not short
Total
117
209
Degrees of Freedom
Rows = 2
Columns = 2
Ever Bullied
Height
Yes
No
Total
Short
42
(31.69)
30
(40.30)
72
50
(60.30)
87
(76.69)
137
92
Not short
Total
117
209
Degrees of Freedom
• To determine the degrees of freedom you
use the number of rows (R) and the
number of columns (C)
• df = (R - 1)(C - 1)
• df = (2 - 1)(2 - 1) = 1
Significance
• Look on page 736
• df = 1
•  = .05
• 2critical = 3.84
Decision
• Thus, if 2 > than 2critical
– Reject H0, and accept H1
• If 2 < or = to 2critical
– Fail to reject H0
Current Example
• 2 = 9.13
• 2critical = 3.84
• Thus, reject H0, and accept H1
Current Example
• H1: There is a relationship between the the
two variables
– A persons size is significantly (alpha = .05)
related to if they were bullied