Transcript 2 - Interpersonal Research Laboratory

Extra Brownie Points!
• Lottery
• To Win:
• choose the 5 winnings numbers
– from 1 to 49
• AND
• Choose the "Powerball" number
– from 1 to 42
• What is the probability you will win?
– Use combinations to answer this question
• p of winning jackpot
• Total number of ways to win / total number
of possible outcomes
Total Number of Outcomes
49!
 1,906,884
5!(49  5)!
42!
 42
1!(42  1)!
Different 5 number combinations
Different Powerball outcomes
Thus, there are 1,906,884 * 42 = 80,089,128 ways the drawing can
occur
Total Number of Ways to Win
Only one way to win –
1
 .000000012
80,089,128
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
2
(O - E) (O - E)
E
2
O
42
50
30
87
E
O-E
2
2
(O - E) (O - E)
E
2
O
E
42
31.69
50
60.30
30
40.30
87
76.69
O-E
2
2
(O - E) (O - E)
E
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
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
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 691
• 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
Seven Steps for Doing 2
•
•
•
•
•
•
•
1) State the hypothesis
2) Create data table
3) Find 2 critical
4) Calculate the expected frequencies
5) Calculate 2
6) Decision
7) Put answer into words
Example
• With whom do you find it easiest to
make friends?
• Subjects were either male and female.
• Possible responses were: “opposite
sex”, “same sex”, or “no difference”
• Is there a significant (.05) relationship
between the gender of the subject and
their response?
Results
Opposite Sex
Same Sex
No Difference
Females
58
16
63
Males
15
13
40
Step 1: State the Hypothesis
• H1: There is a relationship between gender
and with whom a person finds it easiest to
make friends
• H0:Gender and with whom a person finds
it easiest to make friends are independent
of each other
Step 2: Create the Data Table
Opposite Sex
Same Sex
No Difference
Females
58
16
63
Males
15
13
40
Step 2: Create the Data Table
Add “total” columns and rows
Opposite Sex
Same Sex
No Difference
Total
Females
58
16
63
137
Males
15
13
40
68
Total
73
29
103
205
Step 3: Find 2 critical
• df = (R - 1)(C - 1)
Step 3: Find 2 critical
• df = (R - 1)(C - 1)
• df = (2 - 1)(3 - 1) = 2
•  = .05
• 2 critical = 5.99
Step 4: Calculate the Expected
Frequencies
• Two steps:
• 4.1) Calculate values
• 4.2) Put values on your data table
Step 4: Calculate the Expected
Frequencies
E = (73 * 137) /205 = 48.79
Opposite Sex
Same Sex
No Difference
Total
58
(48.78)
16
63
137
Males
15
13
40
68
Total
73
29
103
205
Females
Step 4: Calculate the Expected
Frequencies
E = (73 * 68) /205 = 24.21
Opposite Sex
Same Sex
No Difference
Total
Females
58
(48.78)
16
63
137
Males
15
(24.21)
13
40
68
Total
73
29
103
205
Step 4: Calculate the Expected
Frequencies
E = (29 * 137) /205 = 19.38
Opposite Sex
Same Sex
No Difference
Total
Females
58
(48.78)
16
(19.38)
63
137
Males
15
(24.21)
13
40
68
Total
73
29
103
205
Step 4: Calculate the Expected
Frequencies
Opposite Sex
Same Sex
No Difference
Total
Females
58
(48.78)
16
(19.38)
63
(68.83)
137
Males
15
(24.21)
13
(9.62)
40
(34.17)
68
Total
73
29
103
205
Step 5: Calculate 2
O = observed frequency
E = expected frequency
2
O
E
58
48.79
15
24.21
16
19.38
13
9.62
63
68.83
40
34.17
O-E
(O - E)2 (O - E)2
E
2
O
E
O-E
58
48.79
9.21
15
24.21
16
19.38
13
9.62
63
68.83
40
34.17
(O - E)2 (O - E)2
E
84.82
1.74
2
O
E
O-E
58
48.79
9.21
15
24.21
-9.21
16
19.38
13
9.62
63
68.83
40
34.17
(O - E)2 (O - E)2
E
84.82
1.74
84.82
3.50
2
2
2
O
E
O-E
(O - E) (O - E)
E
84.82
1.74
58
48.79
9.21
15
24.21
-9.21
84.82
3.50
16
19.38
-3.38
11.42
.59
13
9.62
3.38
11.42
1.19
63
68.83
-5.83
33.99
.49
40
34.17
5.83
33.99
.99
2 8.5
2
2
O
E
O-E
(O - E) (O - E)
E
84.82
1.74
58
48.79
9.21
15
24.21
-9.21
84.82
3.50
16
19.38
-3.38
11.42
.59
13
9.62
3.38
11.42
1.19
63
68.83
-5.83
33.99
.49
40
34.17
5.83
33.99
.99
Step 6: Decision
• Thus, if 2 > than 2critical
– Reject H0, and accept H1
• If 2 < or = to 2critical
– Fail to reject H0
Step 6: Decision
• Thus, if 2 > than 2critical
– Reject H0, and accept H1
• If 2 < or = to 2critical
– Fail to reject H0
2 = 8.5
2 crit = 5.99
Step 7: Put it answer into words
• H1: There is a relationship between gender
and with whom a person finds it easiest to
make friends
• A persons gender is significantly (.05)
related with whom it is easiest to make
friends.
Practice
• Is there a significant ( = .01) relationship between
opinions about the death penalty and opinions
about the legalization of marijuana?
• 933 Subjects responded yes or no to:
• “Do you favor the death penalty for persons
convicted of murder?”
• “Do you think the use of marijuana should be made
legal?”
Results
Death
Penalty ?
Marijuana ?
Yes
No
Yes
152
561
No
61
159
Step 1: State the Hypothesis
• H1: There is a relationship between
opinions about the death penalty and the
legalization of marijuana
• H0:Opinions about the death penalty and
the legalization of marijuana are
independent of each other
Step 2: Create the Data Table
Death
Penalty ?
Marijuana ?
Yes
No
Total
Yes
152
561
713
No
61
159
220
Total
213
720
933
Step 3: Find 2 critical
• df = (R - 1)(C - 1)
• df = (2 - 1)(2 - 1) = 1
•  = .01
• 2 critical = 6.64
Step 4: Calculate the Expected
Frequencies
Death
Penalty ?
Marijuana ?
Yes
No
Total
Yes
No
Total
152
(162.77)
61
(50.23)
213
561
(550.23)
159
(169.78)
720
713
220
933
Step 5: Calculate 2
2
O-E
2
O
E
152
162.77
(O - E) (O - E)
E
-10.77 115.99
.71
61
50.23
10.77
115.99
2.31
561
550.23
10.77
115.99
.21
159
169.78
-10.78
115.99
.68
 =
3.91
2
Step 6: Decision
• Thus, if 2 > than 2critical
– Reject H0, and accept H1
• If 2 < or = to 2critical
– Fail to reject H0
Step 6: Decision
• Thus, if 2 > than 2critical
– Reject H0, and accept H1
• If 2 < or = to 2critical
– Fail to reject H0
2 = 3.91
2 crit = 6.64
Step 7: Put it answer into words
• H0:Opinions about the death penalty and
the legalization of marijuana are
independent of each other
• A persons opinion about the death penalty
is not significantly (p > .01) related with
their opinion about the legalization of
marijuana
Effect Size
• Chi-Square tests are null hypothesis tests
• Tells you nothing about the “size” of the
effect
• Phi (Ø)
– Can be interpreted as a correlation coefficient.
Phi
• Use with 2x2 tables

N = sample size

2
N
Practice
• Is there a significant ( = .01) relationship
between opinions about the death penalty
and opinions about the legalization of
marijuana?
• 933 Subjects responded yes or no to:
• “Do you favor the death penalty for persons
convicted of murder?”
• “Do you think the use of marijuana should be
made legal?”
Results
Death
Penalty ?
Marijuana ?
Yes
No
Yes
152
561
No
61
159
Step 6: Decision
• Thus, if 2 > than 2critical
– Reject H0, and accept H1
• If 2 < or = to 2critical
– Fail to reject H0
2 = 3.91
2 crit = 6.64
Phi
• Use with 2x2 tables
3.91

 .06
933
Bullied Example
Height
Ever Bullied
Yes
No
Short
42
50
92
Not short
30
87
117
Total
72
137
209
Total
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
Phi
• Use with 2x2 tables
9.13

 .21
209