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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
Ingredients:N = total number of eventsp = the probability of a success on any one trialq = (1 – p) = the probability of a failure on any one trialX = number of successful events
)(
)!(!
!)( XNXqp
XNX
NXp
Binomial Distribution
Ingredients:N = total number of eventsp = the probability of a success on any one trialq = (1 – p) = the probability of a failure on any one trialX = number of successful events
)810(8 502.498.)!810(!8
!10)(
Xp
Binomial Distribution
Ingredients:N = total number of eventsp = the probability of a success on any one trialq = (1 – p) = the probability of a failure on any one trialX = number of successful events
)810(8 502.498.)!810(!8
!10)(
Xp
p = .0429
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
• p = .498, N = 10Number of Wins p
0
1
2
3
4
5
6
7
8
9
10
Probability of Winning Blackjack
• p = .498, N = 10Number of Wins p
0 .001
1
2
3
4
5
6
7
8
9
10
Probability of Winning Blackjack
• p = .498, N = 10Number 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 = 10Number 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
0.05
0.1
0.15
0.2
0.25
0.3
0 1 2 3 4 5 6 7 8 9 10
Games Won
p
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
0.05
0.1
0.15
0.2
0.25
0.3
0 1 2 3 4 5 6 7 8 9 10
Games Won
p
Binomial Distribution
0
0.05
0.1
0.15
0.2
0.25
0.3
0 1 2 3 4 5 6 7 8 9 10
Games Won
p
Probability of Winning Blackjack
• p = .498, N = 10Number 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
77 .115.115
88 .044.044
99 .009.009
1010 .001.001
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 SlotNumber 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
00.050.1
0.150.2
0.250.3
0.350.4
0.450.5
0 1 2 3 4 5 6 7 8 9 10
Games Won
p
Probability of Winning Slot
• p of winning at least 4 games
• .005+.001+.000 . . . .000 = .006
• p< .05
• Winning at least 4 games is significantly better than chance
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
• 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
0.05
0.1
0.15
0.2
0.25
0.3
0 1 2 3 4 5 6 7 8 9 10
Games Won
p
Binomial Distribution
• Statistical Slot Machine; p = .08, N = 10
• M = .8
• SD = .86
Binomial Distribution
00.050.1
0.150.2
0.250.3
0.350.4
0.450.5
0 1 2 3 4 5 6 7 8 9 10
Games Won
p
Note: as N gets bigger, distributions will approach normal
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
Practice
• In the past you have had a 5% success rate at getting someone to accept a date from you.
• What is the probability that at least 1 of the next 10 people you ask out will accept?
• Note: N isn’t big enough in these problems to use the Z-score formula
Bullied as a child?
Are you tall or short?
6’ 4”
5’ 10”
4’
2’ 4”
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
Height Yes No
Short 42 50
Not short 30 87
Ever Bullied
Results
Height Yes No Total
Short 42 50 92
Not short 30 87 117
Total 72 137 209
Ever Bullied
Results
Height Yes No Total
Short 42 50 92
Not short 30 87 117
Total 34% 66% 209
Ever Bullied
Results
Height Yes No Total
Short 42 50 44%
Not short 30 87 56%
Total 72 137 209
Ever Bullied
Results
Height Yes No Total
Short 42 50 92
Not short 30 87 117
Total 72 137 209
Ever Bullied
Results
Height Yes No Total
Short 45% 55% 92
Not short 26% 74% 117
Total 72 137 209
Ever Bullied
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
Height Yes No Total
Short 42 50 92
Not short 30 87 117
Total 72 137 209
Ever Bullied
Observed Frequencies
Height Yes No Total
Short 42 50 92
Not short 30 87 117
Total 72 137 209
Ever Bullied
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 Yes No Total
Short 42 50 92
Not short 30 87 117
Total 72 137 209
Ever Bullied
Expected Frequencies
Height Yes No Total
Short 42 50 92
Not short 30 87 117
Total 72 137 209
Ever Bullied
Expected Frequencies
Height Yes No Total
Short 42 50 92
Not short 30 87 117
Total 72 137 209
Ever Bullied
Row total = 92
Expected Frequencies
Height Yes No Total
Short 42 50 92
Not short 30 87 117
Total 72 137 209
Ever Bullied
Row total = 92
Column total = 72
Expected Frequencies
Height Yes No Total
Short 42 50 92
Not short 30 87 117
Total 72 137 209
Ever Bullied
Row total = 92 N = 209
Column total = 72
Expected Frequencies
Height Yes No Total
Short 42 50 92
Not short 30 87 117
Total 72 137 209
Ever Bullied
E = (92 * 72) /209 = 31.69
Expected Frequencies
Height Yes No Total
Short 42(31.69)
50 92
Not short 30 87 117
Total 72 137 209
Ever Bullied
Expected Frequencies
Height Yes No Total
Short 42(31.69)
50 92
Not short 30 87 117
Total 72 137 209
Ever Bullied
Expected Frequencies
Height Yes No Total
Short 42(31.69)
50(60.30)
92
Not short 30 87 117
Total 72 137 209
Ever Bullied
E = (92 * 137) /209 = 60.30
Expected Frequencies
Height Yes No Total
Short 42(31.69)
50(60.30)
92
Not short 30(40.30)
87(76.69)
117
Total 72 137 209
Ever Bullied
E = (117 * 72) / 209 = 40.30
E = (117 * 137) / 209 = 76.69
Expected Frequencies
Height Yes No Total
Short 42(31.69)
50(60.30)
92
Not short 30(40.30)
87(76.69)
117
Total 72 137 209
Ever Bullied
The expected frequencies are what you would expect if there was no relationship between the two variables!
How do the expected frequencies work?
Height Yes No Total
Short 42(31.69)
50(60.30)
92
Not short 30(40.30)
87(76.69)
117
Total 72 137 209
Ever Bullied
Looking only at:
How do the expected frequencies work?
Height Yes No Total
Short 42(31.69)
50(60.30)
92
Not short 30(40.30)
87(76.69)
117
Total 72 137 209
Ever Bullied
If you randomly selected a person from these 209 people what is the probability you would select a person who is short?
How do the expected frequencies work?
Height Yes No Total
Short 42(31.69)
50(60.30)
92
Not short 30(40.30)
87(76.69)
117
Total 72 137 209
Ever Bullied
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
How do the expected frequencies work?
Height Yes No Total
Short 42(31.69)
50(60.30)
92
Not short 30(40.30)
87(76.69)
117
Total 72 137 209
Ever Bullied
If you randomly selected a person from these 209 people what is the probability you would select a person who was bullied?
How do the expected frequencies work?
Height Yes No Total
Short 42(31.69)
50(60.30)
92
Not short 30(40.30)
87(76.69)
117
Total 72 137 209
Ever Bullied
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
How do the expected frequencies work?
Height Yes No Total
Short 42(31.69)
50(60.30)
92
Not short 30(40.30)
87(76.69)
117
Total 72 137 209
Ever Bullied
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?
How do the expected frequencies work?
Height Yes No Total
Short 42(31.69)
50(60.30)
92
Not short 30(40.30)
87(76.69)
117
Total 72 137 209
Ever Bullied
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
How do the expected frequencies work?
Height Yes No Total
Short 42(31.69)
50(60.30)
92
Not short 30(40.30)
87(76.69)
117
Total 72 137 209
Ever Bullied
How many people do you expect to have been bullied and short?
How do the expected frequencies work?
Height Yes No Total
Short 42(31.69)
50(60.30)
92
Not short 30(40.30)
87(76.69)
117
Total 72 137 209
Ever Bullied
How many people would you expect to have been bullied and short? (.15 * 209) = 31.35 (difference due to rounding)
Back to Chi-Square
O = observed frequency
E = expected frequency
2
O E O - E (O - E)2 (O - E)2
E
2
O E O - E (O - E)2 (O - E)2
E42
50
30
87
2
O E O - E (O - E)2 (O - E)2
E42 31.69
50 60.30
30 40.30
87 76.69
2
O E O - E (O - E)2 (O - E)2
E42 31.69 10.31
50 60.30 -10.30
30 40.30 -10.30
87 76.69 10.31
2
O E O - E (O - E)2 (O - E)2
E42 31.69 10.31 106.30
50 60.30 -10.30 106.09
30 40.30 -10.30 106.09
87 76.69 10.31 106.30
2
O E O - E (O - E)2 (O - E)2
E42 31.69 10.31 106.30 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
O E O - E (O - E)2 (O - E)2
E42 31.69 10.31 106.30 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 1.392 = 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
Height Yes No Total
Short 42(31.69)
50(60.30)
92
Not short 30(40.30)
87(76.69)
117
Total 72 137 209
Ever Bullied
Rows = 2
Degrees of Freedom
Height Yes No Total
Short 42(31.69)
50(60.30)
92
Not short 30(40.30)
87(76.69)
117
Total 72 137 209
Ever Bullied
Rows = 2
Columns = 2
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
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
Opposite Sex Same Sex No Difference Total
Females 58 16 63 137
Males 15 13 40 68
Total 73 29 103 205
Add “total” columns and rows
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
Opposite Sex Same Sex No Difference Total
Females 58(48.78)
16 63 137
Males 15 13 40 68
Total 73 29 103 205
E = (73 * 137) /205 = 48.79
Step 4: Calculate the Expected Frequencies
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
E = (73 * 68) /205 = 24.21
Step 4: Calculate the Expected Frequencies
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
E = (29 * 137) /205 = 19.38
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 O - E (O - E)2 (O - E)2
E58 48.79
15 24.21
16 19.38
13 9.62
63 68.83
40 34.17
2
O E O - E (O - E)2 (O - E)2
E58 48.79 9.21 84.82 1.74
15 24.21
16 19.38
13 9.62
63 68.83
40 34.17
2
O E O - E (O - E)2 (O - E)2
E58 48.79 9.21 84.82 1.74
15 24.21 -9.21 84.82 3.50
16 19.38
13 9.62
63 68.83
40 34.17
2
O E O - E (O - E)2 (O - E)2
E58 48.79 9.21 84.82 1.74
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
O E O - E (O - E)2 (O - E)2
E58 48.79 9.21 84.82 1.74
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
8.5
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 Thus, if 22 > than > than 22criticalcritical
– Reject HReject H00, and accept H, and accept H11
• 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
• 6.15
• 6.16