3. 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

4. p of winning jackpot
Total number of ways to win / total number of possible outcomes

5. Total Number of Outcomes
Different 5 number combinations
Different Powerball outcomes
Thus, there are 1,906,884 * 42 = 80,089,128 ways the drawing can occur

6. Total Number of Ways to Win
Only one way to win –

8. 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?

9. Binomial Distribution
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

10. Binomial Distribution
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

11. Binomial Distribution
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
p = .0429

13. 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

14. Probability of Winning Blackjack
Number of Wins p 1 2 3 4 5 6 7 8 9 10
p = .498, N = 10

15. Probability of Winning Blackjack
Number of Wins p
.001 1 2 3 4 5 6 7 8 9 10
p = .498, N = 10

16. Probability of Winning Blackjack
Number of Wins p
.001 1
.010 2
.045 3
.119 4
.207 5
.246 6
.203 7
.115 8
.044 9
.009 10
p = .498, N = 10

17. Probability of Winning Blackjack
Number of Wins p
.001 1
.010 2
.045 3
.119 4
.207 5
.246 6
.203 7
.115 8
.044 9
.009 10
1.00
p = .498, N = 10

18. Binomial Distributionp
Games Won

19. 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?

20. Binomial Distributionp
Games Won

21. Binomial Distributionp
Games Won

22. Probability of Winning Blackjack
Number of Wins p
.001 1
.010 2
.045 3
.119 4
.207 5
.246 6
.203 7
.115 8
.044 9
.009 10
1.00
p = .498, N = 10

23. Probability of Winning Blackjack
Number of Wins p
.001 1
.010 2
.045 3
.119 4
.207 5
.246 6
.203 7
.115 8
.044 9
.009 10
1.00
p = .498, N = 10
p of winning 7 or more games
= .169
p > .05
Not better than chance

24. 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.

25. Probability of Winning Slot
Number of Wins p
.434 1
.378 2
.148 3
.034 4
.005 5
.001 6
.000 7 8 9 10
1.00

26. Binomial Distributionp
Games Won

27. Probability of Winning Slot
Number of Wins p
.434 1
.378 2
.148 3
.034 4
.005 5
.001 6
.000 7 8 9 10
1.00
p of winning at least 4 games
= .006
p< .05
Winning at least 4 games is significantly better than chance

28. Binomial Distribution
These distributions can be described with means and SD.
Mean = Np
SD =

29. Binomial Distribution
Black Jack; p = .498, N =10
M = 4.98
SD = 1.59

30. Binomial Distributionp
Games Won

31. Binomial Distribution
Statistical Slot Machine; p = .08, N = 10
M = .8
SD = .86

32. Binomial Distribution
Note: as N gets bigger, distributions will approach normal p
Games Won

33. 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?

34. 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

35. Distribution
Mean = 26.7
SD = 4.32
X = 39

36. Z-score

37. Results
(39 – 26.7) / 4.32 = 2.85
p = .0021
p < .05
.44 is significantly bigger than 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)

38. 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%

39. Blingoo M = 45 SD = 5.61 A) did better than you?
(57 – 45) / 5.61 = 2.14
p = .0162
2,642 * = 42.8 or 43 people

40. Blingoo M = 45 SD = 5.61 A) did worse than you?
(57 – 45) / 5.61 = 2.14
p = .9838
2,642 * = 2,599.2 or 2,599 people

41. 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 = ; p = .3577
= .8415
2,642 * = 2,223.2 or 2, 223 people

42. 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
(1.28) = games or 52 games

44. 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)

45. Results
Ever Bullied

46. Results
Ever Bullied

47. Results
Ever Bullied

48. Results
Ever Bullied

49. Results
Ever Bullied

50. Results
Ever Bullied

51. Is this difference in proportion due to chance?
To test this you use a Chi-Square (2)
Notice you are using nominal data

52. 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

53. 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

54. Chi-Square
O = observed frequency
E = expected frequency

55. Results
Ever Bullied

56. Observed Frequencies
Ever Bullied

57. 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

58. Expected frequencies To calculate a cells expected frequency:
For each cell you do this formula

59. Expected Frequencies
Ever Bullied

60. Expected Frequencies
Ever Bullied

61. Expected Frequencies
Row total = 92
Ever Bullied

62. Expected Frequencies
Row total = 92
Column total = 72
Ever Bullied

63. Expected Frequencies Ever Bullied Row total = 92 N = 209
Column total = 72
Ever Bullied

64. Expected Frequencies
E = (92 * 72) /209 = 31.69
Ever Bullied

65. Expected Frequencies
Ever Bullied

66. Expected Frequencies
Ever Bullied

67. Expected Frequencies
E = (92 * 137) /209 = 60.30
Ever Bullied

68. Expected Frequencies Ever Bullied E = (117 * 72) / 209 = 40.30

69. Expected Frequencies Ever Bullied
The expected frequencies are what you would expect if there was no relationship between the two variables!
Ever Bullied

70. How do the expected frequencies work?
Looking only at:
Ever Bullied

71. 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

72. 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

73. 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

74. 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

75. 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

76. 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

77. How do the expected frequencies work?
How many people do you expect to have been bullied and short?
Ever Bullied

78. How do the expected frequencies work?
How many people would you expect to have been bullied and short? (.15 * 209) = (difference due to rounding)
Ever Bullied

79. Back to Chi-Square
O = observed frequency
E = expected frequency

80. 2

81. 2

82. 2

83. 2

84. 2

85. 2

86. 2

87. Significance Is a 2 of 9.13 significant at the .05 level?
To find out you need to know df

88. 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)

89. Degrees of Freedom
Rows = 2
Ever Bullied

90. Degrees of Freedom
Rows = 2
Columns = 2
Ever Bullied

91. 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

92. Significance
Look on page 736
df = 1
 = .05
2critical = 3.84

93. Decision Thus, if 2 > than 2critical
Reject H0, and accept H1
If 2 < or = to 2critical
Fail to reject H0

94. Current Example 2 = 9.13 2critical = 3.84
Thus, reject H0, and accept H1

95. 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