3. 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?

4. 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

5. 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

6. 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

8. 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

9. Probability of Winning Blackjack p =.498, N = 10 Number of Winsp 0 1 2 3 4 5 6 7 8 9 10

10. Probability of Winning Blackjack p =.498, N = 10 Number of Winsp 0.001 1 2 3 4 5 6 7 8 9 10

11. Probability of Winning Blackjack p =.498, N = 10 Number of Winsp 0.001 1.010 2.045 3.119 4.207 5.246 6.203 7.115 8.044 9.009 10.001

12. Probability of Winning Blackjack p =.498, N = 10 Number of Winsp 0.001 1.010 2.045 3.119 4.207 5.246 6.203 7.115 8.044 9.009 10.001 1.00

13. Binomial Distribution Games Won p

14. Hypothesis Testing You wonder if winning at least 7 games of blackjack is significantly (.05) better than what would be expected due to chance. H 1 = Games won > 6 H 0 = Games won < or equal to 6 What is the probability of winning 7 or more games?

15. Binomial Distribution Games Won p

16. Binomial Distribution Games Won p

17. Probability of Winning Blackjack p =.498, N = 10 Number of Winsp 0.001 1.010 2.045 3.119 4.207 5.246 6.203 7.115 8.044 9.009 10.001 1.00

18. 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 Winsp 0.001 1.010 2.045 3.119 4.207 5.246 6.203 7.115 8.044 9.009 10.001 1.00

19. 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.

20. Probability of Winning Slot Number of Winsp 0.434 1.378 2.148 3.034 4.005 5.001 6.000 7 8 9 10.000 1.00

21. Binomial Distribution Games Won p

22. 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 Winsp 0.434 1.378 2.148 3.034 4.005 5.001 6.000 7 8 9 10.000 1.00

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

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

25. Binomial Distribution Games Won p

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

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

28. 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?

29. Hypothesis H 1 =.44 >.30 H 0 =.44 < or equal to.30 Or H 1 = 39 wins > 26.7 wins H 0 = 39 wins < or equal to 26.7 wins

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

31. Z-score

32. 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)

33. 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%

34. 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

35. 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

36. 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

37. 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

38. 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

40. Bullied as a child?

41. Are you tall or short? 6’ 4” 5’ 10” 4’ 2’ 4”

42. 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)

43. Results Ever Bullied

44. Results Ever Bullied

45. Results Ever Bullied

46. Results Ever Bullied

47. Results Ever Bullied

48. Results Ever Bullied

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

50. Hypothesis H 1: There is a relationship between the two variables –i.e., a persons size is related to if they were bullied H 0 :The two variables are independent of each other –i.e., there is no relationship between a persons size and if they were bullied

51. 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

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

53. Results Ever Bullied

54. Observed Frequencies Ever Bullied

55. 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

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

57. Expected Frequencies Ever Bullied

58. Expected Frequencies Ever Bullied

59. Expected Frequencies Ever Bullied Row total = 92

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

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

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

63. Expected Frequencies Ever Bullied

64. Expected Frequencies Ever Bullied

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

66. Expected Frequencies Ever Bullied E = (117 * 72) / 209 = 40.30 E = (117 * 137) / 209 = 76.69

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

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

69. How do the expected frequencies work? Ever Bullied If you randomly selected a person from these 209 people what is the probability you would select a person who is short?

70. How do the expected frequencies work? 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

71. How do the expected frequencies work? 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. How do the expected frequencies work? 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

73. How do the expected frequencies work? 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?

74. How do the expected frequencies work? 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

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

76. How do the expected frequencies work? Ever Bullied How many people would you expect to have been bullied and short? (.15 * 209) = 31.35 (difference due to rounding)

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

78. 22

79. 22

80. 22

81. 22

82. 22

83. 22

84. 22

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

86. 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)

87. Degrees of Freedom Ever Bullied Rows = 2

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

89. 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

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

91. Decision Thus, if  2 > than  2 critical –Reject H 0, and accept H 1 If  2 < or = to  2 critical –Fail to reject H 0

92. Current Example  2 = 9.13  2 critical = 3.84 Thus, reject H 0, and accept H 1

93. Current Example H 1: There is a relationship between the the two variables –A persons size is significantly (alpha =.05) related to if they were bullied

95. 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

96. 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?

97. Results

98. Step 1: State the Hypothesis H 1: There is a relationship between gender and with whom a person finds it easiest to make friends H 0 :Gender and with whom a person finds it easiest to make friends are independent of each other

99. Step 2: Create the Data Table

100. Add “total” columns and rows

101. Step 3: Find  2 critical df = (R - 1)(C - 1)

102. Step 3: Find  2 critical df = (R - 1)(C - 1) df = (2 - 1)(3 - 1) = 2  =.05  2 critical = 5.99

103. Step 4: Calculate the Expected Frequencies Two steps: 4.1) Calculate values 4.2) Put values on your data table

104. Step 4: Calculate the Expected Frequencies E = (73 * 137) /205 = 48.79

105. Step 4: Calculate the Expected Frequencies E = (73 * 68) /205 = 24.21

106. Step 4: Calculate the Expected Frequencies E = (29 * 137) /205 = 19.38

107. Step 4: Calculate the Expected Frequencies

108. Step 5: Calculate  2 O = observed frequency E = expected frequency

109. 22

110. 22

111. 22

112. 22

113. 22 8.5

114. Step 6: Decision Thus, if  2 > than  2 critical –Reject H 0, and accept H 1 If  2 < or = to  2 critical –Fail to reject H 0

115. Step 6: Decision Thus, if  2 > than  2 criticalThus, if  2 > than  2 critical –Reject H 0, and accept H 1 If  2 < or = to  2 critical –Fail to reject H 0  2 = 8.5  2 crit = 5.99

116. Step 7: Put it answer into words H 1: 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.

117. Practice 6.15 6.16