2. 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?”

3. Results Marijuana ? Death Penalty ?

4. Step 1: State the Hypothesis H 1: There is a relationship between opinions about the death penalty and the legalization of marijuana H 0 :Opinions about the death penalty and the legalization of marijuana are independent of each other

5. Step 2: Create the Data Table Marijuana ? Death Penalty ?

6. Step 3: Find  2 critical df = (R - 1)(C - 1) df = (2 - 1)(2 - 1) = 1  =.01  2 critical = 6.64

7. Step 4: Calculate the Expected Frequencies Marijuana ? Death Penalty ?

8. Step 5: Calculate  2

9. 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

10. Step 6: Decision Thus, if  2 > than  2 critical –Reject H 0, and accept H 1 If  2 < or = to  2 criticalIf  2 < or = to  2 critical –Fail to reject H 0  2 = 3.91  2 crit = 6.64

11. Step 7: Put it answer into words H 0 :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

12. 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.

13. Phi Use with 2x2 tables N = sample size

14. 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?”

15. Results Marijuana ? Death Penalty ?

16. Step 6: Decision Thus, if  2 > than  2 critical –Reject H 0, and accept H 1 If  2 < or = to  2 criticalIf  2 < or = to  2 critical –Fail to reject H 0  2 = 3.91  2 crit = 6.64

17. Phi Use with 2x2 tables

18. Bullied Example Ever Bullied

19. 22

20. Phi Use with 2x2 tables

21. Practice –Page 170 #6.10 –How strong is the relationship?

22. Results English ADD X 2 = 5.38 X 2 crit = 3.83

23. Phi Use with 2x2 tables

26. Practice In the 1930’s 650 boys participated in the Cambridge-Somerville Youth Study. Half of the participants were randomly assigned to a delinquency-prevention pogrom and the other half to a control group. At the end of the study, police records were examined for evidence of delinquency. In the prevention program 114 boys had a police record and in the control group 101 boys had a police record. Analyze the data and write a conclusion.

27. Chi Square = 1.17 Chi Square observed = 3.84 Phi =.04 –Note the results go in the opposite direction that was expected!

29.  2 as a test for goodness of fit But what if: You have a theory or hypothesis that the frequencies should occur in a particular manner?

30. Example M&Ms claim that of their candies: 30% are brown 20% are red 20% are yellow 10% are blue 10% are orange 10% are green

31. Example Based on genetic theory you hypothesize that in the population: 45% have brown eyes 35% have blue eyes 20% have another eye color

32. To solve you use the same basic steps as before (slightly different order) 1) State the hypothesis 2) Find  2 critical 3) Create data table 4) Calculate the expected frequencies 5) Calculate  2 6) Decision 7) Put answer into words

33. Example M&Ms claim that of their candies: 30% are brown 20% are red 20% are yellow 10% are blue 10% are orange 10% are green

34. Example Four 1-pound bags of plain M&Ms are purchased Each M&Ms is counted and categorized according to its color Question: Is M&Ms “theory” about the colors of M&Ms correct?

36. Step 1: State the Hypothesis H 0 : The data do fit the model –i.e., the observed data does agree with M&M’s theory H 1: The data do not fit the model –i.e., the observed data does not agree with M&M’s theory –NOTE: These are backwards from what you have done before

37. Step 2: Find  2 critical df = number of categories - 1

38. Step 2: Find  2 critical df = number of categories - 1 df = 6 - 1 = 5  =.05  2 critical = 11.07

39. Step 3: Create the data table

40. Add the expected proportion of each category

41. Step 4: Calculate the Expected Frequencies

42. Expected Frequency = (proportion)(N)

43. Step 4: Calculate the Expected Frequencies Expected Frequency = (.30)(2081) = 624.30

44. Step 4: Calculate the Expected Frequencies Expected Frequency = (.20)(2081) = 416.20

45. Step 4: Calculate the Expected Frequencies Expected Frequency = (.20)(2081) = 416.20

46. Step 4: Calculate the Expected Frequencies Expected Frequency = (.10)(2081) = 208.10

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

48. 22

49. 22

50. 22

51. 22

52. 22

53. 22 15.52

54. 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

55. 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 = 15.52  2 crit = 11.07

56. Step 7: Put it answer into words H 1: The data do not fit the model M&M’s color “theory” did not significantly (.05) fit the data

57. Practice Among women in the general population under the age of 40: 60% are married 23% are single 4% are separated 12% are divorced 1% are widowed

58. Practice You sample 200 female executives under the age of 40 Question: Is marital status distributed the same way in the population of female executives as in the general population (  =.05)?

60. Step 1: State the Hypothesis H 0 : The data do fit the model –i.e., marital status is distributed the same way in the population of female executives as in the general population H 1: The data do not fit the model –i.e., marital status is not distributed the same way in the population of female executives as in the general population

61. Step 2: Find  2 critical df = number of categories - 1

62. Step 2: Find  2 critical df = number of categories - 1 df = 5 - 1 = 4  =.05  2 critical = 9.49

63. Step 3: Create the data table

64. Step 4: Calculate the Expected Frequencies

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

66. 22 19.42

67. 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

68. 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 = 19.42  2 crit = 9.49

69. Step 7: Put it answer into words H 1: The data do not fit the model Marital status is not distributed the same way in the population of female executives as in the general population (  =.05)

71. Practice In the past you have had a 20% success rate at getting someone to accept a date from you. What is the probability that at least 2 of the next 10 people you ask out will accept?

72. Practice p zero will accept =.11 p one will accept =.27 p zero OR one will accept =.38 p two or more will accept = 1 -.38 =.62

74. Practice IQ –Mean = 100 –SD = 15 What is the probability that the stranger you just bumped into on the street has an IQ between 95 and 110?

75. Step 1: Sketch out question -3  -2  -1   1  2  3  11095 ?

76. Step 2: Calculate Z scores for both values Z = (X -  ) /  Z = (95 - 100) / 15 = -.33 Z = (110 - 100) / 15 =.67

77. Step 3: Look up Z scores -3  -2  -1   1  2  3 .67-.33.1293.2486

78. Step 4: Add together the two values -3  -2  -1   1  2  3 .67-.33.3779

80. Practice A professor would like to determine if there has been a change in grading practices over the years. In the past, the overall grade distribution was 14% As, 26% Bs, 31% Cs, 19% Ds, and 10% Fs. A sample of 200 students this years had the following grades

81. Practice A = 32 B = 61 C = 64 D = 31 F = 12 Do the data indicate a significant change in the grade distribution? Test at the.05 level.

82. Step 1: State the Hypothesis H 0 : The data do fit the model –i.e., the grades are distributed the same H 1: The data do not fit the model –i.e., the grades are not distributed the same

83. Practice A = 3228 B = 6152 C = 6462 D = 3138 F = 1220 Chi square = 6.68 Critical Chi square (4) = 9.49

84. Step 6: Decision Thus, if  2 > than  2 critical –Reject H 0, and accept H 1 If  2 < or = to  2 criticalIf  2 < or = to  2 critical –Fail to reject H 0  2 = 6.68  2 crit = 9.49

85. Step 7 H 0 : The data do fit the model –i.e., the grades are distributed the same There is no evidence that the grades have changed

86. Practice An early hypothesis of schizophrenia was that it has a simple genetic cause. In accordance with the theory 25% of the offspring of a selected group of parents would be expected to be diagnosed as schizophrenic. Suppose that of 140 offspring, 19.3% were schizophrenic. Test this theory.

87. Goodness of fit chi-square Make sure you compute the Chi square with the frequencies. Chi square = 2.439 Observed = 3.84 These data are consistent with the theory!

89. Practice In 1693, Samuel Pepys asked Isaac Newton whether it is more likely to get at least one ace in 6 rolls of a die or at least two aces in 12 rolls of a die. This problems is known a Pepys' problem.

90. Binomial Distribution Aces p p =.67

91. Binomial Distribution Aces p p =.62

92. Practice In 1693, Samuel Pepys asked Isaac Newton whether it is more likely to get at least one ace in 6 rolls of a die or at least two aces in 12 rolls of a die. This problems is known a Pepys' problem. It is more likely to get at least one ace in 6 rolls of a die!

94. Practice Which is more likely: at least one ace with 4 throws of a fair die or at least one double ace in 24 throws of two fair dice? This is known as DeMere's problem, named after Chevalier De Mere. Blaise Pascal later solved this problem.

95. Binomial Distribution p =.482 of zero aces 1 -.482 =.518 at least one ace will occur

96. Binomial Distribution p =.508 of zero double aces 1 -.508 =.492 at least one double ace will occur

97. Practice Which is more likely: at least one ace with 4 throws of a fair die or at least one double ace in 24 throws of two fair dice? This is known as DeMere's problem, named after Chevalier De Mere. More likely at least one ace with 4 throws will occur