2. 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)

3. Results
Ever Bullied

4. Results
Ever Bullied

5. Results
Ever Bullied

6. Results
Ever Bullied

7. Results
Ever Bullied

8. Results
Ever Bullied

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

10. 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

11. 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

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

13. Results
Ever Bullied

14. Observed Frequencies
Ever Bullied

15. 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

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

17. Expected Frequencies
Ever Bullied

18. Expected Frequencies
Ever Bullied

19. Expected Frequencies
Row total = 92
Ever Bullied

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

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

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

23. Expected Frequencies
Ever Bullied

24. Expected Frequencies
Ever Bullied

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

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

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

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

29. 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

30. 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

31. 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

32. 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

33. 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

34. 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

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

36. 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

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

38. 2

39. 2

40. 2

41. 2

42. 2

43. 2

44. 2

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

46. 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)

47. Degrees of Freedom
Rows = 2
Ever Bullied

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

49. 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

50. Significance
Look on page 691
df = 1
 = .05
2critical = 3.84

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

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

53. 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

55. 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

56. 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?

57. Results

58. 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

59. Step 2: Create the Data Table

60. Step 2: Create the Data Table
Add “total” columns and rows

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

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

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

64. Step 4: Calculate the Expected Frequencies

65. Step 4: Calculate the Expected Frequencies

66. Step 4: Calculate the Expected Frequencies

67. Step 4: Calculate the Expected Frequencies

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

69. 2

70. 2

71. 2

72. 2

73. 2
8.5

74. Step 6: Decision Thus, if 2 > than 2critical
Reject H0, and accept H1
If 2 < or = to 2critical
Fail to reject H0

75. Step 6: Decision Thus, if 2 > than 2critical
2 = 8.5
2 crit = 5.99
Thus, if 2 > than 2critical
Reject H0, and accept H1
If 2 < or = to 2critical
Fail to reject H0

76. 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.

78. 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.

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

80. Bullied Example
Ever Bullied

81. 2

82. Phi
Use with 2x2 tables

83. SPSS

85. 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?

86. 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

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

88. 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

89. 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

90. 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?

92. Step 1: State the Hypothesis
H0: The data do fit the model
i.e., the observed data does agree with M&M’s theory
H1: 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

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

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

95. Step 3: Create the data table

96. Step 3: Create the data table
Add the expected proportion of each category

97. Step 4: Calculate the Expected Frequencies

98. Step 4: Calculate the Expected Frequencies
Expected Frequency = (proportion)(N)

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

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

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

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

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

104. 2

105. 2

106. 2

107. 2

108. 2

109. 2
15.52

110. Step 6: Decision Thus, if 2 > than 2critical
Reject H0, and accept H1
If 2 < or = to 2critical
Fail to reject H0

111. Step 6: Decision Thus, if 2 > than 2critical
2 = 15.52
2 crit = 11.07
Thus, if 2 > than 2critical
Reject H0, and accept H1
If 2 < or = to 2critical
Fail to reject H0

112. Step 7: Put it answer into words
H1: The data do not fit the model
M&M’s color “theory” did not significantly (.05) fit the data

113. 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

114. 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)?

116. Step 1: State the Hypothesis
H0: 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
H1: 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

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

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

119. Step 3: Create the data table

120. Step 4: Calculate the Expected Frequencies

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

122. 2
19.42

123. Step 6: Decision Thus, if 2 > than 2critical
Reject H0, and accept H1
If 2 < or = to 2critical
Fail to reject H0

124. Step 6: Decision Thus, if 2 > than 2critical
2 = 19.42
2 crit = 9.49
Thus, if 2 > than 2critical
Reject H0, and accept H1
If 2 < or = to 2critical
Fail to reject H0

125. Step 7: Put it answer into words
H1: 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)

127. 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?

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

132. 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.

133. Binomial Distribution
p = .67 p
Aces

134. Binomial Distribution
p = .62 p
Aces

135. 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!

137. 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.