This document provides an overview of chi-square tests, including chi-square goodness of fit and chi-square test of independence. It uses examples from a hypothetical New York City mayoral election poll to demonstrate how to perform each test. The chi-square goodness of fit test determines if the distribution of proportions in a sample fits the expected distribution. The chi-square test of independence determines if there is a relationship between two categorical variables, like voter gender and candidate preference. Both tests use a chi-square calculation and degrees of freedom to obtain a p-value, and Cramér's V can estimate effect size.

1. Statistics One
Lecture 19
Chi-square tests

2. Two segments
• Chi-square goodness of fit
• Chi-square test of independence
2

3. Lecture 19 ~ Segment 1
Chi-square goodness of fit

4. Chi-square tests
• All of the analyses covered thus far in the
course have assumed that the outcome
variable is a normally distributed continuous
variable
• Interval variable
• Ratio variable
4

5. Chi-square tests
• What if the outcome variables is
categorical?
– For example, nominal variables
• Diagnosis (positive, negative)
• Verdict (guilty, innocent)
• Vote (candidate A, candidate B, candidate C)
5

6. Chi-square tests
• Chi-square goodness of fit statistic
• Chi-square test of independence
• Both can be used in either experimental or
correlational research
6

7. Chi-square tests
• Chi-square goodness of fit statistic
– Determines how well a distribution of
proportions “fits” an expected distribution
– In election polls, is there a statistically significant
difference in voter preference among candidates?
7

8. Chi-square tests
• Chi-square test of independence
– Determines whether there is a relationship
between two categorical variables
– In election polls, is there a relationship between
voter gender and preference among candidates?
8

9. Chi-square goodness of fit
• New York City mayoral election
– Assume a small poll was conducted (N=60)
– Do you intend to vote for:
• Christine Quinn
• Joseph Lhota
• Other
9

10. Chi-square goodness of fit
Quinn
Lhota
23
Other
12
25
10

11. Chi-square goodness of fit
• Null hypothesis
– Equal proportions
• Alternative hypothesis
– Unequal proportions
11

12. Chi-square goodness of fit
χ2 = Σ [(O - E)2 / E]
O = Observed
E = Expected
df = # of categories – 1
p-value depends on χ2 and df
12

13. Chi-square goodness of fit
13

14. Chi-square goodness of fit
To estimate effect size
Cramér’s V (or Phi)
Φc = SQRT(χ2 / N(k – 1))
N = sample size
k = # of categories
14

15. Chi-square goodness of fit
Quinn
Lhota
Other
20 (E)
20 (E)
20 (E)
23 (O)
12 (O)
25 (O)
15

16. Chi-square goodness of fit
χ2 = Σ [(O - E)2 / E]
df = # of categories – 1
16

17. Chi-square goodness of fit
O
E
(O – E)2
(O – E)
(O – E)2 / E
Quinn
23
20
3
9
0.45
Lhota
12
20
-8
64
3.20
Other
25
20
5
25
1.25
Total
60
60
0
98
4.90
17

18. Chi-square goodness of fit
χ2 = Σ [(O - E)2 / E]
χ2 = 4.90, df = 2
p = .09
∴ Retain the null hypothesis and conclude that the
slight preferences observed here are not statistically
18

19. Chi-square goodness of fit
To estimate effect size
Cramer’s V (or Phi)
Φc = SQRT(χ2 / N(k – 1))
Φc = SQRT(4.90 / 60(3 – 1)) = 0.20
19

20. Dataframe in R (Election)
Voter.ID
Candidate
Gender
1
Quinn
M
2
Quinn
F
3
Other
F
4
Lhota
M
5
Other
M
….
…
…
20

21. Chi-square goodness of fit in R
> Observed <-- table(Election$Candidate)
> chisq.test(Observed)
21

22. Chi-square goodness of fit in R
22

23. Segment summary
• Chi-square tests are used when outcome
and predictor variables are all categorical
• Chi-square goodness of fit is an NHST
• Cramér’s V estimates effect size
23

24. END SEGMENT

25. Lecture 19 ~ Segment 2
Chi-square test of independence

26. Chi-square test of independence
• Determines whether there is a relationship
between two categorical variables
– In election polls, is there a relationship between
voter gender and preference among candidates?
26

27. Chi-square test of independence
• New York City mayoral election
– Assume a small poll was conducted (N=200)
– More males than females (n = 140, n = 60)
– Do you intend to vote for:
• Christine Quinn
• Joseph Lhota
• Other
27

28. Chi-square test of independence
Female
Male
Quinn
40
90
Lhota
10
40
Other
10
10
28

29. Chi-square test of independence
• Null hypothesis
– There is no relationship between voter gender
and voter preference
• Alternative hypothesis
– There is a relationship between voter gender and
voter preference
29

30. Chi-square test of independence
χ2 = Σ [(O - E)2 / E]
df = (# of rows - 1) * (# of columns - 1)
p-value depends on χ2 and df
30

31. Chi-square test of independence
31

32. Chi-square test of independence
To estimate effect size
Cramér’s V (or Phi)
Φc = SQRT(χ2 / N(k – 1))
N = sample size
k = # of rows or # of categories (whichever is less)
32

33. Chi-square test of independence
• Compute the expected frequencies
– The proportion of male and female voters for
each candidate should be the same as the
overall voter preference rates
33

34. Chi-square test of independence
• Compute the expected frequencies
E = (R/N)*C
E: Expected frequency
R: # of entries in the cell’s row
N: total # of entries
C: # of entries in the cell’s column
34

35. Chi-square test of independence
Quinn
Female
Male
Sum (C)
Lhota
40
90
130
Other
10
40
50
Sum (R)
10
10
20
60
140
200
35

36. Chi-square test of independence
E = (R/N)*C
Quinn
Lhota
Other
Sum (R)
Female
(60/200)*130
39
(60/200)*50
15
(60/200)*20
6
60
Male
(140/200)*130
91
(140/200)*50
35
(140/200)*20
14
140
Sum (C)
130
50
20
200
36

37. Chi-square test of independence
O
E
(O – E)2
(O – E)
(O – E)2 / E
F / Quinn
40
39
1
1
F / Lhota
10
15
-­‐5
25
F / Other
10
6
4
16
M / Quinn
90
91
1
1
M / Lhota
40
35
5
25
M / Other
10
14
-4
16
200
200
0
84
Sum
0.03
1.67
2.67
0.01
0.71
1.14
6.23
37

38. Chi-square test of independence
χ2 = Σ [(O - E)2 / E]
χ2 = 6.23, df = 2
p = .04
∴ Reject the null hypothesis and conclude that the
there is a significant relationship between gender of
38
the defendant and verdict

39. Chi-square test of independence
To estimate effect size
Cramér’s V (or Phi)
Φc = SQRT(χ2 / N(k – 1))
Φc = SQRT(6.23 / 200(2 – 1)) = .18
39

40. Dataframe in R (Election)
Voter.ID
Candidate
Gender
1
Quinn
M
2
Quinn
F
3
Other
F
4
Lhota
M
5
Other
M
….
…
…
40

41. Chi-square test in R
> Observed = table(Election$Candidate, Election$Gender)
> chisq.test(Observed)
41

42. Chi-square test in R
42

43. Assumptions
• Adequate expected cell counts
– A common rule is 5 or more in all cells of a 2by-2 table, and 5 or more in 80% of cells in
larger tables, and no cells with zero.
– When this assumption is not met, Fisher’s exact
test, a non-parametric test, is recommended.
43

44. Assumptions
• Independence
– The observations are assumed to be independent of
each other.
– This means chi-squared cannot be used to test
correlated data (like matched pairs or panel data).
– In such cases McNemar’s test of dependent
proportions is recommended.
44

45. Segment summary
• Chi-square tests are used when outcome
and predictor variables are all categorical
• Chi-square test of independence is an NHST
• Cramér’s V estimates effect size
• Assumptions
– Adequate expected cell counts
– Independence
45

46. END SEGMENT

47. END LECTURE 19