1. Chapter 14
Chi Square 2
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Chi square

2. Chi Square
Chi Square is a non-parametric statistic used to test the null hypothesis.
It is used for nominal data.
It is equivalent to the F test that we used for single factor and factorial analysis.
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3. … Chi Square
Nominal data puts each participant in a category. Categories are best when mutually exclusive and exhaustive. This means that each and every participant fits in one and only one category
Chi Square looks at frequencies in the categories.
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4. Expected frequencies and the null hypothesis ...
Chi Square compares the expected frequencies in categories to the observed frequencies in categories.
“Expected frequencies”are the frequencies in each cell predicted by the null hypothesis
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5. … Expected frequencies and the null hypothesis ...
H0: fo = fe
There is no difference between the observed frequency and the frequency predicted (expected) by the null.
The experimental hypothesis:
H1: fo  fe
The observed frequency differs significantly from the frequency predicted (expected) by the null.
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6. Calculating 2 For each cell:
Calculate the deviations of the observed from the expected.
Square the deviations.
Divide the squared deviations by the expected value.
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7. Calculating 2 Add ‘em up. Then, look up 2 in Chi Square Table
df = k - 1 (one sample 2)
OR df= (Columns-1) * (Rows-1)
(2 or more samples)
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8. Critical values of 2 df df df df

9. Critical values of 2 df df df df
Degrees of
freedom

10.  = .05 Critical values of 2 df 1 2 3 4 5 6 7 8 df df df
Critical values
 = .05

11.  = .01 Critical values of 2 df 1 2 3 4 5 6 7 8 df df df
Critical values
 = .01

12. Example If there were 5 degrees of freedom, how big would 2
have to be for significance at the .05 level?
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13. Critical values of 2 df df df df

14. Another example If there were 2 degrees of freedom, how big would 2
have to be for significance at the .05 level?
Note: Unlike most other tables you have seen, the critical
values for Chi Square get larger as df increase. This is
because you are summing over more cells, each of which
usually contributes to the total observed value of chi square.
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15. Critical values of 2 df df df df

16. One sample example from the cpe: Party: 75% male, 25% female There are 40 swimmers. Since 75% of people at party are male, 75% of swimmers should be male. So expected value for males is .750 X 40 = 30. For women it is .250 x 40 = 10.00
Observed 20
Expected 30 10
O-E
-10 10
(O-E)2
100
(O-E)2/E
3.33 10
Male
Female
2 = 13.33
df = k-1 = 2-1 = 1
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17. 2 (1, n=40)= 13.33 Critical values of 2 df 1 2 3 4 5 6 7 8 df df df
Exceeds critical value at  = .01
Reject the null hypothesis.
Gender does affect who goes
swimming.
Women go swimming
more than expected.
Men go swimming
less than expected.

18. Freshman and sophomores who like horror movies.
2 sample example
Freshman and sophomores who like horror movies.
Freshmen
Sophomores
Likes horror films
150 50
Dislikes horror films
100
200
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19. … CPE 15.2.1 Freshman and sophomores and horror movies.
There are 500 altogether. 200 (or a proportion of .400 are freshmen, 300 (.600) are sophmores. (Proportions appear in parentheses in the margins.) Multiplying by row totals yield the following expected frequency for the first cell. (This time we use the formula: (Proprowncol)=Expected Frequency).
(EF appears in parentheses in each cell.)
Freshmen
Sophomores
Likes horror films
150
(100)
50 (100)
200
(.400)
Dislikes horror films
(150)
200 (150)
300
(.600)
250
250
500
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20. Computing 2 2 = 83.33 Observed 150 100 50 200 Expected 100 150 O-E
-50
(O-E)2
2500
(O-E)2/E
25.00
16.67
Fresh Likes
Fresh Dislikes
Soph Likes
Soph Dislikes
2 = 83.33
df = (C-1)(R-1) = (2-1)(2-1) = 1
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21. 2 (1, n=500)= 83.33 Critical values of 2 df 1 2 3 4 5 6 7 8 df df df
Critical at  = .01
Reject the null hypothesis.
Fresh/Soph dimension does affect
liking for horror movies.
Proportionally, more freshman
than sophomores
like horror movies

22. The only (slightly) hard part is computing expected frequencies
In one sample case, multiply n by hypothetical proportion based on random model.
Random model says that proportion in population in each category should be same as in the sample.
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23. Simple Example - 100 teenagers listen to radio stations
H1: Some stations are more popular with teenagers than others.
H0: Radio station do not differ in popularity with teenagers.
NOTE: YOU ALWAYS TEST H0
Expected frequencies are the frequencies predicted by the null hypothesis. In this case, the problem is simple because the null predicts an equal proportion of teenagers will prefer each of the four radio stations.
Station Station Station Station
A B C D 25
Expected Values 40 10 20 30
Observed Values
Is the observed
significantly different
from the expected?
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24. 2 = 20.00 2(3, n=100) = 20.00, p<.01 (O-E)2 O-E (O-E)2/E Observed
Expected 40 30 20 10 25 15 5 -5
225 25
9.00
1.00
Closeness to final exam
Category 1
Station 2
Station 3
Station 4
2 = 20.00
df = k-1 = (4-1) = 3
2(3, n=100) = 20.00, p<.01

25. Example - Admissions to Psychiatric Hospitals Close to a once/year final
H1: More students are admitted to psychiatric
hospitals when it is near their final exam.
H0: Time from final exam does not have an
effect on hospital admissions. .
Category 1: Within 7 days of final. (11 admitted)
Category 2: Between 8 and 30 days. (24 admitted)
Category 3: Between 31 and 90 days. (69 admitted)
Category 4: More than 90 days. (96 admitted)
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26. Psychiatric Admissions
Expected frequency=expected proportion of days*n
There are 365 days and 1 final and 200 patients admitted each year.
Proportion of each kind of day computed below:
Number of days
Category 1 (within 7):
Category 2 (8-30):
Category 3 (31-90):
Category 4 (rest of year):
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27. Expected Frequencies
To obtain expected frequencies with 200 admissions: multiply proportion of days of each type by n=200. This time the proportions are not equal.
Days: Category 1 (within 7):
Category 2 (8-30):
Category 3 (31-90 ):
Category 4 (rest of year):
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28. 2 = 1.57 2(3, n=200) = 1.57, n.s. (O-E)2 O-E (O-E)2/E Observed
Expected 11 24 69 96 8 26 66
100 3 -2 -4 9 4 16
1.12
0.15
0.14
0.16
Closeness to final exam
Category 1
Category 2
Category 3
Category 4
2 = 1.57
df = k-1 = (4-1) = 3
2(3, n=200) = 1.57, n.s.

29. The only (slightly)hard part is computing expected frequencies
In the multi-sample case, multiply the proportion in each row by n in each column to obtain EF in each cell.
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30. Vit C and flu study Sixty randomly chosen participants.
Thirty get Vitamin C.
Of that 30, 10 get the flu, 20 do not
Thirty get placebo
Of that 30, 15 get the flu, 15 do not
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31. Expected frequency = proportionROW nCOL
got flu no flu row n (prop.)
Vit C (.500)
No Vit C (.500)
Col. Totals n=60
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32. Expected frequencies 10 15 20 (12.50) (17.50) Vitamin C Placebo
Multiply the proportion in each row times the number in each column. Here Vitamin C row has 30 research participants. Total n = 60.
So proportion in that row =30/60=.500. Same for placebo group.
Number in each column: Twenty-five got influenza. So (25 X .500=12.50 should come from the Vitamin C group. Same for placebo.
Thirty five did not get influenza, so 35X.500 = 17.5 of each group should not have gotten the flu.
Vitamin C
Placebo
Had Influenza.
No influenza. 10 15 20
(Expected)
Values
(12.50)
(17.50)
Observed
Values
Are the observed
significantly different
from the expected?
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33. Computing 2 2 = 1.72 Observed 10 20 15 Expected 12.50 17.50 O-E
-2.50
2.50
(O-E)2
6.25
(O-E)2/E
.50
.36
VitC-got flu
VitC-no flu
Placebo-got flu
Placebo-no flu
2 = 1.72
df = (C-1)(R-1) = (2-1)(2-1) = 1
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34. Differences are not significant
2 (1, n=60) = 1.72, n.s.
Vit C consumption not significantly related to getting the flu in this study.
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35. A 3 x 4 Chi Square Women, stress, and seating preferences.
(and perimeter vs. interior, front vs. back
Front Front Back Back
Perim Inter Perim Inter
Very Stressed Females
Moderately Stressed Females
Control Group Females 10 70 5 15
100 15 50 10 25
100 35 30 15 20
100 30 60
n=300 60
150
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36. Proportion in each row
nROW/n=100/300=.333
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37. Expected frequencies Women, stress, and perimeter versus interior
seating preferences.
Front Front Back Back
Perim Inter Perim Inter
Very Stressed Females
Moderately Stressed Females
Control Group Females 10
(20) 70 5 15
100 15
(20) 50 10 25
100 35
(20) 30 15 20
100 30 60
300 60
150
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38. Column 2 Women, stress, and perimeter versus interior
seating preferences.
Front Front Back Back
Perim Inter Perim Inter
Very Stressed Females
Moderately Stressed Females
Control Group Females 10
(20) 70
(50) 5 15
100 15
(20) 50
(50) 10 25
100 35
(20) 30
(50) 15 20
100 30 60
300 60
150
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39. Column 3 Women, stress, and perimeter versus interior
seating preferences.
Front Front Back Back
Perim Inter Perim Inter
Very Stressed Females
Moderately Stressed Females
Control Group Females 10
(20) 70
(50) 5
(10) 15
100 15
(20) 50
(50) 10
(10) 25
100 35
(20) 30
(50) 15
(10) 20
100 30 60
300 60
150
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40. All the expected frequencies
Women, stress, and perimeter versus interior
seating preferences.
Front Front Back Back
Perim Inter Perim Inter
Very Stressed Females
Moderately Stressed Females
Control Group Females 10
(20) 70
(50) 5
(10) 15
(20)
100 15
(20) 50
(50) 10
(10) 25
(20)
100 35
(20) 30
(50) 15
(10) 20
(20)
100 30 60
300 60
150
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41. 2 = 41.00 Observed 10 70 5 15 Expected 20 50 10 O-E -10 20 -5 (O-E)2
100
400 25
(O-E)2/E
5.00
8.00
2.50
1.25
Very Stressed
FrontP
FrontI
BackP
BackI 15 50 10 25 20 50 10 -5 5 25
1.25
0.00
Moderately Stressed
FrontP
FrontI
BackP
BackI 35 30 15 20 20 50 10 15
-20 5
225
400 25
11.25
8.00
2.50
0.00
Control Group
FrontP
FrontI
BackP
BackI
2 = 41.00
df = (C-1)(R-1) = (4-1)(3-1) = 6

42. 2 (6, N=300)= 41.00 Critical values of 2 df 1 2 3 4 5 6 7 8 df df df
Critical at  = .01
Reject the null hypothesis.
There is an effect
between stressed women and
seating position.

43. 2 = 41.00 Observed 10 70 5 15 Expected 20 50 10 O-E -10 20 -5 (O-E)2
100
400 25
(O-E)2/E
5.00
8.00
2.50
1.25
Very Stressed
FrontP
FrontI
BackP
BackI 15 50 10 25 20 50 10 -5 5 25
1.25
0.00
Moderately Stressed
FrontP
FrontI
BackP
BackI
Very stressed women avoid
the perimeter and
prefer the front interior.
The control group prefers
the perimeter and avoids
the front interior. 35 30 15 20 20 50 10 15
-20 5
225
400 25
11.25
8.00
2.50
0.00
Control Group
FrontP
FrontI
BackP
BackI
2 = 41.00
df = (C-1)(R-1) = (4-1)(3-1) = 6

44. One sample Multi-sample
Summary: Different Ways of Computing the Frequencies Predicted by the Null Hypothesis
One sample
Expect subjects to be distributed equally in each cell. OR
Expect subjects to be distributed proportionally in each cell. OR
Expect subjects to be distributed in each cell based on prior knowledge, such as, previous research.
Multi-sample
Expect subjects in different conditions to be distributed similarly to each other. Find the proportion in each row and multiply by the number in each column to do so.
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45. Conclusion - Chi Square
Chi Square is a non-parametric statistic,used for nominal data.
It is equivalent to the F test that we used for single factor and factorial analysis.
Chi Square compares the expected frequencies in categories to the observed frequencies in categories.
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Chi square

46. … Conclusion - Chi Square
The null hypothesis:
H0: fo = fe
There is no difference between the observed frequency and frequency predicted by the null hypothesis.
The experimental hypothesis:
H1: fo  fe
The observed frequency differs significantly from the frequency expected by the null hypothesis.
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Chi square

47. The end.
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