1. The Analysis of Categorical Data and Goodness of Fit Tests
Chapter 12
The Analysis of Categorical Data and Goodness of Fit Tests

2. We could record the color of each candy in the bag.
Suppose we wanted to determine if the proportions for the different colors in a large bag of M&M candies matches the proportions that the company claims is in their candies.
k is used to denote the number of categories for a categorical variable
We could record the color of each candy in the bag.
Create a jar with different types of coins . . .
There are six colors – so k = 6.
This would be univariate, categorical data.
How many categories for color would there be?

3. We could count how many candies of each color are in the bag.
M&M Candies Continued . . .
We could count how many candies of each color are in the bag.
Red
Blue
Green
Yellow
Orange
Brown 23 28 21 19 22 25
A goodness-of-fit test will allow us to determine if these observed counts are consistent with what we expect to have.
A one-way frequency table is used to display the observed counts for the k categories.
Create a jar with different types of coins . . .

4. Goodness-of-Fit Test Procedure
Null Hypothesis:
H0: p1 = hypothesized proportion for Category 1
pk = hypothesized proportion for Category k
Ha: H0 is not true
Test Statistic:
The goodness-of-fit statistic, denoted by X2, is a quantitative measure to the extent to which the observed counts differ from those expected when H0 is true.
The goodness-of-fit test is used to analysze univariate categorical data from a single sample.
. . .
Read “chi-squared”
The X2 value can never be negative.

5. Goodness-of-Fit Test Procedure Continued . . .
P-values: When H0 is true and all expected counts are at least 5, X2 has approximately a chi-square distribution with df = k – 1. Therefore, the P-value associated with the computed test statistic value is the area to the right of X2 under the df = k – 1 chi-square curve.
Assumptions:
Observed cell counts are based on a random sample
The sample size is large enough as long as every expected cell count is at least 5

6. Facts About c2 distributions
Different df have different curves
c2 curves are skewed right
As df increases, the c2 curve shifts toward the right and becomes more like a normal curve
df=3
df=5
df=10

7. There are eight phases so k = 8.
A common urban legend is that more babies than expected are born during certain phases of the lunar cycle, especially near the full moon.
The table below shows the number of days in the eight lunar phases with the number of births in each phase for 24 lunar cycles.
There are eight phases so k = 8.
Lunar Phase
Number of Days
Number of Births
New Moon 24
7680
Waxing Crescent
152
48,442
First Quarter
7579
Waxing Gibbous
149
47,814
Full Moon
7711
Waning Gibbous
150
47,595
Last Quarter
7733
Waning Crescent
48,230

8. The hypothesis statements would be:
Lunar Phases Continued . . .
Let:
p1 = proportion of births that occur during the new moon
p2 = proportion of births that occur during the waxing crescent moon
p3 = proportion of births that occur during the first quarter moon
p4 = proportion of births that occur during the waxing gibbous moon
p5 = proportion of births that occur during the full moon
p6 = proportion of births that occur during the waning gibbous moon
p7 = proportion of births that occur during the last quarter moon
p8 = proportion of births that occur during the waning crescent moon
There is a total of 699 days in the 24 lunar cycles. If there is no relationship between the number of births and lunar phase, then the expected proportions equal the number of days in each phase out of the total number of days.
The hypothesis statements would be:
H0: p1 = .0343, p2 = .2175, p3 = .0343, p4 = .2132, p5 = .0343, p6 = .2146, p7 = .0343, p8 = .2175
Ha: H0 is not true
p1 = p2 = p3 = p4 = .2132
P5 = p6 = p7 = p8 = .2175

9. Lunar Phases Continued . . .
H0: p1 = .0343, p2 = .2175, p3 = .0343, p4 = .2132, p5 = .0343, p6 = .2146, p7 = .0343, p8 = .2175
Ha: H0 is not true
Lunar Phase
Observed Number of Births
Expected Number of Births
New Moon
7680
Waxing Crescent
48,442
First Quarter
7579
Waxing Gibbous
47,814
47,497.55
Full Moon
7711
Waning Gibbous
47,595
Last Quarter
7733
Waning Crescent
48,230
48,455.52
There is a total of 222,784 births in the sample. If there is no relationship between the number of births and lunar phase, then the expected counts for each category would equal n(hypothesized proportion).

10. Lunar Phases Continued . . .
H0: p1 = .0343, p2 = .2175, p3 = .0343, p4 = .2132, p5 = .0343, p6 = .2146, p7 = .0343, p8 = .2175
Ha: H0 is not true
What type of error could we have potentially made with this decision?
Type II
Test Statistic:
P-value > .10 df = 7 a = .05
Since the P-value > a, we fail to reject H0. There is not sufficient evidence to conclude that lunar phases and number of births are related.
The X2 test statistic is smaller than the smallest entry in the df = 7 column of Appendix Table 8.

11. These values in green are the observed counts.
A study was conducted to determine if collegiate soccer players had in increased risk of concussions over other athletes or students. The two-way frequency table below displays the number of previous concussions for students in independently selected random samples of 91 soccer players, 96 non-soccer athletes, and 53 non-athletes.
If there were no difference between these 3 populations in regards to the number of concussions, how many soccer players would you expect to have no concussions?
We would expect (158/240)(91).
These values in green are the observed counts.
Also called a contingency table.
Number of Concussions 1 2
3 or more
Total
Soccer Players 45 25 11 10 91
Non-Soccer Players 68 15 8 5 96
Non-Athletes 3 53
158 22
240
This is univariate categorical data - number of concussions - from 3 independent samples.
These values in blue are the marginal totals.
This value in red is the grand total.

12. X2 Test for Homogeneity Null Hypothesis:
H0: the true category proportions are the same for all the populations or treatments
Alternative Hypothesis:
Ha: the true category proportions are not all the same for all the populations or treatments
Test Statistic:
The c2 Test for Homogeneity is used to analyze univariate categorical data from 2 or more independent samples.

13. X2 Test for Homogeneity Continued . . .
Expected Counts: (assuming H0 is true)
P-value: When H0 is true and all expected counts are at least 5, X2 has approximately a chi-square distribution with df = (number of rows – 1)(number of columns – 1). The P-value associated with the computed test statistic value is the area to the right of X2 under the appropriate chi-square curve.

14. X2 Test for Homogeneity Continued . . .
Assumptions:
Data are from independently chosen random samples or from subjects who were assigned at random to treatment groups.
The sample size is large: all expected cell counts are at least 5. If some expected counts are less than 5, rows or columns of the table may be combined to achieve a table with satisfactory expected counts.

15. df = (number of rows – 1)(number of columns – 1)
Soccer Players Continued . . .
State the hypotheses.
Number of Concussions 1 2
3 or more
Total
Soccer Players 45 25 11 10 91
Non-Soccer Players 68 15 8 5 96
Non-Athletes 3 53
158 22
240
H0: Proportions in each response category (number of concussions) are the same for all three groups
Ha: Category proportions are not all the same for all three groups
Df = (2)(3) = 6
To find df count the number of rows and columns – not including the totals!
df = (number of rows – 1)(number of columns – 1)
Another way to find df – you can also cover one row and one column, then count the number of cells left (not including totals)

16. Soccer Players Continued . . .
Number of Concussions 1
2 or more
Total
Soccer Players
45 (59.9)
25 (17.1)
21 (14.0) 91
Non-Soccer Players
68 (63.2)
15 (18.0)
13 (14.8) 96
Non-Athletes
45 (34.9)
5 (10.0)
3 (8.2) 53
158 45 22
240
Number of Concussions 1 2
3 or more
Total
Soccer Players
45 (59.9)
25 (17.1)
11 (8.3
10 (5.7) 91
Non-Soccer Players
68 (63.2)
15 (18.0)
8 (8.8)
5 (6.0) 96
Non-Athletes
45 (34.9)
5 (10.0)
3 (4.9)
0 (3.3) 53
158 45 22 15
240
df = 4
Test Statistic:
Notice that NOT all the expected counts are at least 5.
So combine the column for 2 concussions and the column for 3 or more concussions.
This combined table has a df = (2)(2) = 4.
Expected counts are shown in the parentheses next to the observed counts.
P-value < a = .05

17. These cells had the largest contributions to the X2 test statistic.
Soccer Players Continued . . .
Number of Concussions 1
2 or more
Total
Soccer Players
45 (59.9)
25 (17.1)
21 (14.0) 91
Non-Soccer Players
68 (63.2)
15 (18.0)
13 (14.8) 96
Non-Athletes
45 (34.9)
5 (10.0)
3 (8.2) 53
158 45 22
240
Since the P-value < a, we reject H0. There is strong evidence to suggest that the category proportions for the number of concussions is not the same for the groups.
We can look at the chi-square contributions – which of the cells above have the greatest contributions to the value of the X2 statistic?
These cells had the largest contributions to the X2 test statistic.
Is that all I can say – that there is a difference in proportions for the groups?

18. X2 Test for Independence
Null Hypothesis:
H0: The two variables are independent
Alternative Hypothesis:
Ha: The two variables are not independent
Test Statistic:
The c2 Test for Independence is used to analyze bivariate categorical data from a single sample.

19. X2 Test for Independence Continued . . .
Expected Counts: (assuming H0 is true)
P-value: When H0 is true and assumptions for X2 test are satisfied, X2 has approximately a chi-square distribution with df = (number of rows – 1)(number of columns – 1). The P-value associated with the computed test statistic value is the area to the right of X2 under the appropriate chi-square curve.

20. X2 Test for Independence Continued . . .
Assumptions:
The observed counts are based on data from a random sample.
The sample size is large: all expected cell counts are at least 5. If some expected counts are less than 5, rows or columns of the table may be combined to achieve a table with satisfactory expected counts.

21. Both Body Piercing and Tattoos
The paper “Contemporary College Students and Body Piercing” (Journal of Adolescent Health, ) described a survey of 450 undergraduate students at a state university in the southwestern region of the United States. Each student in the sample was classified according to class standing (freshman, sophomore, junior, senior) and body art category (body piercing only, tattoos only, both tattoos and body piercing, no body art). Is there evidence that there is an association between class standing and response to the body art question? Use a = .01.
State the hypotheses.
Body Piercing Only
Tattoos Only
Both Body Piercing and Tattoos
No Body Art
Freshman 61 7 14 86
Sophomore 43 11 10 64
Junior 20 9
Senior 21 17 23 54

22. H0: class standing and body art category are independent
Body Art Continued . . .
Body Piercing Only
Tattoos Only
Both Body Piercing and Tattoos
No Body Art
Freshman 61 7 14 86
Sophomore 43 11 10 64
Junior 20 9
Senior 21 17 23 54
Body Piercing Only
Tattoos Only
Both Body Piercing and Tattoos
No Body Art
Freshman
61 (49.7)
7 (15.1)
14 (18.5)
86 (84.7)
Sophomore
43 (37.9)
11 (11.5)
10 (14.1)
64 (64.5)
Junior
20 (23.4)
9 (7.1)
7 (8.7)
43 (39.8)
Senior
21 (34.0)
17 (10.3)
23 (12.7)
54 (58.0)
H0: class standing and body art category are independent
Ha: class standing and body art category are not independent
df = 9
Assuming H0 is true, what are the expected counts?
How many degrees of freedom does this two-way table have?

23. Both Body Piercing and Tattoos
Body Art Continued . . .
Body Piercing Only
Tattoos Only
Both Body Piercing and Tattoos
No Body Art
Freshman
61 (49.7)
7 (15.1)
14 (18.5)
86 (84.7)
Sophomore
43 (37.9)
11 (11.5)
10 (14.1)
64 (64.5)
Junior
20 (23.4)
9 (7.1)
7 (8.7)
43 (39.8)
Senior
21 (34.0)
17 (10.3)
23 (12.7)
54 (58.0)
Test Statistic:
P-value < a = .01

24. Which cell contributes the most to the X2 test statistic?
Body Art Continued . . .
Body Piercing Only
Tattoos Only
Both Body Piercing and Tattoos
No Body Art
Freshman
61 (49.7)
7 (15.1)
14 (18.5)
86 (84.7)
Sophomore
43 (37.9)
11 (11.5)
10 (14.1)
64 (64.5)
Junior
20 (23.4)
9 (7.1)
7 (8.7)
43 (39.8)
Senior
21 (34.0)
17 (10.3)
23 (12.7)
54 (58.0)
Since the P-value < a, we reject H0. There is sufficient evidence to suggest that class standing and the body art category are associated.
Seniors having both body piercing and tattoos contribute the most to the X2 statistic.
Which cell contributes the most to the X2 test statistic?