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

2. Chi-Square Tests for Univariate Data
Section 12.1
Chi-Square Tests for Univariate Data

3. 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?

4. 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 . . .

5. Notation k = number of categories of a categorical variable
p1 = true proportion for category 1
p2 = true proportion for category 2 
pk = true proportion for category k
(note: p1 + p2 +  + pk = 1)

6. Hypotheses H0: p1 = hypothesized proportion for category 1
pk = hypothesized proportion for category k
Ha: H0 is not true, so at least one of the true category proportions differs from the corresponding hypothesized value.

7. Expected Counts
For each category, the expected count for that category is the product of the total number of observations with the hypothesized proportion for that category.

8. Expected Counts Example
Consider the sample of faculty from a large university system and recall that the newsperson wanted to test to see if each of the groups occurred with equal frequency.
Categories
Full Professor
Associate Professor
Assistant Professor
Instructor
Adjunct/Part Time
Total
Frequency 22 31 25 35 41
Hypothesized Proportion
0.2
Expected Count
154 1
30.8
30.8
30.8
30.8
30.8
154

9. Goodness-of-Fit Test Procedure
Hypotheses:
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 c2, 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 analyze univariate categorical data from a single sample.
. . .
Read “chi-squared”
The c2 value can never be negative.

10. 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

11. 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

12. Chi-square distributions

13. Upper-tail Areas for Chi-square Distributions

14. Faculty Distribution Example
Consider the newsperson’s desire to determine if the faculty of a large university system were equally distributed. Let us test this hypothesis at a significance level of 0.05.
Let p1, p2, p3, p4, and p5 denote the proportions of all faculty in this university system that are full professors, associate professors, assistant professors, instructors and adjunct/part time respectively.
H0: p1 = 0.2, p2 = 0.2, p3 = 0.2, p4= 0.2, p5 = 0.2
Ha: H0 is not true

15. Faculty Distribution Example
Assumptions: As we saw in an earlier slide, the expected counts were all 30.8 which is greater than 5. Although we do not know for sure how the sample was obtained for the purposes of this example, we shall assume selection procedure generated a random sample.
Significance level:  = 0.05
Test statistic:

16. Faculty Distribution Example
Calculation:

17. Faculty Distribution Example
P-value:
The P-value is based on a chi-squared distribution with df = = 4. The computed value of 2, 7.56 is smaller than 7.77, the lowest value of 2 in the table for df = 4, so that the P-value is greater than
Conclusion:
Since the P-value > 0.05 = , H0 cannot be rejected. There is insufficient evidence to refute the claim that the proportion of faculty in each of the different categories is the same.

18. 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

19. 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

20. 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
Full Moon
7711
Waning Gibbous
47,595
Last Quarter
7733
Waning Crescent
48,230
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).

21. 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 c2 test statistic is smaller than the smallest entry in the df = 7 column of Appendix Table 8.

22. Tests for Homogeneity and Independence in a Two-way Table
Section 12.2
Tests for Homogeneity and Independence in a Two-way Table

23. Tests for Homogeneity and Independence in a Two-Way Table
Data resulting from observations made on two different categorical variables can be summarized using a tabular format. For example, consider the student data set giving information from a sample of 79 students taking elementary statistics. The table is on the next slide.

24. Tests for Homogeneity and Independence in a Two-Way Table
This is an example of a two-way frequency table, or contingency table.
The numbers in the 6 cells with clear backgrounds are the observed cell counts.

25. Tests for Homogeneity and Independence in a Two-Way Table
Marginal totals are obtained by adding the observed cell counts in each row and also in each column.
Contacts
Glasses
None
Row Marginal Total
Female 5 9 11
Male 22 27
Column Marginal Total 25 54 10 31 38 79
The sum of the column marginal totals (or the row marginal totals) is called the grand total.

26. Tests for Homogeneity in a Two-Way Table
Typically, with a two-way table used to test homogeneity, the rows indicate different populations and the columns indicate different categories or vice versa.
For a test of homogeneity, the central question is whether the category proportions are the same for all of the populations

27. Tests for Homogeneity in a Two-Way Table
When the row indicates the population, the expected count for a cell is simply the overall proportion (over all populations) that have the category times the number in the population.
To illustrate:
= overall proportion of students using contacts
54 = total number of male students
= expected number of males that use contacts as primary vision correction

28. Tests for Homogeneity in a Two-Way Table
The expected values for each cell represent what would be expected if there is no difference between the groups under study can be found easily by using the following formula.

29. Tests for Homogeneity in a Two-Way Table
Calculate the expected count for each cell.
Contacts
Glasses
None
Row Marginal Totals
Female 5 9 11 25
Male 22 27 54
Column Marginal Totals 10 31 38 79
25 • 10 79
25 • 31 79
25 • 38 79
54 • 10 79
54 •31 79
54 • 38 79

30. Tests for Homogeneity in a Two-Way Table
Expected counts are in parentheses.
Contacts
Glasses
None
Row Marginal Totals
Female 5 9 11 25
Male 22 27 54
Column Marginal Totals 10 31 38 79
(3.16)
(9.81)
(12.03)
(6.84)
(21.19)
(25.97)

31. 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.

32. 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.

33. 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.

34. These values in green are the observed counts.
A study was conducted to determine if collegiate soccer players had an 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.

35. 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)

36. 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

37. 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?

38. Drug Example
The following data came from a clinical trial of a drug regime used in treating a type of cancer, lymphocytic lymphoma. A sample of 273 patients were randomly divided into two groups, with one group of patients receiving cytoxan plus prednisone (CP) and the other receiving BCNU plus prednisone (BP). The responses to treatment were graded on a qualitative scale. The two-way table summary of the results is on the following slide.

39. Drug Example
Set up and perform an appropriate hypothesis test at the 0.05 level of significance.
Complete Response
Partial Response
No Change
Progression
Row Marginal Total BP 26 51 21 40 CP 31 59 11 34
Column Marginal Total
138
135 57
110 32 74
273

40. Drug Example (28.81) (55.60) (16.18) (37.41) (28.19) (54.40) (15.82)
Assumptions:
Samples were chosen randomly
All expected cell counts are at least 5
Complete Response
Partial Response
No Change
Progression
Row Marginal Total BP 26 51 21 40
138 CP 31 59 11 34
135
Column Marginal Total 57
110 32 74
273
(28.81)
(55.60)
(16.18)
(37.41)
(28.19)
(54.40)
(15.82)
(36.59)

41. Drug Example Hypotheses:
H0: The true response to treatment proportions are the same for both treatments (homogeneity of populations).
Ha: The true response to treatment proportions are not all the same for both treatments.
Significance level:  = 0.05
Test statistic:

42. Drug Example Calculations:
The two-way table for this example has 2 rows and 4 columns, so the appropriate df is (2-1)(4-1) = 3. Since 4.60 < 6.25, the P-value > 0.10 >  = 0.05 so H0 is not rejected. There is insufficient evidence to conclude that the response rates are different for the two treatments.

43. Shopping Example
A student decided to study the shoppers in Wegman’s, a local supermarket, to see if males and females exhibited the same behavior patterns with regard to the device used to carry items.
He observed 57 shoppers (presumably randomly) and obtained the results that are summarized in the table on the next slide.

44. Shopping Example
Determine if the carrying device proportions are the same for both genders using a 0.05 level of significance.
Cart
Basket
Nothing
Row Marginal Total
Male 9 21 5
Female 7 8
Column Marginal Total 35 22 16 28 13 57

45. Shopping Example Using Minitab, we get the following output:
Chi-Square Test: Basket, Cart, Nothing
Expected counts are printed below observed counts
Basket Cart Nothing Total
Total
Chi-Sq =
= 5.251
DF = 2, P-Value = 0.072

46. Shopping Example Test statistic: Hypotheses:
H0: The true proportions of the device used are the same for both genders.
Ha: The true proportions of the device used are not the same for both genders.
Significance level:  = 0.05
Test statistic:

47. Shopping Example Conclusion: P-value = 0.072 a = 0.05
Since P-value > a, we fail to reject H0. there is insufficient evidence to support a claim that males and females are not the same in terms of proportionate use of carrying devices at Wegman’s supermarket.

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

49. c2 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 c2 under the appropriate chi-square curve.

50. 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.

51. Both Body Piercing and Tattoos
The paper “Contemporary College Students and Body Piercing” (Journal of Adolescent Health, 2004) 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.
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
State the hypotheses.

52. 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?

53. 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

54. 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?

55. Vision Correction Example
Consider the two categorical variables, gender and principle form of vision correction for the sample of students used earlier in this presentation.
Contacts
Glasses
None
Row Marginal Total
Female 5 9 11
Male 22 27
Column Marginal Total 25 54 10 31 38 79
We shall now test to see if the gender and the principle form of vision correction are independent.

56. Vision Correction Example
Assumptions:
Sample of students was randomly & independently chosen.
All expected cell counts are at least 5,
Contacts
Glasses
None
Row Marginal Total
Female 5 9 11 25
Male 22 27 54
Column Marginal Total 10 31 38 79
(3.16)
(9.81)
(12.03)
(6.84)
(21.19)
(25.97)

57. Vision Correction Example
Assumptions:
Notice that the expected count is less than 5 in the
cell corresponding to Female and Contacts. So we
should combine the columns for Contacts and
Glasses to get
Contacts or Glasses
None
Row Marginal Total
Female 11 25
Male 27 54
Column Marginal Total 38 79 14
(12.97)
(12.03) 27
(28.03)
(25.97) 41

58. Vision Correction Example
Hypotheses:
H0: Gender and principle method of vision correction are independent.
Ha: Gender and principle method of vision correction are not independent.
Significance level: We have not chosen one, so we shall look at the practical significance level.
Test statistic:

59. Vision Correction Example
Calculations:
The contingency table for this example has 2 rows and 2 columns, so the appropriate df is (2-1)(2-1) = 1. Since < 2.70, the P-value is substantially greater than H0 would not be rejected for any reasonable significance level. There is not sufficient evidence to conclude that the gender and vision correction are related.
(I.e., For all practical purposes, one would find it reasonable to assume that gender and need for vision correction are independent.

60. Vision Correction Example
Minitab would provide the following output if the frequency table was input as shown.
Chi-Square Test: Contacts or Glasses, None
Expected counts are printed below observed counts
Contacts None Total
Total
Chi-Sq =
= 0.246
DF = 1, P-Value = 0.620