Chapter 12

Chapter 12

The Analysis of Categorical Data and

Goodness-of-Fit Tests

2 Copyright (c) 2001 Brooks/Cole, a division of Thomson Learning, Inc.

Univariate Categorical Data

Univariate categorical data is best summarized in a one-way frequency table. For example, consider the following observations of sample of faculty status for faculty in a large university system.

Full Professor

Associate Professor

Assistant Professor Instructor

Adjunct/Part time

Frequency 22 31 25 35 41

Category


Univariate Categorical Data

A local newsperson might be interested in testing hypotheses about the proportion of the population that fall in each of the categories. For example, the newsperson might want to test to see if the five categories occur with equal frequency throughout the whole university system.

To deal with this type of question we need to establish some notation.


Notation

k = number of categories of a categorical variable

1 = true proportion for category 1

2 = true proportion for category 2 k = true proportion for category k

(note: 1 + 2 + + k = 1)


Hypotheses

H0: 1 = hypothesized proportion for category 1

2 = hypothesized proportion for category 2

k = 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.


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.


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.

Full Professor

Associate Professor

Assistant Professor

InstructorAdjunct/Part time Total

Frequency 22 31 25 35 41 154Hypothesized

Proportion0.2 0.2 0.2 0.2 0.2 1

Expected Count

30.8 30.8 30.8 30.8 30.8 154

Category


Goodness-of-fit statistic, 2

22 (observed cell count - expected cell count)expected cell count

The value of the 2 statistic is the sum of these terms.


The value of the 2 statistic is the sum of these terms.

The goodness-of-fit statistic, 2, results from first computing the quantity

for each cell.

2(observed cell count - expected cell count)expected cell count

The goodness-of-fit statistic, 2, results from first computing the quantity

for each cell.

2(observed cell count - expected cell count)expected cell count


Chi-square distributions

Chi-square Distributions

0 5 10 15 20 25x

df = 1

df = 2

df = 3

df = 4

df = 5

df = 8

df = 10

df = 15


Upper-tail Areas for Chi-squared DistributionsRight-tail area df = 1 df = 2 df = 3 df = 4 df = 5

> .100 < 2.70 < 4.60 < 6.25 < 7.77 < 9.230.100 2.70 4.60 6.25 7.77 9.230.095 2.78 4.70 6.36 7.90 9.370.090 2.87 4.81 6.49 8.04 9.520.085 2.96 4.93 6.62 8.18 9.670.080 3.06 5.05 6.75 8.33 9.830.075 3.17 5.18 6.90 8.49 10.000.070 3.28 5.31 7.06 8.66 10.190.065 3.40 5.46 7.22 8.84 10.380.060 3.53 5.62 7.40 9.04 10.590.055 3.68 5.80 7.60 9.25 10.820.050 3.84 5.99 7.81 9.48 11.070.045 4.01 6.20 8.04 9.74 11.340.040 4.21 6.43 8.31 10.02 11.640.035 4.44 6.70 8.60 10.34 11.980.030 4.70 7.01 8.94 10.71 12.370.025 5.02 7.37 9.34 11.14 12.830.020 5.41 7.82 9.83 11.66 13.380.015 5.91 8.39 10.46 12.33 14.090.010 6.63 9.21 11.34 13.27 15.080.005 7.87 10.59 12.83 14.86 16.740.001 10.82 13.81 16.26 18.46 20.51

< .001 > 10.82 > 13.81 > 16.26 > 18.46 > 20.51

Right-tail area df = 6 df = 7 df = 8 df = 9 df = 10 > .100 < 10.64 < 12.01 < 13.36 < 14.68 < 15.980.100 10.64 12.01 13.36 14.68 15.980.095 10.79 12.17 13.52 14.85 16.160.090 10.94 12.33 13.69 15.03 16.350.085 11.11 12.50 13.87 15.22 16.540.080 11.28 12.69 14.06 15.42 16.750.075 11.46 12.88 14.26 15.63 16.970.070 11.65 13.08 14.48 15.85 17.200.065 11.86 13.30 14.71 16.09 17.440.060 12.08 13.53 14.95 16.34 17.710.055 12.32 13.79 15.22 16.62 17.990.050 12.59 14.06 15.50 16.91 18.300.045 12.87 14.36 15.82 17.24 18.640.040 13.19 14.70 16.17 17.60 19.020.035 13.55 15.07 16.56 18.01 19.440.030 13.96 15.50 17.01 18.47 19.920.025 14.44 16.01 17.53 19.02 20.480.020 15.03 16.62 18.16 19.67 21.160.015 15.77 17.39 18.97 20.51 22.020.010 16.81 18.47 20.09 21.66 23.200.005 18.54 20.27 21.95 23.58 25.180.001 22.45 24.32 26.12 27.87 29.58

< .001 > 22.45 > 24.32 > 26.12 > 27.87 > 29.58


Goodness-of-Fit Test ProcedureHypotheses:

H0:1 = hypothesized proportion for category 1

2 = hypothesized proportion for category 2

k = hypothesized proportion for category k

Ha:H0 is not true

Test statistic:22 (observed cell count - expected cell count)

expected cell count Test statistic:



Goodness-of-Fit Test Procedure

P-values: When H0 is true and all expected counts are at least 5, 2 has approximately a chi-squared distribution with df = k-1.

Therefore, the P-value associated with the computed test statistic value is the area to the right of 2 under the df = k-1 chi-squared curve.


Goodness-of-Fit Test Procedure

Assumptions:

1. Observed cell counts are based on a random sample.

2. The sample size is large. The sample size is large enough for the chi-squared test to be appropriate as long as every expected count is at least 5.


ExampleConsider 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 1, 2, 3, 4, and 5 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: 1 = 0.2, 2 = 0.2, 3 = 0.2, 4= 0.2, 5 = 0.2

Ha: H0 is not true


ExampleSignificance level: = 0.05

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.

Test statistic:22 (observed cell count - expected cell count)

expected cell count Test statistic:



ExampleCalculation:

recallFull

ProfessorAssociate Professor

Assistant Professor

InstructorAdjunct/Part time Total

Frequency 22 31 25 35 41 154Hypothesized

Proportion0.2 0.2 0.2 0.2 0.2 1

Expected Count

30.8 30.8 30.8 30.8 30.8 154

Category

2 2 2 2 2

2 22 30.8 31 30.8 25 30.8 35 30.8 41 30.8

30.8 30.8 30.8 30.8 30.82.514 0.001 1.092 0.573 3.378

7.56


ExampleP-value: The P-value is based on a chi-

squared distribution with df = 5 - 1 =4. The compute 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 0.100.

Conclusion: Since the P-value > 0.05 = , H0 cannot be rejected. There is not sufficient evidence to refute the claim that the proportion of faculty in each of the different categories is the same.


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 on 79 student dataset that was obtained from a sample of 79 students taking elementary statistics. The table is on the next slide.


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.

Contacts Glasses NoneFemale 5 9 11Male 5 22 27


Tests for Homogeneity and Independence in a Two-Way TableMarginal totals are obtained by adding the observed cell counts in each row and also in each column.

The sum of the column marginal total (or the row marginal totals) is called the grand total.

Contacts Glasses NoneRow Marginal

Total

Female 5 9 11 25Male 5 22 27 54

Column Marginal Total

10 31 38 79


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


Tests for Homogeneity in a Two-Way TableWhen 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:


Total

Female 5 9 11 25Male 5 22 27 54


10 31 38 7910

79= overall proportion of students using contacts

54 = total number of male students

1054 6.83

79 = expected number of males that use

contacts as primary vision correction

1054 6.83

79 = expected number of males that use

contacts as primary vision correction



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.

(Row total)(Column total)Expected cell count =

Grand total



Contacts Glasses None

Row Marginal

Total

5 9 11

5 22 27

Column Marginal

Total10 31 38 79

Female

Male

25

54

25 10

79

25 31

79

25 38

79

54 10

79

54 31

79

54 38

79



Expected counts are in parentheses.


Row Marginal

Total

5 9 11(3.16) (9.81) (12.03)

5 22 27(6.84) (21.19) (25.97)

Column Marginal

Total10 31 38 79

Female25

Male54


Comparing Two or More Populations Using the 2 Statistic

Hypotheses:

H0: The true category proportions are the same for all of the populations (homogeneity of populations).

Ha: The true category proportions are not all the same for all of the populations.



Test statistic:


Test statistic:



Grand total

The expected cell counts are estimated from the sample data (assuming that H0 is true) using the formula


Grand total




P-value: When H0 is true, 2 has approximately a chi-squared 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 chi-squared curve with the appropriate df.



Assumptions:

1. The data consists of independently chosen random samples.

2. The sample size is large: all expected 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.


Example

The following data come from a clinical trial of a drug regime used in treating a type of cancer, lymphocytic lymphomas.* Patients (273) 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.

* Ezdinli, E., S., Berard, C. W., et al. (1976) Comparison of intensive versus moderate chemotherapy of lympocytic lymphomas: a progress report. Cancer, 38, 1060-1068.


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

26 51 21 4031 59 11 34

Column Marginal

Total57 110 32 74 273

BPCP

138135


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.

Example

Significance level: = 0.05

Test statistic:


Test statistic:



Example

Assumptions: All expected cell counts are at least 5, and samples were chosen independently so the 2 test is appropriate.

Complete Response

Partial Response

No Change Progression

Row Marginal

Total

26 51 21 40(28.81) (55.60) (16.18) (37.41)

31 59 11 34(28.19) (54.40) (15.82) (36.59)

Column Marginal

Total57 110 32 74 273

Female138

Male135


ExampleCalculations:

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 not sufficient evidence to conclude that the responses are different for the two treatments.

2 2 2 2

2

2 2 2 2

26 28.81 51 55.60 21 16.18 40 37.41

28.81 55.60 16.18 37.41

31 28.19 59 54.40 11 15.82 34 36.59

28.19 54.40 15.82 36.590.275+0.381+1.439+0.180+0.281+0.390+1.471+0.184

4.60






Grand total



Grand total



Example II

The following data come from a study of the length of stay of 336 patients in psychiatric observation ward prior to be sent to another location. The observed variables are length of stay and four categories combining gender and status of admission.


Example II

Male voluntary

Female voluntary

Male certified

Fermal Certified Total

1 5 9 4 11 292 16 25 18 18 773 20 34 20 28 1024 10 17 6 8 415 5 15 1 12 336 3 8 0 5 167 3 7 0 5 158 5 11 1 6 23

Total 67 126 50 93 336

Nu

mb

er

of d

ays

in

the

wa

rd

Type of patient


Example III

The following data come from a study of Hodgkin's disease, a cancer of the lymph nodes. The observed variables are histological type and response to treatment 3 months after it had begun.

LP = lymphocyte dominanceNS = Nodular sclerosisMC = mixed cellurarityLD = lymphocyte depletion Positive Partial None

LP 74 18 12 104NS 68 16 12 96MC 154 54 58 266LD 18 10 44 72

Total 314 98 126 538

Response

His

tolo

gic

al t

ype


Hypotheses:

H0: The two variables are independent.

Ha: The two variables are not independent.

2 Test for Independence

The 2 test statistic and procedures can also be used to investigate the association between tow categorical variable in a single population.



Test statistic:


Test statistic:



Grand total



Grand total







Grand total



Grand total



Assumptions:

1. The observed counts are from a random sample.

2. The sample size is large: all expected 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.



Example

Consider the two categorical variables, gender and principle form of vision correction for the sample of students used earlier in this presentation.

We shall now test to see if the gender and the principle form of vision correction are independent.


Total

Female 5 9 11 25Male 5 22 27 54


10 31 38 79


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:


Test statistic:



Example

Assumptions:We are assuming that the sample of students was randomly chosen.

All expected cell counts are at least 5, and samples were chosen independently so the 2 test is appropriate.


Row Marginal

Total

5 9 11(3.16) (9.81) (12.03)

5 22 27(6.84) (21.19) (25.97)

Column Marginal

Total10 31 38 79

Female25

Male54


Example

Assumptions:Notice that the expected count is less than 5 in the cell corresponding to Female and Contacts. So that we should combine the columns for Contacts and Glasses to get

Contacts or Glasses None

Row Marginal

Total

14 11

27 27

Column Marginal

Total41 38 79

Female25

Male54

4125

79

38 25

79

4154

79

38 54

79

Contacts or Glasses None

Row Marginal

Total

14 11(12.97) (12.03)

27 27(28.03) (25.97)

Column Marginal

Total41 38 79

Female 25

Male 54


ExampleCalculations:

The contingency table for this example has 2 rows and 2 columns, so the appropriate df is (2-1)(2-1) = 1. Since 0.246 < 2.70, the P-value is substantially greater than 0.10. 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.

2 2 2 2

2 14 12.97 11 12.03 27 28.03 27 25.97

12.97 12.03 28.03 25.970.081+0.087+0.038+0.040

0.246


ExampleMinitab 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 1 14 11 25 12.97 12.03

2 27 27 54 28.03 25.97

Total 41 38 79

Chi-Sq = 0.081 + 0.087 + 0.038 + 0.040 = 0.246DF = 1, P-Value = 0.620

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