chapter12
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1 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Chapter 12
The Analysis of Categorical Data and
Goodness-of-Fit Tests
2 Copyright © 2005 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
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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.
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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)
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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.
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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.
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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
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Goodness-of-fit statistic, 2
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
22 (observed cell count - expected cell count)expected cell count
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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
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Upper-tail Areas for Chi-square Distributions
Right-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
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Goodness-of-Fit Test Procedure
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
Test statistic:22 (observed cell count - expected cell count)
expected cell count
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Goodness-of-Fit Test Procedure
P-values: When H0 is true and all expected counts are at least 5, 2 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 2 under the df = k-1 chi-square curve.
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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.
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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 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
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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
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Example
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
Calculation:recall
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
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ExampleP-value:
The P-value is based on a chi-squared distribution with df = 5 - 1 = 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 0.100.
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.
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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.
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Tests for Homogeneity and Independence in a Two-Way Table
Contacts Glasses NoneFemale 5 9 11Male 5 22 27
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.
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Tests for Homogeneity and Independence in a Two-Way Table
Contacts Glasses NoneRow Marginal
Total
Female 5 9 11 25Male 5 22 27 54
Column Marginal Total
10 31 38 79
Marginal 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.
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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
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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: Contacts Glasses None
Row Marginal Total
Female 5 9 11 25Male 5 22 27 54
Column Marginal Total
10 31 38 79
54 = total number of male students
= overall proportion of students using contacts10
79
= expected number of males that use contacts as primary vision correction
1054 6.83
79
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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.
(Row total)(Column total)Expected cell count =
Grand total
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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
Tests for Homogeneity in a Two-Way Table
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Contacts Glasses None
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
Tests for Homogeneity in a Two-Way Table
Expected counts are in parentheses.
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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.
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Comparing Two or More Populations Using the 2 Statistic
The expected cell counts are estimated from the sample data (assuming that H0 is true) using the formula
(Row total)(Column total)Expected cell count =
Grand total
Test statistic:22 (observed cell count - expected cell count)
expected cell count
28 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Comparing Two or More Populations Using the 2 Statistic
P-value:When H0 is true, 2 has approximately a chi-square distribution with
The P-value associated with the computed test statistic value is the area to the right of 2 under the chi-square curve with the appropriate df.
df = (number of rows - 1)(number of columns - 1)
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Comparing Two or More Populations Using the 2 Statistic
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.
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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.
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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
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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:22 (observed cell count - expected cell count)
expected cell count
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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
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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 insufficient 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
28.81 55.60 16.18
40 37.41 31 28.19
37.41 28.19
59 54.40 11 15.82 34 36.59
54.40 15.82 36.59
0.275+0.381+1.439+0.180+0.281+0.390+1.471+0.184 4.60
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Comparing Two or More Populations Using the 2 Statistic
P-value: When H0 is true, 2 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 2 under the chi-square curve with the appropriate df.
(Row total)(Column total)Expected cell count =
Grand total
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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 use to carry items.
He observed 57 shoppers (presumably randomly) and obtained the results that are summarized in the table on the next slide.
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Example
Determine if the carrying device proportions are the same for both genders using a 0.05 level of significance.
Device
Gender Cart Basket Nothing
Row Marginal
Total
Male 9 21 5 35Female 7 7 8 22
Column Marginal
Total 16 28 13 57
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Hypotheses:
H0: The true proportions of the device used are the same for both genders.
Ha: The true proportions of the device used are the same for both genders.
Example
Significance level: = 0.05
Test statistic:22 (observed cell count - expected cell count)
expected cell count
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Using Minitab, we get the following output:
Example
Chi-Square Test: Basket, Cart, Nothing
Expected counts are printed below observed counts
Basket Cart Nothing Total 1 9 21 5 35 9.82 17.19 7.98
2 7 7 8 22 6.18 10.81 5.02
Total 16 28 13 57
Chi-Sq = 0.069 + 0.843 + 1.114 + 0.110 + 1.341 + 1.773 = 5.251DF = 2, P-Value = 0.072
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We draw the following conclusion.
Example
With a P-value of 0.072, there is insufficient evidence at the 0.05 significance level to support a claim that males and females are not the same in terms of proportionate use of carrying devices at Wegman’s supermarket.
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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.
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The expected cell counts are estimated from the sample data (assuming that H0 is true) using the formula
2 Test for Independence
Test statistic:
22 (observed cell count - expected cell count)expected cell count
(Row total)(Column total)Expected cell count =
Grand total
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2 Test for Independence
The P-value associated with the computed test statistic value is the area to the right of 2 under the chi-square curve with the appropriate df.
(Row total)(Column total)Expected cell count =
Grand total
P-value:When H0 is true, 2 has approximately a chi-square distribution with
df = (number of rows - 1)(number of columns - 1)
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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.
2 Test for Independence
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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.
Contacts Glasses NoneRow Marginal
Total
Female 5 9 11 25Male 5 22 27 54
Column Marginal Total
10 31 38 79
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ExampleHypotheses:
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:22 (observed cell count - expected cell count)
expected cell count
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ExampleAssumptions:
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.
Contacts Glasses None
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
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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(12.97) (12.03)
27 27(28.03) (25.97)
Column Marginal
Total41 38 79
Female 25
Male 54
Contacts or Glasses None
Row Marginal
Total
14 11
27 27
Column Marginal
Total41 38 79
Female25
Male54
41 25
79
38 25
79
41 54
79
38 54
79
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Example
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.
Calculations: 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
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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 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