© 2010 Pearson Prentice Hall. All rights reserved Chapter Inference on Categorical Data 12

© 2010 Pearson Prentice Hall. All rights reserved

Chapter

Inference on Categorical Data

12


Section

Goodness-of-Fit Test

12.1

© 2010 Pearson Prentice Hall. All rights reserved 12-3

Objective

• Perform a goodness-of-fit test


Characteristics of the Chi-Square Distribution

1. It is not symmetric.


1. It is not symmetric.2. The shape of the chi-square distribution

depends on the degrees of freedom, just like Student’s t-distribution.





3. As the number of degrees of freedom increases, the chi-square distribution becomes more nearly symmetric.





3. As the number of degrees of freedom increases, the chi-square distribution becomes more nearly symmetric.

4. The values of 2 are nonnegative, i.e., the values of 2 are greater than or equal to 0.




A goodness-of-fit test is an inferential procedure used to determine whether a frequency distribution follows a specific distribution.


Expected Counts

Suppose that there are n independent trials of an

experiment with k ≥ 3 mutually exclusive possible

outcomes. Let p1 represent the probability of observing

the first outcome and E1 represent the expected count of

the first outcome; p2 represent the probability of

observing the second outcome and E2 represent the

expected count of the second outcome; and so on. The

expected counts for each possible outcome are given by

Ei = i = npi for i = 1, 2, …, k


A sociologist wishes to determine whether the distribution for the number of years care-giving grandparents are responsible for their grandchildren is different today than it was in 2000. According to the United States Census Bureau, in 2000, 22.8% of grandparents have been responsible for their grandchildren less than 1 year; 23.9% of grandparents have been responsible for their grandchildren for 1 or 2 years; 17.6% of grandparents have been responsible for their grandchildren 3 or 4 years; and 35.7% of grandparents have been responsible for their grandchildren for 5 or more years. If the sociologist randomly selects 1,000 care-giving grandparents, compute the expected number within each category assuming the distribution has not changed from 2000.

Parallel Example 1: Finding Expected Counts


Step 1: The probabilities are the relative frequencies from the 2000 distribution:

p<1yr = 0.228 p1-2yr = 0.239

p3-4yr = 0.176 p ≥5yr = 0.357

Solution


Step 2: There are n=1,000 trials of the experiment so the expected counts are:

E<1yr = np<1yr = 1000(0.228) = 228

E1-2yr = np1-2yr = 1000(0.239) = 239

E3-4yr = np3-4yr =1000(0.176) = 176

E≥5yr = np ≥5yr = 1000(0.357) = 357

Solution


Test Statistic for Goodness-of-Fit Tests

Let Oi represent the observed counts of category i, Ei

represent the expected counts of category i, k representthe number of categories, and n represent the number ofindependent trials of an experiment. Then the formula

approximately follows the chi-square distribution withk-1 degrees of freedom, provided that• all expected frequencies are greater than or equal to 1 (all Ei ≥ 1)

and• no more than 20% of the expected frequencies are less than 5.

2 Oi E i 2

E i

i 1, 2,, k


CAUTION!

Goodness-of-fit tests are used to test hypotheses regarding the distribution of a variable based on a single population. If you wish to compare two or more populations, you must use the tests for homogeneity presented in Section 12.2.


Step 1: Determine the null and alternative hypotheses. H0: The random variable follows a

certain distribution H1: The random variable does not

follow a certain distribution

The Goodness-of-Fit Test

To test the hypotheses regarding a distribution, we use the steps that follow.


Step 2: Decide on a level of significance, , depending on the seriousness of making a Type I error.


Step 3: a) Calculate the expected counts for each

of the k categories. The expected counts are Ei=npi for i = 1, 2, … , k where n is the number of trials and pi is the probability of the ith category, assuming that the null hypothesis is true.


Step 3: b) Verify that the requirements for the goodness-

of-fit test are satisfied.1. All expected counts are greater than or

equal to 1 (all Ei ≥ 1).2. No more than 20% of the expected counts

are less than 5.c) Compute the test statistic:

Note: Oi is the observed count for the ith category.

02

Oi E i 2

E i


CAUTION!

If the requirements in Step 3(b) are not satisfied, one option is to combine two or more of the low-frequency categories into a single category.


Step 4: Determine the critical value. All goodness-of-fit tests are right-tailed tests, so the critical value is with k-1 degrees of freedom.

Classical Approach

2


Step 5: Compare the critical value to the test statistic. If reject the null hypothesis.

Classical Approach

02

2,


Step 4: Use Table VII to obtain an approximate P-value by determining the area under the chi-square distribution with k-1 degrees of freedom to the right of the test statistic.

P-Value Approach


Step 5: If the P-value < , reject the null hypothesis.

P-Value Approach


Step 6: State the conclusion.


A sociologist wishes to determine whether the distribution forthe number of years care-giving grandparents are responsiblefor their grandchildren is different today than it was in 2000.

According to the United States Census Bureau, in 2000, 22.8%of grandparents have been responsible for their grandchildrenless than 1 year; 23.9% of grandparents have been responsiblefor their grandchildren for 1 or 2 years; 17.6% of grandparentshave been responsible for their grandchildren 3 or 4 years; and35.7% of grandparents have been responsible for theirgrandchildren for 5 or more years. The sociologist randomlyselects 1,000 care-giving grandparents and obtains thefollowing data.

Parallel Example 2: Conducting a Goodness-of -Fit Test


Test the claim that the distribution is different today than it was in 2000 at the = 0.05 level of significance.


Step 1: We want to know if the distribution today is different than it was in 2000. The hypotheses are then:

H0: The distribution for the number of years care-giving grandparents are responsible for their grandchildren is the same today as it was in 2000

H1: The distribution for the number of years care-giving grandparents are responsible for their grandchildren is different today than it was in 2000

Solution


Step 2: The level of significance is =0.05.

Step 3:

(a) The expected counts were computed in Example 1.

Solution

Number of Years

Observed Counts

Expected Counts

<1 252 228

1-2 255 239

3-4 162 176

≥5 331 357


Step 3:

(b) Since all expected counts are greater than or equal to 5, the requirements for the goodness-of-fit test are satisfied.

(c) The test statistic is

Solution

02

252 228 2

228

255 239 2

239

162 176 2

176

331 357 2

3576.605


Step 4: There are k = 4 categories, so we find the critical value using 4-1=3 degrees of freedom. The critical value is

Solution: Classical Approach

0.052 7.815


Step 5: Since the test statistic, is less than the critical value , we fail to reject the null hypothesis.


02 6.605

0.052 7.815


Step 4: There are k = 4 categories. The P-value is the area under the chi-square distribution with 4-1=3 degrees of freedom to the right of . Thus, P-value ≈ 0.09.

Solution: P-Value Approach

02 6.605


Step 5: Since the P-value ≈ 0.09 is greater than the level of significance = 0.05, we fail to reject the null hypothesis.



Step 6: There is insufficient evidence to conclude that the distribution for the number of years care-giving grandparents are responsible for their grandchildren is different today than it was in 2000 at the = 0.05 level of significance.

Solution


Section

Tests for Independence and the Homogeneity of Proportions

12.2


1. Perform a test for independence

2. Perform a test for homogeneity of proportions

Objectives


Objective 1

• Perform a Test for Independence


The chi-square test for independence is used to determine whether there is an association between a row variable and column variable in a contingency table constructed from sample data. The null hypothesis is that the variables are not associated; in other words, they are independent. The alternative hypothesis is that the variables are associated, or dependent.


“In Other Words”

In a chi-square independence test, the null

hypothesis is always

H0: The variables are independent

The alternative hypothesis is always

H0: The variables are not independent


The idea behind testing these types of claims is to compare actual counts to the counts we would expect if the null hypothesis were true (if the variables are independent). If a significant difference between the actual counts and expected counts exists, we would take this as evidence against the null hypothesis.


If two events are independent, thenP(A and B) = P(A)P(B)

We can use the Multiplication Principle for independent events to obtain the expected proportion of observations within each cell under the assumption of independence and multiply this result by n, the sample size, in order to obtain the expected count within each cell.


In a poll, 883 males and 893 females were asked “If you could have only one of the following, which would you pick: money, health, or love?” Their responses are presented in the table below. Determine the expected counts within each cell assuming that gender and response are independent.

Source: Based on a Fox News Poll conducted in January, 1999

Parallel Example 1: Determining the Expected Counts in a Test for Independence


Step 1: We first compute the row and column totals:

Solution

Money Health Love Row Totals

Men 82 446 355 883

Women 46 574 273 893

Column totals 128 1020 628 1776


Step 2: Next compute the relative marginal frequencies for the row variable and column variable:

Solution

Money Health Love Relative Frequency

Men 82 446 355 883/1776

≈ 0.4972

Women 46 574 273 893/1776

≈0.5028

Relative Frequency

128/1776

≈0.0721

1020/1776

≈0.5743

628/1776

≈0.3536 1


Step 3: Assuming gender and response are independent, we use the Multiplication Rule for Independent Events to compute the proportion of observations we would expect in each cell.

Solution

Money Health Love

Men 0.0358 0.2855 0.1758

Women 0.0362 0.2888 0.1778


Step 4: We multiply the expected proportions from step 3 by 1776, the sample size, to obtain the expected counts under the assumption of independence.

Solution

Money Health Love

Men 1776(0.0358)

≈ 63.5808

1776(0.2855)

≈ 507.048

1776(0.1758)

≈ 312.2208

Women

1776(0.0362)

≈ 64.2912

1776(0.2888)

≈ 512.9088

1776(0.1778)

≈ 315.7728


Expected Frequencies in a Chi-Square Test for Independence

To find the expected frequencies in a cell when performing a chi-square independence test, multiply the row total of the row containing the cell by the column total of the column containing the cell and divide this result by the table total. That is,

Expected frequency = (row total)(column total)

table total


Test Statistic for the Test of Independence

Let Oi represent the observed number of counts in the

ith cell and Ei represent the expected number of countsin the ith cell. Then

approximately follows the chi-square distribution with (r-1)(c-1) degrees of freedom, where r is the number of rowsand c is the number of columns in the contingency table,provided that (1) all expected frequencies are greater than orequal to 1 and (2) no more than 20% of the expectedfrequencies are less than 5.

2 Oi E i 2

E i


Step 1: Determine the null and alternative hypotheses. H0: The row variable and column

variable are independent. H1: The row variable and column

variables are dependent.

Chi-Square Test for Independence

To test the association (or independence of) two variables in a contingency table:


Step 2: Choose a level of significance, , depending on the seriousness of making a Type I error.


Step 3: a) Calculate the expected frequencies

(counts) for each cell in the contingency table.

b) Verify that the requirements for the chi-square test for independence are satisfied:1. All expected frequencies are greater

than or equal to 1 (all Ei ≥ 1).2. No more than 20% of the expected

frequencies are less than 5.


Step 3: c) Compute the test statistic:

Note: Oi is the observed count for the ith category.

02

Oi E i 2

E i


Step 4: Determine the critical value. All chi-square tests for independence are right-tailed tests, so the critical value is with (r-1)(c-1) degrees of freedom, where r is the number of rows and c is the number of columns in the contingency table.

Classical Approach

2



Step 5: Compare the critical value to the test statistic. If reject the null hypothesis.

Classical Approach

02

2,


Step 4: Use Table VII to determine an approximate P-value by determining the area under the chi-square distribution with (r-1)(c-1) degrees of freedom to the right of the test statistic.

P-Value Approach


Step 5: If the P-value < , reject the null hypothesis.

P-Value Approach


Step 6: State the conclusion.


In a poll, 883 males and 893 females were asked “If you could have only one of the following, which would you pick: money, health, or love?” Their responses are presented in the table below. Test the claim that gender and response are independent at the = 0.05 level of significance.


Parallel Example 2: Performing a Chi-Square Test for Independence


Step 1: We want to know whether gender and response are dependent or independent so the hypotheses are:

H0: gender and response are independent

H1: gender and response are dependent


Solution


Step 3:

(a) The expected frequencies were computed in Example 1 and are given in parentheses in the table below, along with the observed frequencies.

Solution

Money Health Love

Men 82

(63.5808)

446

(507.048)

355

(312.2208)

Women 46

(64.2912)

574

(512.9088)

273

(315.7728)


Step 3:

(b) Since none of the expected frequencies are less than 5, the requirements for the goodness-of-fit test are satisfied.


Solution

02

82 63.5808 2

63.5808

446 507.048 2

507.048

273 315.7728 2

315.772836.82


Step 4: There are r = 2 rows and c =3 columns, so we find the critical value using (2-1)(3-1) = 2 degrees of freedom. The critical value is .


0.052 5.99


Step 5: Since the test statistic, is greater than the critical value , we reject the null hypothesis.


02 36.82

0.052 5.99


Step 4: There are r = 2 rows and c =3 columns so we find the P-value using (2-1)(3-1) = 2 degrees of freedom. The P-value is the area under the chi-square distribution with 2 degrees of freedom to the right of which is approximately 0.


02 36.82


Step 5: Since the P-value is less than the level of significance = 0.05, we reject the null hypothesis.



Step 6: There is sufficient evidence to conclude that gender and response are dependent at the = 0.05 level of significance.

Solution


To see the relation between response and gender, we draw bar graphs of the conditional distributions of response by gender. Recall that a conditional distribution lists the relative frequency of each category of a variable, given a specific value of the other variable in a contingency table.


Find the conditional distribution of response by gender for the data from the previous example, reproduced below.


Parallel Example 3: Constructing a Conditional Distribution and Bar Graph


We first compute the conditional distribution of response by gender.

Solution

Money Health Love

Men 82/883

≈ 0.0929

446/883

≈ 0.5051

355/883

≈ 0.4020

Women 46/893

≈ 0.0515

574/893

≈ 0.6428

273/893

≈ 0.3057


Solution


Objective 2

• Perform a Test for Homogeneity of Proportions


In a chi-square test for homogeneity of proportions, we test whether different populations have the same proportion of individuals with some characteristic.


The procedures for performing a test of homogeneity are identical to those for a test of independence.


The following question was asked of a random sample of individuals in 1992, 2002, and 2008: “Would you tell me if you feel being a teacher is an occupation of very great prestige?” The results of the survey are presented below:

Test the claim that the proportion of individuals that feel being a teacher is an occupation of very great prestige is the same for each year at the = 0.01 level of significance.

Source: The Harris Poll

Parallel Example 5: A Test for Homogeneity of Proportions

1992 2002 2008

Yes 418 479 525

No 602 541 485


Step 1: The null hypothesis is a statement of “no difference” so the proportions for each year who feel that being a teacher is an occupation of very great prestige are equal. We state the hypotheses as follows:

H0: p1= p2= p3

H1: At least one of the proportions is different from the others.


Solution


Step 3:

(a) The expected frequencies are found by multiplying the appropriate row and column totals and then dividing by the total sample size. They are given in parentheses in the table below, along with the observed frequencies.

Solution

1992 2002 2008

Yes418

(475.554)

479

(475.554)

525

(470.892)

No602

(544.446)

541

(544.446)

485

(539.108)


Step 3:

(b) Since none of the expected frequencies are less than 5, the requirements are satisfied.


Solution

02

418 475.554 2

475.554

479 475.554 2

475.554

485 539.108 2

539.10824.74


Step 4: There are r = 2 rows and c =3 columns, so we find the critical value using (2-1)(3-1) = 2 degrees of freedom. The critical value is .


0.012 9.210


Step 5: Since the test statistic, is greater than the critical value , we reject the null hypothesis.


02 24.74

0.012 9.210


Step 4: There are r = 2 rows and c =3 columns so we find the P-value using (2-1)(3-1) = 2 degrees of freedom. The P-value is the area under the chi-square distribution with 2 degrees of freedom to the right of which is approximately 0.


02 24.74


Step 5: Since the P-value is less than the level of significance = 0.01, we reject the null hypothesis.



Step 6: There is sufficient evidence to reject the null hypothesis at the = 0.01 level of significance. We conclude that the proportion of individuals who believe that teaching is a very prestigious career is different for at least one of the three years.

Solution

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© 2010 Pearson Prentice Hall. All rights reserved Chapter Inference on Categorical Data 12