STATISTICS - Islamic University of Gaza

STATISTICSINFORMED DECISIONS USING DATAFifth Edition

Chapter 12

Inference on Categorical Data

12.1 Goodness-of-Fit TestLearning Objective

1. Perform a goodness-of-fit test

12.1 Goodness-of-Fit Test12.1.1 Perform a goodness-of-fit test (1 of 32)

Characteristics of the Chi-Square Distribution

1. It is not symmetric.

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

3. As the number of degrees of freedom increases, it becomes more nearly symmetric.

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 = μi = npi for i = 1, 2, …, k

Parallel Example 1: Finding Expected Counts

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.

Solution

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

Solution

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

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 represent the number of categories, and n represent the number of independent trials of an experiment. Then the formula

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.

The Goodness-of-Fit Test

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

Step 1: Determine the null and alternative hypotheses.

H0: The random variable follows a certain distribution

H1: The random variable does not follow the distribution in the null hypothesis

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

Step 3:

a) Calculate the expected counts, Ei , for each of the kcategories. 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.

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.

Classical Approach

Step 3 (continued):

c) Compute the test statistic:

Classical Approach

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

Compare the critical value to the test statistic.

P-value Approach

By Hand Step 3 (continued):

P-value Approach

d) Use Table VIII to approximate the 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

Technology Step 3 (continued):

c) Use a statistical spreadsheet or calculator with statistical capabilities to obtain the P-value. The directions for obtaining the P-value using the TI-84 Plus graphing calculator, MINITAB, Excel, and StatCrunch are in the Technology Step-by-Step in the text.

P-value Approach

Step 4: If the P-value < α, reject the null hypothesis.

P-value Approach

Step 5: State the conclusion.

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

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. The sociologist randomly selects 1,000 care-giving grandparents and obtains the following data.

Number of Years Frequency

Less than 1 year 252

1 or 2 years 255

3 or 4 years 162

5 or more years 331

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

Solution

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 α = 0.05.

Step 3:

a) The expected counts were computed in Example 1.

Number of Years

Observed Counts

Expected Counts

<1 252 228

1-2 255 239

3-4 162 176

≥5 331 357

Solution

Step 3:

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

Solution: Classical Approach

Solution: P-value Approach

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

Solution

Step 5: 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 α = 0.05 level of significance.

12.2 Tests for Independence and the Homogeneity of ProportionsLearning Objective

1. Perform a test for independence2. Perform a test for homogeneity of proportions

12.2 Tests for Independence and the Homogeneity of Proportions12.2.1 Perform a Test for Independence (1 of 36)

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

H1: 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, then

P(E and F) = P(E)P(F)

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.

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

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.

blank Money Health Love

Men 82 446 355

Women 46 574 273

Solution

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

blank Money Health Love Row Totals

Men 82 446 355 883

Women 46 574 273 893

Column totals 128 1020 628 1776

Solution

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

Solution

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.

Men 0.0358 0.2855 0.1758

Women 0.0362 0.2888 0.1778

Solution

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.

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 cell’s row total by its column total and divide this result by the table total. That is,

approximately follows the chi-square distribution 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, provided that (1) all expected frequencies are greater than or equal to 1 and (2) no more than 20% of the expected frequencies are less than 5.

Chi-Square Test for Independence

To test the hypothesis regarding the association between (or independence of) two variables in a contingency table, we use the steps that follow:

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.

Step 2: Choose a level of significance, α, 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.

Classical Approach

Step 3:

Classical Approach

P-value Approach

By Hand Step 3:

P-value Approach

d) Use Table VIII 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

Technology Step 3:

c) Use a statistical spreadsheet or calculator with statistical capabilities to obtain the P-value. The directions for obtaining the P-value using the TI-83/84 Plus graphing calculator, MINITAB, Excel, and StatCrunch are in the Technology Step-by-Step in the text.

P-value Approach

Step 4: If the P-value < α, reject the null hypothesis.

P-value Approach

Step 5: State the conclusion.

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

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 α = 0.05 level of significance.

Men 82 446 355

Women 46 574 273

Solution

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.

Men 82

(63.5808)

(507.048)

(312.2208)

Women 46

(64.2912)

(512.9088)

(315.7728)

Solution

Step 3:

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

Step 4: Since the P-value is less than the level of significance α = 0.05, we reject the null hypothesis.

Solution

Step 5: There is sufficient evidence to conclude that gender and response are dependent at the α = 0.05 level of significance.

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.

Parallel Example 3: Constructing a Conditional Distribution and Bar Graph

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

Men 82 446 355

Women 46 574 273

Solution

We first compute the conditional distribution of response by gender.

Solution

12.2 Tests for Independence and the Homogeneity of Proportions12.2.2 Perform a Test for Homogeneity of Proportions (1 of 13)

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.

Parallel Example 5: A Test for Homogeneity of Proportions

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:

blank 1992 2002 2008

Yes 418 479 525

No 602 541 485

Solution

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.

blank 1992 2002 2008

Yes418

(475.554)

(470.892)

(544.446)

(539.108)

Solution

Step 3:

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

Step 3:

c) The test statistic is

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.

By Hand Step 3:

c) The test statistic is

Step 4: Because the P-value is less than the level of significance α = 0.01, we reject the null hypothesis.

Solution

Step 5: There is sufficient evidence to reject the null hypothesis at the α = 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.

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