aron chpt 11 ed (2)

Chi-Square Tests and Strategies Chi-Square Tests and Strategies When Population Distributions Are When Population Distributions Are Not NormalNot Normal

Chapter 11

Copyright © 2011 by Pearson Education, Inc. All rights reserved

Chapter OutlineChapter OutlineChi-Square TestsThe Chi-Square Statistic and the Chi-Square

Test for Goodness of FitThe Chi-Square Test for IndependenceAssumptions of the Chi-Square TestsEffect Size and Power for the Chi-Square Tests

for IndependenceStrategies for Hypothesis Testing When

Population Distributions Are Not NormalData TransformationsRank-Order TestsComparison of MethodsChi-Square Tests, Data Transformation, and

Rank-Order Tests in Research ArticlesCopyright © 2011 by Pearson Education, Inc. All rights reserved

What If You Have Variables Whose What If You Have Variables Whose Values Are Categories?Values Are Categories?

t Tests and the ANOVA require:◦ the measured variable to have scores that are

quantitative e.g., ratings on a scale of stress that range from 0–

10, numerical scores on a test of intelligence, scores on a measure of gastrointestinal symptoms

◦ the populations to follow a normal curve.

Categorical variables require an alternative hypothesis-testing procedure.


Chi-Square TestsChi-Square TestsAre used when the variable of interest is

a nominal variable◦ The values of a nominal variable are categories.◦ The scores of a nominal variable represent

frequencies.

Chi-square tests examine how well the observed breakdown of people or observations over categories fits an expected breakdown.◦ Chi-square test of goodness of fit involves levels of a

single nominal variable.◦ Chi-square test for independence is used when there

are two nominal variables each with several categories.


Steps for Figuring the Chi-Square Steps for Figuring the Chi-Square Statistic Statistic Determine the actual observed frequencies in each

category. Determine the expected frequency in each

category. In each category, take the observed minus

expected frequencies. Square each of these differences.Divide each squared difference by the

expected frequency for its category.Add up the these results for all the categories.

The Chi-Square DistributionThe Chi-Square DistributionEstimating the distribution of chi-square

statistics that would arise by chanceThe exact shape of the chi-square distribution

depends on degrees of freedom, but they are all skewed to the right because the chi-square statistic cannot be less than 0 but can have very high values.

The degrees of freedom for a chi-square test are the number of categories that are free to vary, given the total.◦ For example, if there are two categories,

there is one degree of freedom. df = Ncategories – 1


The Chi-Square Table The Chi-Square Table (pg. 443)(pg. 443) The cutoff for a chi-square to be extreme enough to

reject the null hypothesis is determined using a chi-square table.◦ To use this table, you need to determine the degrees

of freedom and the significance level that you will use for your study.

Summary: Hypothesis Testing Chi-Summary: Hypothesis Testing Chi-Square Test for Goodness of FitSquare Test for Goodness of Fit

Restate the question as a research hypothesis and a null hypothesis about the population.◦ The research hypothesis is that the observations over categories in the two

populations are different.◦ The null hypothesis is that the observations over categories in the two

populations are the same. Determine the characteristics of the comparison distribution.

◦ chi-squared distribution; df = number of categories – 1 Determine the cutoff on the comparison distribution at which the null

hypothesis should be rejected. Determine your sample’s score on the comparison distribution.

◦ Determine the actual observed frequencies in each category.◦ Determine the expected frequencies in each category.◦ In each category, take observed minus expected frequencies.◦ Square each of these differences.◦ Divide each squared difference by the expected frequency for its category.◦ Add up these results for all categories.

Decide whether to reject the null hypothesis.◦ Compare your sample’s score to the cutoff score.


Example of Hypothesis Testing Chi-Example of Hypothesis Testing Chi-Square Test for Goodness of FitSquare Test for Goodness of Fit In the example from Chapter 11 of the text

(pgs 366-372), the researchers looked at the gender of characters on cereal boxes.

“In order to test their hypothesis that male characters would appear more often than female characters on cereal boxes, researchers coded the gender of the characters on every cereal box in a large grocery superstore”

Of the 1,386 characters whose gender was determined, 996 (72%) were male and 390 (28%) were female characters.


Example of Hypothesis Testing Chi-Example of Hypothesis Testing Chi-Square Test for Goodness of FitSquare Test for Goodness of Fit

Step 1: Restate the question as a HStep 1: Restate the question as a Haa and Hand HooPopulation 1: characters on cereal boxes like

those in the studyPopulation 2: characters on cereal boxes who

are equally likely to be male and female

Ha = Male characters will appear more often than females characters on cereal boxes.

(the distribution of observations over categories in the 2 populations is different)

Ho = There will be no difference is the appearance of male and female characters on cereal boxes.

(the distribution of observations over categories in the 2 populations is not different)


Step 2: Determine the Step 2: Determine the characteristics of the comparison characteristics of the comparison distribution.distribution.

◦Chi-squared distribution

◦Degrees of Freedom = Number of Categories –1 2 categories (male & female characters) -1 = 1


Step 3: Determine the cutoff on the Step 3: Determine the cutoff on the comparison distribution at which the null comparison distribution at which the null hypothesis should be rejected.hypothesis should be rejected.

◦Look at the cutoff on the chi-square table for your significance level and the study’s degrees of freedom.

◦Using a significance level of .05 and 1 degree of freedom the cutoff from the chi-square table is 3.841.


Step 4: Determine your sample’s Step 4: Determine your sample’s score on the comparison score on the comparison distribution.distribution.

◦ Determine the actual observed frequencies in each category.• males = 996, females = 390

◦ Determine the expected frequencies in each category.• males = 693, females = 693

◦ In each category, take observed minus expected frequencies.• for males: O – E = 996 – 693= 303• for females: O – E = 390 – 693= -303

◦ Square each of these differences.• for males: (O – E)2 = (303)2 = 91,809• for females: (O – E)2 = (-303)2 = 91,809

◦ Divide each squared difference by the expected frequency for its category.• for males: (O-E)2 /E = 132.48• for females: (O-E)2 / E = 132.48 ◦ Add up these results for all categories.• X2 = 132.48 + 132.48 = 264.9

Step 4: Determine your sample’s Step 4: Determine your sample’s score on the comparison distribution.score on the comparison distribution.

◦ Determine the actual observed frequencies in each category.• males = 996, females = 390

O E (O-E) (O-E)2

Male 996 693 303 91,809 91809/693 132.48

Female 390 693 -303 91,809 91809/693 132.48

264.96


◦ Determine the expected frequencies in each category.• males = 693, females = 693

O E (O-E) (O-E)2

Male 996 693 303 91,809 91809/693 132.48

Female 390 693 -303 91,809 91809/693 132.48

264.96


◦ In each category, take observed minus expected frequencies.

O E (O-E) (O-E)2

Male 996 693 303 91,809 91809/693 132.48

Female 390 693 -303 91,809 91809/693 132.48

264.96


◦ Square each of these differences.

O E (O-E) (O-E)2

Male 996 693 303 91,809 91809/693 132.48

Female 390 693 -303 91,809 91809/693 132.48

264.96


◦ Divide each squared difference by the expected frequency for its category.

O E (O-E) (O-E)2

Male 996 693 303 91,809 91809/693 132.48

Female 390 693 -303 91,809 91809/693 132.48

264.96


◦ Add up these results for all categories.

O E (O-E) (O-E)2

Male 996 693 303 91,809 91809/693 132.48

Female 390 693 -303 91,809 91809/693 132.48

264.96


Example of Hypothesis Testing Chi-Example of Hypothesis Testing Chi-Square Test for Goodness of Fit: Step Square Test for Goodness of Fit: Step 55Decide whether to reject the null

hypothesis.◦ The chi-square of the sample—264.96—is

greater than the cutoff to reject the null hypothesis, which is 3.841.

◦ The researchers can reject the null hypothesis.

Second Example of Hypothesis Second Example of Hypothesis Testing Testing Chi-Square Test for Goodness of FitChi-Square Test for Goodness of Fit

Chapter 11 Study Guide Problem 1 (pg. 170)The table that follows includes the primary method of conflict resolution used by 20 students.

(a) Following the five steps of hypothesis testing, conduct the appropriate X2 test to determine whether the observed frequencies are significantly different from the frequencies expected at the .05 level of significance. (b) Explain your results.

Method Aggressive

Manipulative

Passive Assertive

N of students

8 2 2 8

Step 1: Restate the question as a HStep 1: Restate the question as a Haa and Hand HooPopulation 1: the population of students who use each method of conflict resolution like those observed.Population 2: the population of students who use each method of conflict resolution equally.

Ha = Students will use the methods of conflict resolution differently. (the distribution of observations over categories in the 2 populations is different)

Ho = Students will not use the methods of conflict resolution differently. (the distribution of observations over categories in the 2 populations is not different)


Step 2: Determine the Step 2: Determine the characteristics of the comparison characteristics of the comparison distribution.distribution.

Chi-squared distribution

Degrees of Freedom = Number of Categories –1 4 categories (methods of conflict resolution) -1 = 3


Step 3: Determine the cutoff on the Step 3: Determine the cutoff on the comparison distribution at which the null comparison distribution at which the null hypothesis should be rejected.hypothesis should be rejected.

◦Look at the cutoff on the chi-square table for your significance level and the study’s degrees of freedom.

◦Using a significance level of .05 and 3 degree of freedom the cutoff from the chi-square table is 7.815.


◦ Determine the actual observed frequencies in each category.• Aggressive = 8• Manipulative = 2• Passive = 2• Assertive = 8

O E (O-E) (O-E)2

Aggressive 8 5 3 9 9/5 1.8Manipulative 2 5 -3 9 9/6 1.8Passive 2 5 3 9 9/7 1.8Assertive 8 5 -3 9 9/8 1.8

7.2


◦ Determine the expected frequencies in each category.• Aggressive = 5• Manipulative = 5• Passive = 5• Assertive = 5

O E (O-E) (O-E)2


7.2

O E (O-E) (O-E)2


7.2


◦ In each category, take observed minus expected frequencies.

O E (O-E) (O-E)2


7.2


◦ Square each of these differences.

O E (O-E) (O-E)2


7.2


◦ Divide each squared difference by the expected frequency for its category.

O E (O-E) (O-E)2


7.2


◦ Add up these results for all categories.


Example of Hypothesis Testing Chi-Example of Hypothesis Testing Chi-Square Test for Goodness of Fit: Step Square Test for Goodness of Fit: Step 55Decide whether to reject the null

hypothesis.◦ The chi-square of the sample—7.2—is not more

extreme than the cutoff to reject the null hypothesis, which is 7.815.

◦ The researchers cannot reject the null hypothesis.

The Chi-Square Test for The Chi-Square Test for IndependenceIndependence

• Used when there are two nominal variables, each with several categories

• Contingency Table• table in which the distributions of two

nominal variables are set up so that you have the frequencies of their contributions as well as the totals


Gender TOTALAge Male Female

Child 28 30 58 (26.1%)Adult 125 39 164 (73.9%)TOTAL 153 69 222 (100.0%)

IndependenceIndependenceNo relationship between the variables

in the contingency tableIt is important to determine whether

the lack of independence in the sample is large enough to reject the null hypothesis of independence of the population. Gender TOTAL

Age Male Female Child 28 (18.3%) 30 (43.5%) 58 (26.1%)Adult 125 (81.7%) 39 (56.5%) 128 (73.9%)TOTAL 153 69 222 100.0%)

Determining Expected Determining Expected FrequenciesFrequencies

Figuring the Chi-SquareFiguring the Chi-SquareFigure the weighted squared difference

for each cell and add these up


Degrees of FreedomDegrees of FreedomWith a chi-square test for independence, the

degrees of freedom are the number of categories free to vary once the totals are known.◦ The number of categories is the number of cells.◦ The totals include the row and column totals.◦ df = (NColumns – 1)(NRows – 1)


Steps of Hypothesis Steps of Hypothesis TestingTesting Restate the question as a research hypothesis and a null

hypothesis about the population. Determine the characteristics of the comparison distribution.

◦ chi-square distribution◦ degrees of freedom = (NColumns – 1)(NRows – 1)

Determine the cutoff sample score on the comparison distribution at which the null hypothesis should be rejected.◦ Decide the significance level you will use for your study.◦ Use the degrees of freedom you calculated and the significance level you

will use for this study to find the cutoff score on a chi-square table. Determine your sample’s score on the comparison

distribution.◦ Determine the actual observed frequencies in each cell.◦ Determine the expected frequencies in each cell.

Find each row’s percentage of the total. For each row, multiply its rows percentage by its column’s total. In each cell, take the observed minus the expected frequencies. Square each of these differences. Divide each squared difference by the expected frequency for its cell. Add up the results for all the cells.

Decide whether to accept or reject the null hypothesis.


Example of Steps of Hypothesis Example of Steps of Hypothesis Testing: Step 1Testing: Step 1Restate the question as a research

hypothesis and a null hypothesis about the population.◦ Population 1: characters on cereal boxes like

those in the study◦ Population 2: characters on cereal boxes for

which the ages distribution of the characters is independent of the gender of the characters

◦ Ha: The proportion of characters that are children and adults are different for male and female characters.

◦ Ho: The proportion of characters that are children and adults are not different for male and female characters.

Step 2: Determine the Step 2: Determine the characteristics of the characteristics of the comparison distribution.comparison distribution.◦ Chi-square distribution◦ Degrees of freedom = (NColumns – 1)(NRows – 1) df = (NGender – 1)(NAge – 1) df = (2-1)(2-1)df = 1


Child 28 30 58 (26.1%)Adult 125 39 164 (73.9%)TOTAL 153 69 222 (100.0%)

Step 3: Determine the cutoff sample Step 3: Determine the cutoff sample score on the comparison distribution at score on the comparison distribution at which the null hypothesis should be which the null hypothesis should be rejected.rejected.◦Decide the significance level you will

use for your study.◦For 1 degree of freedom and a .05

significance level, using the chi-square table your cutoff is 3.841

Step 4: Determine your sample’s Step 4: Determine your sample’s score on the comparison score on the comparison distribution.distribution.◦ Determine the actual observed frequencies in each cell.

male/child cell O = 28 male/adult cell O = 125 female/child cell O = 30 female/adult cell O= 39

◦ Find each row’s percentage of the total. total for the child row = 58 total for the adult row = 164 total = 222 child row percentage = 58 / 222 = 26.1% adult row percentage = 164 / 222 = 73.9%


Child 28 30 58 (26.1%)Adult 125 39 164 (73.9%)TOTAL 153 69 222


◦ EXPECTED FREQUENCIES◦ For each cell, multiply its row’s percentage by the column’s

total.

Gender TOTAL

Age Male Female

Child 28 (.261)(153)=39.9 30 (.261)(69)=18.0 58 (26.1%)

Adult 125 (.739)(153)=113.1 39 (.739)(69)=51.0 164 (73.9%)

TOTAL 153 69 222

Determine your sample’s score on the comparison distribution.◦Figure the X2


Gender TOTAL

Age Male Female O E O E

Child 28 39.9 30 18.0 58 (26.1%)

Adult 125 113.1 39 51.0 164 (73.9%)

TOTAL 153 69 222


O E O E

Child 28 39.9 30 18.0 58 (26.1%)Adult 125 113.1 39 51.0 164 (73.9%)TOTAL 153 69 222

+ + +

Step 5: Step 5: Decide whether to Decide whether to accept or reject the null accept or reject the null hypothesis.hypothesis.◦Your sample is X2 = 15.62; this is

greater than 3.841, the score needed to reject the null hypothesis.

◦You can reject the null hypothesis.Ha: The proportion of characters that are children and adults are different for male and female characters.

Second Example of Hypothesis Second Example of Hypothesis Testing Testing Chi-Square Test for IndependenceChi-Square Test for Independence

Chapter 11 Study Guide Problem 2 (pg. 170)The behavioral scientists categorized the students based on the primary method of conflict resolution used and whether the student had been suspended from school for misbehavior.

(a) Following the five steps of hypothesis testing, conduct the appropriate X2 test to determine whether the observed frequencies are significantly different from the frequencies expected at the .05 level of significance. (b) Explain your results.

Suspended

Aggressive

Manipulative

Passive

Assertive

TOTAL

Yes 7 1 1 1 10No 1 1 1 7 10

TOTAL 8 2 2 8 20

Step 1: Restate the question as a research Step 1: Restate the question as a research hypothesis and a null hypothesis about the hypothesis and a null hypothesis about the population.population.Population 1: Students for whom the primary method of conflict resolution is associated with being or not being suspended from school for misbehavior.Population 2: Students for whom the primary method of conflict resolution is independent of being or not being suspended from school for misbehavior.

Ha: The primary method of conflict resolution used by students who have been suspended from school are different from the methods of students who have not been suspended.Ho: The primary method of conflict resolution used by students who have been suspended from school are no different from the methods of students who have not been suspended.

Step 2: Determine the Step 2: Determine the characteristics of the characteristics of the comparison distribution.comparison distribution.◦ Chi-square distribution◦ Degrees of freedom = (NColumns – 1)(NRows – 1) df = (NMethod – 1)(NSuspended – 1) df = (4-1)(2-1)df = 3

Suspended

Aggressive

Manipulative

Passive

Assertive

TOTAL

Yes 7 1 1 1 10No 1 1 1 7 10

TOTAL 8 2 2 8 20

Step 3: Determine the cutoff sample Step 3: Determine the cutoff sample score on the comparison distribution at score on the comparison distribution at which the null hypothesis should be which the null hypothesis should be rejected.rejected.◦Decide the significance level you will

use for your study.◦For 1 degree of freedom and a .05

significance level, using the chi-square table your cutoff is 7.815

Step 4: Determine your sample’s Step 4: Determine your sample’s score on the comparison score on the comparison distribution.distribution.◦ Determine the actual observed frequencies in

each cell.◦ Find each row’s percentage of the total.

Suspended

Aggressive

Manipulative

Passive

Assertive

TOTAL

Yes 7 1 1 1 10 (50%)

No 1 1 1 7 10 (50%)

TOTAL 8 2 2 8 20


◦ EXPECTED FREQUENCIES◦ For each cell, multiply its row’s percentage by the column’s

total.

Suspended

Aggressive

Manipulative

Passive

Assertive

TOTAL

Yes 7 (4) 1 (1) 1 (1) 1 (4) 10 (50%)

No 1 (4) 1 (1) 1 (1) 7 (4) 10 (50%)

TOTAL 8 2 2 8 20

Determine your sample’s score on the comparison distribution.◦Figure the X2


Suspended

Aggressive

Manipulative

Passive

Assertive

TOTAL

Yes 7 (4) 1 (1) 1 (1) 1 (4) 10 (50%)

No 1 (4) 1 (1) 1 (1) 7 (4) 10 (50%)

TOTAL 8 2 2 8 20

Suspended

Aggressive

Manipulative

Passive

Assertive

TOTAL

Yes 7 (4) 1 (1) 1 (1) 1 (4) 10 (50%)

No 1 (4) 1 (1) 1 (1) 7 (4) 10 (50%)

TOTAL 8 2 2 8 20

Step 5: Step 5: Decide whether to Decide whether to accept or reject the null accept or reject the null hypothesis.hypothesis.◦Your sample is X2 = 9; this is greater

than 7.815, the score needed to reject the null hypothesis.

◦You can reject the null hypothesis.Ha: The primary method of conflict resolution used by students who have been suspended from school are different from the methods of students who have not been suspended.

Assumptions for the Chi-Assumptions for the Chi-Square TestsSquare Tests

Each score must not have any special relationship to any other score.◦You cannot use the chi-square tests if

the scores are based on the same people being tested more than once.


aron chpt 11 ed (2)

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