1 Chi-Square Heibatollah Baghi, and Mastee Badii.

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Chi-Square

Heibatollah Baghi, and

Mastee Badii

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Different Scales, Different Measures of Association

Scale of Both Variables

Measures of Association

Nominal Scale Pearson Chi-Square: χ2

Ordinal Scale Spearman’s rho

Interval or Ratio Scale

Pearson r

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Chi-Square (χ2) and Frequency Data

Up to this point, the inference to the population has been concerned with “scores” on one or more variables, such as CAT scores, mathematics achievement, and hours spent on the computer.We used these scores to make the inferences about population means. To be sure not all research questions involve score data.Today the data that we analyze consists of frequencies; that is, the number of individuals falling into categories. In other words, the variables are measured on a nominal scale.The test statistic for frequency data is Pearson Chi-Square. The magnitude of Pearson Chi-Square reflects the amount of discrepancy between observed frequencies and expected frequencies.

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Steps in Test of Hypothesis

1. Determine the appropriate test 2. Establish the level of significance:α3. Formulate the statistical hypothesis4. Calculate the test statistic5. Determine the degree of freedom6. Compare computed test statistic against a

tabled/critical value

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1. Determine Appropriate Test

Chi Square is used when both variables are measured on a nominal scale.It can be applied to interval or ratio data that have been categorized into a small number of groups.It assumes that the observations are randomly sampled from the population.All observations are independent (an individual can appear only once in a table and there are no overlapping categories).It does not make any assumptions about the shape of the distribution nor about the homogeneity of variances.

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2. Establish Level of Significance

α is a predetermined value

The convention• α = .05

• α = .01

• α = .001

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3. Determine The Hypothesis:Whether There is an Association

or NotHo : The two variables are independent

Ha : The two variables are associated

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4. Calculating Test Statistics

Contrasts observed frequencies in each cell of a contingency table with expected frequencies.The expected frequencies represent the number of cases that would be found in each cell if the null hypothesis were true ( i.e. the nominal variables are unrelated).Expected frequency of two unrelated events is product of the row and column frequency divided by number of cases.

Fe= Fr Fc / N

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e

eo

F

FF 22 )(

10


e

eo

F

FF 22 )(

Observed

frequencies

Expe

cted

fre

quen

cy

Expected

frequency

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5. Determine Degrees of Freedom

df = (R-1)(C-1)

Num

ber of

levels in column

variable

Num

ber of levels in row

variable

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6. Compare computed test statistic against a tabled/critical value

The computed value of the Pearson chi- square statistic is compared with the critical value to determine if the computed value is improbable

The critical tabled values are based on sampling distributions of the Pearson chi-square statistic

If calculated 2 is greater than 2 table value, reject Ho

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Example

Suppose a researcher is interested in voting preferences on gun control issues.

A questionnaire was developed and sent to a random sample of 90 voters.

The researcher also collects information about the political party membership of the sample of 90 respondents.

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Bivariate Frequency Table or Contingency Table

Favor Neutral Oppose f row

Democrat 10 10 30 50

Republican 15 15 10 40

f column 25 25 40 n = 90

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f column 25 25 40 n = 90

Observ

ed

frequ

encie

s

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f column 25 25 40 n = 90

Row

frequency

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f column 25 25 40 n = 90Column frequency

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1. Determine Appropriate Test

1. Party Membership ( 2 levels) and Nominal

2. Voting Preference ( 3 levels) and Nominal

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2. Establish Level of Significance

Alpha of .05

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3. Determine The Hypothesis

• Ho : There is no difference between D & R in their opinion on gun control issue.

• Ha : There is an association between responses to the gun control survey and the party membership in the population.

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Democrat fo =10

fe =13.9

fo =10

fe =13.9

fo =30

fe=22.2

50

Republican fo =15

fe =11.1

fo =15

fe =11.1

fo =10

fe =17.8

40

f column 25 25 40 n = 90

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Democrat fo =10

fe =13.9

fo =10

fe =13.9

fo =30

fe=22.2

50

Republican fo =15

fe =11.1

fo =15

fe =11.1

fo =10

fe =17.8

40

f column 25 25 40 n = 90

= 50*25/90

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Democrat fo =10

fe =13.9

fo =10

fe =13.9

fo =30

fe=22.2

50

Republican fo =15

fe =11.1

fo =15

fe =11.1

fo =10

fe =17.8

40

f column 25 25 40 n = 90

= 40* 25/90

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8.17

)8.1710(

11.11

)11.1115(

11.11

)11.1115(

2.22

)2.2230(

89.13

)89.1310(

89.13

)89.1310(

222

2222

= 11.03

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5. Determine Degrees of Freedom

df = (R-1)(C-1) =(2-1)(3-1) = 2

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6. Compare computed test statistic against a tabled/critical value

α = 0.05df = 2Critical tabled value = 5.991Test statistic, 11.03, exceeds critical valueNull hypothesis is rejectedDemocrats & Republicans differ significantly in their opinions on gun control issues

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SPSS Output for Gun Control Example

Chi-Square Tests

11.025a 2 .004

11.365 2 .003

8.722 1 .003

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Pearson Chi-Square

Likelihood Ratio

Linear-by-LinearAssociation

N of Valid Cases

Value dfAsymp. Sig.

(2-sided)

0 cells (.0%) have expected count less than 5. Theminimum expected count is 11.11.

a.

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Additional Information in SPSS Output

Exceptions that might distort χ2 Assumptions– Associations in some but not all categories– Low expected frequency per cell

Extent of association is not same as statistical significance

Demonstratedthrough an example

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Another Example Heparin Lock Placement

Complication Incidence * Heparin Lock Placement Time Group Crosstabulation

9 11 20

10.0 10.0 20.0

18.0% 22.0% 20.0%

41 39 80

40.0 40.0 80.0

82.0% 78.0% 80.0%

50 50 100

50.0 50.0 100.0

100.0% 100.0% 100.0%

Count

Expected Count

% within Heparin LockPlacement Time Group

Count

Expected Count


Count

Expected Count


Had Compilca

Had NO Compilca

ComplicationIncidence

Total

1 2

Heparin LockPlacement Time Group

Total

from Polit Text: Table 8-1

Time:1 = 72 hrs 2 = 96 hrs

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Hypotheses in Heparin Lock Placement

Ho: There is no association between complication incidence and length of heparin lock placement. (The variables are independent).

Ha: There is an association between complication incidence and length of heparin lock placement. (The variables are related).

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More of SPSS Output

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Pearson Chi-Square

Pearson Chi-Square = .250, p = .617

Since the p > .05, we fail to reject the null hypothesis that the complication rate is unrelated to heparin lock placement time.Continuity correction is used in situations in which the expected frequency for any cell in a 2 by 2 table is less than 10.

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More SPSS Output

Symmetric Measures

-.050 .617

.050 .617

-.050 .100 -.496 .621c

-.050 .100 -.496 .621c

100

Phi

Cramer's V

Nominal byNominal

Pearson's RInterval by Interval

Spearman CorrelationOrdinal by Ordinal

N of Valid Cases

ValueAsymp.

Std. Errora

Approx. Tb

Approx. Sig.

Not assuming the null hypothesis.a.

Using the asymptotic standard error assuming the null hypothesis.b.

Based on normal approximation.c.

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Phi Coefficient

Pearson Chi-Square provides information about the existence of relationship between 2 nominal variables, but not about the magnitude of the relationship

Phi coefficient is the measure of the strength of the association

Symmetric Measures

-.050

.050

-.050

-.050

100

Phi

Cramer's V

Nominal byNominal



N of Valid Cases

Value




N

2

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Cramer’s V

When the table is larger than 2 by 2, a different index must be used to measure the strength of the relationship between the variables. One such index is Cramer’s V.If Cramer’s V is large, it means that there is a tendency for particular categories of the first variable to be associated with particular categories of the second variable.

Symmetric Measures

-.050

.050

-.050 .100

-.050 .100

100

Phi

Cramer's V

Nominal byNominal



N of Valid Cases

ValueAsymp.

Std. Error




)1(

2

kNV

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Cramer’s V

When the table is larger than 2 by 2, a different index must be used to measure the strength of the relationship between the variables. One such index is Cramer’s V.If Cramer’s V is large, it means that there is a tendency for particular categories of the first variable to be associated with particular categories of the second variable.

Symmetric Measures

-.050

.050

-.050 .100

-.050 .100

100

Phi

Cramer's V

Nominal byNominal



N of Valid Cases

ValueAsymp.

Std. Error




)1(

2

kNV

Number of cases

Smallest of number of rows or columns

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Take Home Lesson

How to Test Association between Frequency of Two Nominal Variables

1 Chi-Square Heibatollah Baghi, and Mastee Badii.

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