THE CHI – SQUARE TESTFOR TWO INDEPENDENT SAMPLES

FUNCTION

correlation/test of relationship

nominal or categorical scaling

the question arises as to whether the two variables are independent of each other

used to determine the significance of differences between two independent groups

the hypothesis being tested is usually that the two groups differ with respect to some characteristics

APPLICATION

two political groups differ in their agreement or disagreement with some opinion

sexes differ in the frequency with which they choose certain leisure time activities

METHOD: SAMPLE PROBLEM 150 Male and 50 female HS students were asked about their drink preferences. Among the males, 35 preferred cold drinks and 15 preferred regular drinks. Of the females, 40 preferred cold drinks and 10 preferred regular drinks. Is sex related to drink preference?

1) The data are arranged into a frequency or contingency table in which rows represent groups and each column represents a category of the measured variable. The frequencies shown are observed frequencies.

Cold Regular

Male 35 15 50

Female 40 10 50

75 25 100

2X2 COM

METHOD: SAMPLE PROBLEM 1

2) Expected frequencies must be obtained. For any contingency tables of R rows and C columns, the expected frequencies are obtained by multiplying the appropriate row and column marginal totals, and dividing by the total number of observations N.

Cold Regular

Male 37.5 12.5 50

Female 37.5 12.5 50

75 25 100

COM

METHOD: SAMPLE PROBLEM 1I. H₀: Sex is not related to drink preference or Sex and drink preference are

independent H₁: Sex is related to drink preference or Sex and drink preference are dependent

II. Statistical Test: Since the drink preferences (cold and regular) are categorical variables, and because there were two groups (male and female) , and because the categories are mutually exclusive, the chi – square test for independent groups is appropriate to test H₀.

III. Since we want to accept H₀, we use α = .01 and N = 100. The df is determined by df = (r – 1)(c – 1), where r is the number of rows or groups (2) and c is the number of columns or categories (2). Hence, (2 – 1)(2 – 1) = 1. Appendix Table C indicates that the critical value of χ² is 6.64.

IV. Decision Rule: if χ² ≥ 6.64 reject H₀, otherwise if χ² < 6.64 accept H₀

METHOD: SAMPLE PROBLEM 1V. Computation:

VI. Decision: Since 1.34 < 6.64 then we accept H₀

VII. Thus, sex is not related to drink preference or sex and drink preference are independent.

𝜒2 = (𝑂 − 𝐸)2

𝐸

O E O - E (𝑂 − 𝐸)2(𝑂 − 𝐸)2

𝐸

35 37.5 -2.5 6.25 0.17

15 12.5 2.5 6.25 0.5

40 37.5 -2.5 6.25 0.17

10 12.5 2.5 6.25 0.5

100 100 𝝌𝟐 =1.34

O E

METHOD: SAMPLE PROBLEM 2

Suppose we wish to test whether tall and short people differ with respect to leadership qualities. The table below shows frequencies with which 43 short people and 52 tall people are categorized as “leaders,” “followers,” and “unclassifiable.”

Short Tall Combined

Follower 22 14 36

Unclassifiable 9 6 15

Leader 12 32 44

Total 43 52 95

ATC PAR

2 X 2 CONTINGENCY TABLE

𝜒2 =𝑁 𝐴𝐷 − 𝐵𝐶 2

(𝐴 + 𝐵)(𝐶 + 𝐷)(𝐴 + 𝐶)(𝐵 + 𝐷)

where df = 1

Going back to Sample Problem 1:

𝜒2 =100 (35)(10) − (15)(40) 2

(35 + 15)(40 + 10)(35 + 40)(15 + 10)

𝜒2 = 1.33

PARTITIONING OF DEGREES OF FREEDOM IN R X 2 TABLESIf the table is larger than 2 x 2 and if H₀ is rejected, the contingency table may be partitioned into independent subtables to determine just where the differences are in the original table.

Going back to Sample Problem 2:

Short Tall Short Tall

Followers 22 14 36 Follower or unclassifiable

31 20 51

Unclassifiable 9 6 15 Leader 12 32 44

43 52 95 43 52 95

(1) (2)

PARTITIONING OF DEGREES OF FREEDOM IN R X 2 TABLES

𝜒12 =

𝑁2(𝐴𝐷 − 𝐵𝐶)2

𝐶1𝐶2𝑅2𝑅1(𝑅1 + 𝑅2)𝜒22 =𝑁[𝐹(𝐴 + 𝐶) − 𝐸(𝐵 + 𝐷)]2

𝐶1𝐶2𝑅3 (𝑅1 + 𝑅2)

𝜒12 = .005

With df = 1 at α = .01, H₀ is accepted since .005 < 6.64. Thus, there is no relation between stature and people who are either followers or not classifiable in terms of leadership.

𝜒22 = 10.707

With df = 1 at α = .01, H₀ is rejected since 10.707 ≥ 6.64. Thus, the distribution of leaders and nonleaders differs as a function of stature.

ATC

WHEN TO USE

• When N ≤ 20, always use the

Fisher exact test

• When 20 < N < 40, the χ² test may be used if all expected frequencies are 5 or more. If the smallest expected frequency is less

than 5, use the Fisher exact test

• When N > 40, use χ² corrected

for continuity

The

2 x 2

case

WHEN TO USE

• χ² may be used if fewer than 20% of

the cells have an expected

frequency of less than 5 and if no

cell has an expected frequency of

less than 1

• The researcher should combine

adjacent categories to increase the

expected frequencies in the various

cells.

Contingency

tables with

df > 1

WHEN TO USE

• Yates’ correction for

continuitySmall

Expected

Values

To apply this correction we reduce by .5 the obtained frequencies that are greater than expectation and increase by .5 the obtained frequencies that are less than expectation.

𝜒2 =𝑁 𝐴𝐷 − 𝐵𝐶 −

𝑁2

2

(𝐴 + 𝐵)(𝐶 + 𝐷)(𝐴 + 𝐶)(𝐵 + 𝐷)

APPENDIX TABLE C

MSP1 SP2 PAR

REFERENCES

Siegel, S. and Castellan, N. J. (1988). Nonparametric Statistics for Behavioral Sciences. New York: McGraw Hill.

Ferguson, G. A. and Takane, Y. (1989). Statistical Analysis in Psychology and Education. United States: McGraw Hill.