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Chap 12-*

Chapter 12

Sections 12.1 and 12.3

Chi-Square Tests

Basic Business Statistics
10th Edition

Chap 12-*

Learning Objectives

In this chapter, you learn:

How and when to use the chi-square test for contingency tables

Chap 12-*

Contingency Tables

Useful in situations involving two or more variables that are in categories.
Used to classify sample observations according to two or more characteristics
Also called a cross-classification table.

Chap 12-*

Contingency Table Example

Left-Handed vs. Gender

Dominant Hand: Left vs. Right

Gender: Male vs. Female

2 categories for each variable, so

called a 2 x 2 table

Suppose we examine a sample of

size 300

Chap 12-*

Contingency Table Example

Sample results organized in a contingency table:

(continued)

120 Females, 12 were left handed

180 Males, 24 were left handed

sample size = n = 300:

Gender	Hand Preference
Left	Right
Female	12	108	120
Male	24	156	180
36	264	300

Chap 12-*

Hypothesis

If H0 is true, then the proportion of left-handed females should be the same as the proportion of left-handed males
The two proportions above should be the same as the proportion of left-handed people overall

H0: π1 = π2 (Proportion of females who are left

handed is equal to the proportion of

males who are left handed)

H1: π1 ≠ π2 (The two proportions are not the same –

Hand preference is not independent

of gender)

Chap 12-*

Critical Value

2

2U

Decision Rule:

If statistic 2 > 2U, reject H0, otherwise, do not reject H0

The 2 test statistic approximately follows a chi-squared distribution with df = (r-1)(c-1)

r: number of rows

c: number of columns



Reject H0

Do not

reject H0

Chap 12-*

The Chi-Square Test Statistic

where:

fo = observed frequency in a particular cell

fe = expected frequency based on the total sample

2 for the 2 x 2 case has 1 degree of freedom

The Chi-square test statistic is:

Chap 12-*

Observed vs. Expected Frequencies

Gender	Hand Preference
Left	Right
Female	Observed = 12 Expected = 14.4	Observed = 108 Expected = 105.6	120
Male	Observed = 24 Expected = 21.6	Observed = 156 Expected = 158.4	180
36	264	300

Chap 12-*

Finding Expected Frequencies

To obtain the expected frequency for left handed females, multiply the average proportion left handed (56/300 = 0.12) by the total number of females
To obtain the expected frequency for left handed males, multiply the average proportion left handed (0.12) by the total number of males

we expect (.12)(120) = 14.4 females to be left handed

(.12)(180) = 21.6 males to be left handed

Chap 12-*

The Chi-Square Test Statistic

The test statistic is:

Gender	Hand Preference
Left	Right
Female	Observed = 12 Expected = 14.4	Observed = 108 Expected = 105.6	120
Male	Observed = 24 Expected = 21.6	Observed = 156 Expected = 158.4	180
36	264	300

Chap 12-*

Conclusion

Decision Rule:

If 2 > 3.841, reject H0, otherwise, do not reject H0

Here,

2 = 0.7576 < 2U = 3.841,

so we do not reject H0 and conclude that there is not sufficient evidence that the two proportions are different at  = 0.05

2

2U=3.841



Reject H0

Do not

reject H0

Chap 12-*

Example: 2 Test of Independence

Similar to the 2 test for equality of more than two proportions, but extends the concept to contingency tables with r rows and c columns

H0: The two categorical variables are independent

(i.e., there is no relationship between them)

H1: The two categorical variables are dependent

(i.e., there is a relationship between them)

Chap 12-*

2 Test of Independence

where:

fo = observed frequency in a particular cell of the r x c table

fe = expected frequency in a particular cell if H0 is true

2 for the r x c case has (r-1)(c-1) degrees of freedom

(Assumed: each cell in the contingency table has expected frequency of at least 1)

The Chi-square test statistic is:

(continued)

Chap 12-*

Decision Rule

The decision rule is

If 2 > 2U, reject H0, otherwise, do not reject H0

Where 2U is from the chi-squared distribution with (r – 1)(c – 1) degrees of freedom

Chap 12-*

Example

The meal plan selected by 200 students is shown below:

Class Standing	Number of meals per week	Total
20/week	10/week	none
Fresh.	24	32	14	70
Soph.	22	26	12	60
Junior	10	14	6	30
Senior	14	16	10	40
Total	70	88	42	200

Chap 12-*

Example

The hypothesis to be tested is:

(continued)

H0: Meal plan and class standing are independent

(i.e., there is no relationship between them)

H1: Meal plan and class standing are dependent

(i.e., there is a relationship between them)

Chap 12-*

Observed:

Expected cell frequencies if H0 is true:

Example for one cell:

Example:
Expected Cell Frequencies

(continued)

Class Standing	Number of meals per week	Total
20/wk	10/wk	none
Fresh.	24	32	14	70
Soph.	22	26	12	60
Junior	10	14	6	30
Senior	14	16	10	40
Total	70	88	42	200

Class Standing	Number of meals per week	Total
20/wk	10/wk	none
Fresh.	24.5	30.8	14.7	70
Soph.	21.0	26.4	12.6	60
Junior	10.5	13.2	6.3	30
Senior	14.0	17.6	8.4	40
Total	70	88	42	200

Chap 12-*

Example: The Test Statistic

The test statistic value is:

(continued)

2U = 12.592 for  = 0.05 from the chi-squared distribution with (4 – 1)(3 – 1) = 6 degrees of freedom

Chap 12-*

Example:
Decision and Interpretation

(continued)

Decision Rule:

If 2 > 12.592, reject H0, otherwise, do not reject H0

Here,

2 = 0.709 < 2U = 12.592,

so do not reject H0

Conclusion: there is not sufficient evidence that meal plan and class standing are related at  = 0.05

2

2U=12.592



Reject H0

Do not

reject H0

cells

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158.4)

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21.6

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test

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200

total

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total

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709

)

(

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(

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