Chap 11-1 Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall Chapter 11 Chi-Square...

Chap 11-1Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall

Chapter 11

Chi-Square Tests

Business Statistics: A First Course

6th Edition


Learning Objectives

In this chapter, you learn: How and when to use the chi-square test for

contingency tables The 2 test for the difference between two proportions The 2 test for differences in more than two proportions The 2 test for independence


Contingency Tables

Contingency Tables Useful in situations comparing multiple

population proportions Used to classify sample observations according

to two or more characteristics Also called a cross-classification table.

DCOVA


Contingency Table Example

Left-Handed vs. Gender

Dominant Hand: Left vs. Right

Gender: Male vs. Female

2 categories for each variable, so this is called a 2 x 2 table

Suppose we examine a sample of 300 children

DCOVA


Contingency Table Example

Sample results organized in a contingency table:

(continued)

Gender

Hand Preference

Left Right

Female 12 108 120

Male 24 156 180

36 264 300

120 Females, 12 were left handed

180 Males, 24 were left handed

sample size = n = 300:

DCOVA


2 Test for the Difference Between Two Proportions

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)

DCOVA


The Chi-Square Test Statistic

where:

fo = observed frequency in a particular cell

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

(Assumed: each cell in the contingency table has expectedfrequency of at least 5)

cells

22 )(

all e

eoSTAT f

ff

The Chi-square test statistic is:

freedom of degree 1 has case 2x 2 thefor 2STAT

DCOVA


Decision Rule

2

2α

Decision Rule:

If , reject H0, otherwise, do not reject H0

The test statistic approximately follows a chi-squared distribution with one degree of freedom

0

Reject H0Do not reject H0

2STAT

χ

22αSTAT

χ χ

DCOVA


Computing the Average Proportion

Here: 120 Females, 12 were left handed

180 Males, 24 were left handed

i.e., based on all 300 children the proportion of left handers is 0.12, that is, 12%

n

X

nn

XXp

21

21

12.0300

36

180120

2412

p

The average proportion is:

DCOVA


Finding Expected Frequencies

To obtain the expected frequency for left handed females, multiply the average proportion left handed (p) by the total number of females

To obtain the expected frequency for left handed males, multiply the average proportion left handed (p) by the total number of males

If the two proportions are equal, then

P(Left Handed | Female) = P(Left Handed | Male) = .12

i.e., we would expect (.12)(120) = 14.4 females to be left handed(.12)(180) = 21.6 males to be left handed

DCOVA


Observed vs. Expected Frequencies

Gender

Hand Preference

Left Right

FemaleObserved = 12

Expected = 14.4Observed = 108

Expected = 105.6120

MaleObserved = 24


Expected = 158.4180

36 264 300

DCOVA


Gender

Hand Preference

Left Right

FemaleObserved = 12


Expected = 105.6120

MaleObserved = 24


Expected = 158.4180

36 264 300

0.7576158.4

158.4)(156

21.6

21.6)(24

105.6

105.6)(108

14.4

14.4)(12

f

)f(fχ

2222

cells all e

2eo2

STAT


The test statistic is:

DCOVA


Decision Rule

Decision Rule:If > 3.841, reject H0, otherwise, do not reject H0

3.841 d.f. 1 with ; 0.7576 is statistictest The 205.0

2 χχSTAT

Here, =0.7576 < = 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

20.05 = 3.841

0

0.05


2STAT

2STATχ 2

05.0χ

DCOVA


Chi-Square Test In ExcelChi-Square Test

Observed Frequencies

Hand Preference CalculationsGender Left Right Total f0 - feFemale 12 108 120 -2.4 2.4Male 24 156 180 2.4 -2.4Total 36 264 300

Expected Frequencies Hand Preference

Gender Left Right Total (f0 - fe)^2/feFemale 14.4 105.6 120 0.40 0.05Male 21.6 158.4 180 0.27 0.04Total 36 264 300

DataLevel of Significance 0.05Number of Rows 2Number of Columnns 2Degrees of Freedom 1=(B19-1)*(B20-1)

ResultsCritical Value 3.8415=CHIINV(B18,B21)Chi-Square Test Statistic 0.7576=SUM(F13:G14)p-Value 0.3841=CHIDIST(B25,B21)

Do not reject the null hypothesis =IF(B26<B18,"Reject the null hypothesis","Do not reject the null hypothesis")

Expected Frequency assumption is met. =IF(OR(B13<5,C13<5,B14<5,C14<5),

" is violated."," is met.")

DCOVA


Chi-Square Test In Minitab

Tabulated statistics: Gender, Hand

(Expected counts are below observed)

(Chi-Square cell contribution below expected counts)

Rows: Gender Columns: Hand

Left Right All

Female 12 108 120

14.4 105.6 120.0

0.40000 0.05455 *

Male 24 156 180

21.6 158.4 180.0

0.26667 0.03636 *

All 36 264 300

36.0 264.0 300.0

* * *

Pearson Chi-Square = 0.758, DF = 1, P-Value = 0.384

DCOVA


Extend the 2 test to the case with more than two independent populations:

2 Test for Differences Among More Than Two Proportions

H0: π1 = π2 = … = πc

H1: Not all of the πj are equal (j = 1, 2, …, c)

DCOVA



Where:

fo = observed frequency in a particular cell of the 2 x c table


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

cells

22 )(

all e

eoSTAT f

ff


freedom of degrees 1-c 1)-1)(c-(2 has case cx 2 thefor χ 2 STAT

DCOVA


Computing the Overall Proportion

n

X

nnn

XXXp

c

c

21

21 The overall proportion is:

Expected cell frequencies for the c categories are calculated as in the 2 x 2 case, and the decision rule is the same:

Where is from the chi-squared distribution with c – 1 degrees of freedom

Decision Rule:If , reject H0, otherwise, do not reject H0

22αSTAT

χ χ

2α

χ

DCOVA


Example of 2 Test for Differences Among More Than Two Proportions

A University is thinking of switching to a trimester academic calendar. A random sample of 100 administrators, 50 students, and 50 faculty members were surveyed

Opinion Administrators Students Faculty

Favor 63 20 37

Oppose 37 30 13

Totals 100 50 50

Using a 1% level of significance, which groups have a different attitude?

DCOVA


Chi-Square Test Results

Chi-Square Test: Administrators, Students, Faculty

Admin Students Faculty Total

Favor 63 20 37 120

60 30 30

Oppose 37 30 13 80

40 20 20

Total 100 50 50 200

ObservedExpected

H0: π1 = π2 = π3

H1: Not all of the πj are equal (j = 1, 2, 3)

02

0102 Hreject so 9.2103 79212

.STATχ.χ

DCOVA


Excel Output For The ExampleDCOVA

Chi-Square Test

Observed Frequencies Role Calculations

Opinion Admin Students Faculty Total fo - feFavor 63 20 37 120 3.0000 -10.0000 7

Oppose 37 30 13 80 -3.0000 10.0000 -7Total 100 50 50 200

Expected Frequencies Role

Opinion Admin Students Faculty Total (fo - fe)^2/feFavor 60.0 30.0 30.0 120 0.1500 3.3333 1.6333

Oppose 40.0 20.0 20.0 80 0.2250 5.0000 2.4500Total 100 50 50 200

DataLevel of Significance 0.01Number of Rows 2Number of Columns 3Degrees of Freedom 2=($B$19 - 1) * ($B$20 - 1)

ResultsCritical Value 9.2103=CHIINV(B18, B21)Chi-Square Test Statistic 12.7917=SUM(G13:I14)p-Value 0.0017=CHIDIST(B25, B21)

Reject the null hypothesis =IF(B26<B18, "Reject the null hypothesis",Do not reject the null hypothesis")

Expected frequency assumption is met. =IF(OR(B13<1, C13<1, D13<1, B14<1, C14<1, D14<1),

" is violated.", " is met.")


Minitab Output For The ExampleDCOVA

Tabulated statistics: Position, Role

Rows: Position Columns: Role

Admin Faculty Students All

Favor 63 37 20 120

60 30 30 120

0.150 1.633 3.333 *

Oppose 37 13 30 80

40 20 20 80

0.225 2.450 5.000 *

All 100 50 50 200

100 50 50 200

* * * *



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)

DCOVA


2 Test of Independence

where:

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


(Assumed: each cell in the contingency table has expected

frequency of at least 1)

cells

22

)(

all e

eoSTAT f

ffχ


(continued)

freedom of degrees 1)-1)(c-(r has case cr x for the χ 2STAT

DCOVA


Expected Cell Frequencies

Expected cell frequencies:

n

total columntotalrow fe

Where:

row total = sum of all frequencies in the row

column total = sum of all frequencies in the column

n = overall sample size

DCOVA


Decision Rule

The decision rule is

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

If , reject H0,

otherwise, do not reject H0

22αSTAT

χ χ

2α

χ

DCOVA


Example

The meal plan selected by 200 students is shown below:

ClassStanding

Number of meals per weekTotal20/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

DCOVA


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)

DCOVA


ClassStanding

Number of meals per week

Total20/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

ClassStanding

Number of meals per week

Total20/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

Observed:

Expected cell frequencies if H0 is true:

5.10200

7030

n

total columntotalrow fe

Example for one cell:

Example: Expected Cell Frequencies

(continued)

DCOVA


Example: The Test Statistic

The test statistic value is:

709048

4810

830

83032

524

52424 222

cells

22

..

).(

.

).(

.

).(

f

)ff(χ

all e

eoSTAT

(continued)

= 12.592 from the chi-squared distribution with (4 – 1)(3 – 1) = 6 degrees of freedom

2050.

χ

DCOVA


Example: Decision and Interpretation

(continued)

Decision Rule:If > 12.592, reject H0, otherwise, do not reject H0

12.592 d.f. 6 with ; 7090 is statistictest The 2050

2 .STAT

χ.χ

Here, = 0.709 < = 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

20.05=12.592

0

0.05


2STAT

χ

2STAT

χ 2050.

χ

DCOVA


Excel Output Of ExampleDCOVAChi-Square Test of Independence

Observed Frequencies Meals Per Week Calculations

Class Standing 20/wk 10/wk None Total fo - feFresh. 24 32 14 70 -0.5000 1.2000 -0.7000Soph. 22 26 12 60 1.0000 -0.4000 -0.6000Junior 10 14 6 30 -0.5000 0.8000 -0.3000Senior 14 16 10 40 0.0000 -1.6000 1.6000Total 70 88 42 200

Expected Frequencies Meals Per Week

Class Standing 20/wk 10/wk None Total (fo - fe)^2/feFresh. 24.5000 30.8000 14.7000 70 0.0102 0.0468 0.0333Soph. 21.0000 26.4000 12.6000 60 0.0476 0.0061 0.0286Junior 10.5000 13.2000 6.3000 30 0.0238 0.0485 0.0143Senior 14.0000 17.6000 8.4000 40 0.0000 0.1455 0.3048Total 70 88 42 200

DataLevel of Significance 0.05Number of Rows 4Number of Columns 3Degrees of Freedom 6=($B$23 - 1) * ($B$24 - 1)

ResultsCritical Value 12.5916=CHIINV(B22, B25)Chi-Square Test Statistic 0.7093=SUM($G$15:$I$18)p-Value 0.9943=CHIDIST(B29, B25)

Do not reject the null hypothesis =IF(B30<B22, "Reject the null hypothesis",Do not reject the null hypothesis)

Expected frequency assumption is met. =IF(OR(B15<1, C15<1, D15<1, B16<1, C16<1, D16<1, B17<1, C17<1,

D17<1, B18<1, C18<1, D18<1), " is violated.", " is met.")


Minitab Output Of ExampleDCOVA

Tabulated statistics: Class, Meals

Rows: Class Columns: Meals

0 10 20 All

Fresh 14 32 24 70

14.70 30.80 24.50 70.00

0.03333 0.04675 0.01020 *

Junior 6 14 10 30

6.30 13.20 10.50 30.00

0.01429 0.04848 0.02381 *

Senior 10 16 14 40

8.40 17.60 14.00 40.00

0.30476 0.14545 0.00000 *

Soph 12 26 22 60

12.60 26.40 21.00 60.00

0.02857 0.00606 0.04762 *

All 42 88 70 200

42.00 88.00 70.00 200.

* * * *



Chapter Summary

Developed and applied the 2 test for the difference between two proportions

Developed and applied the 2 test for differences in more than two proportions

Examined the 2 test for independence


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