© 2004 Prentice-Hall, Inc.Chap 12-1 Basic Business Statistics (9 th Edition) Chapter 12 Tests for Two or More Samples with Categorical Data

© 2004 Prentice-Hall, Inc. Chap 12-1

Basic Business Statistics(9th Edition)

Chapter 12Tests for Two or More

Samples with Categorical Data


Chapter Topics

Z Test for Differences in Two Proportions(Independent Samples)

2 Test for Differences in Two Proportions(Independent Samples)

2 Test for Differences in More than Two Proportions (Independent Samples)

Marascuilo Procedure 2 Test of Independence


Z Test for Differences in Two Proportions

What is It Used For? To determine whether there is a difference

between 2 population proportions or whether one is larger than the other

Assumptions: Independent samples Population follows binomial distribution Sample size large enough: np 5 and n(1-p)

5 for each population


Z Test Statistic

1 2 1 2

1 2

( ) ( )

1 1(1 )

s sp p p pZ

p pn n

1 2

1 2

X Xp

n n

where

X1 = Number of Successes in Sample 1

X2 = Number of Successes in Sample 2

Pooled Estimate of the Population Proportion


The Hypotheses for the Z Test

Research Questions

Hypothesis No DifferenceAny Difference

Prop 1 Prop 2Prop 1 < Prop 2

Prop 1 Prop 2Prop 1 > Prop 2

H0 p1 - p2 p1 - p2 0 p1 - p2 0

H1 p1 - p2 0 p1 - p2 < 0 p1 - p2 > 0


1

2

1

2

63.808 78

7849

.598 8282

5 1 5

for both pop.

S

S

P n

P n

np n p

Z Test for Differences in Two Proportions: Example

As personnel director, you want to test the perception of fairness of two methods of performance evaluation. 63 of 78 employees rated Method 1 as fair. 49 of 82 rated Method 2 as fair. At the 0.01 significance level, is there a difference in perceptions?


Calculating the Test Statistic

1 2 1 2

1 2

( ) ( )

1 1(1 )

(.808 .598) 02.90

1 1(.70)(.30)

78 82

s sp p p pZ

p pn n

1 2

1 2

63 49.70

78 82

X Xp

n n


Z Test for Differences in Two Proportions: Solution

H0: p1 - p2 = 0

H1: p1 - p2 0

= 0.01n1 = 78 n2 = 82

Critical Value(s):

Test Statistic:

Decision:

Conclusion:

Reject at = 0.01.

There is evidence of a difference in proportions.

Z 290.

Z0 2.58-2.58

.005

Reject H0 Reject H0

.005


Z Test for Differences in Two Proportions in PHStat

PHStat | Two-Sample Tests | Z Test for Differences in Two Proportions …

Example Solution in Excel Spreadsheet

Microsoft Excel Worksheet


Confidence Interval for Differences in Two

Proportions

The Confidence Interval for Differences in Two Proportions

100 1 %

1 1 2 2

1 2 / 21 2

1 1S S S S

S S

p p p pp p Z

n n


1

2

1

2

63.808 78

7849

.598 8282

5 1 5

for both pop.

S

S

P n

P n

np n p

Confidence Interval for Differences in Two Proportions:

Example

As personnel director, you want to find out the perception of fairness of two methods of performance evaluation. 63 of 78 employees rated Method 1 as fair. 49 of 82 rated Method 2 as fair. Construct a 99% confidence interval for the difference in two proportions.


Confidence Interval for Differences in Two Proportions:

Solution

1 1 2 2

1 2 / 21 2

1 2

1 1

0.8077 1 0.8077 0.5976 1 0.59760.8077 0.5976 2.5758

78 820.0294 0.3909

S S S S

S S

p p p pp p Z

n n

p p

We are 99% confident that the difference between two proportions is somewhere between 0.0294 and 0.3909.


2 Test for Two Proportions: Basic Idea

Compares Observed to Expected Frequencies if Null Hypothesis is True

The Closer Observed Frequencies are to Expected Frequencies, the More Likely the H0 is True Measured by squared difference relative to

expected frequency Sum of relative squared differences is the

test statistic


2 Test for Two Proportions: Contingency Table

Evaluation MethodPerception 1 2 Total

Fair 63 49 112Unfair 15 33 48

Total 78 82 160

Contingency Table (Observed Frequencies) for Comparing Fairness of Performance Evaluation Methods

2 Populations

Levels of Variable


2 Test for Two Proportions: Expected Frequencies

112 of 160 Total are “Fair” ( ) 78 Used Evaluation Method 1 Expect (78 112/160) = 54.6 to be

“Fair”

112 /160p

Evaluation MethodPerception 1 2 Total

Fair 63 49 112Unfair 15 33 48

Total 78 82 160


The 2 Test Statistic

2

02

All Cells

: Observed frequency in cell

: Theoretical or expected frequency

= row total column total / n

e

e

o

e

f f

f

f

f


Computation of the 2

Test Statistic

f0 fe (f0 - fe) (f0 - fe)2 (f0 - fe)2 / fe

63 54.6 8.4 70.56 1.293

49 57.4 -8.4 70.56 1.229

15 23.4 -8.4 70.56 3.015

33 24.6 8.4 70.56 2.868

Sum = 8.405Observed Frequencies Expected Frequencies


2 Test for Two Proportions:Finding the Critical Value

0 6.635

Rejectr = 2 (# rows in contingency table)

c = 2 (# columns)

= .01= .01

df = (r - 1)(c - 1) = 1

2 Table (Portion)

Upper Tail AreaDF .995 ….95

…

.051 ... …0.004 3.8412 0.010 0.103 5.991

.025 .015.0247.378

6.6359.210…

……


2 Test for Two Proportions:Solution

H0: p1 - p2 = 0 H1: p1 - p2 0

Test Statistic = 8.405Decision:

Conclusion:

6.635 0

Reject

= .01

Reject at = 0.01.

There is evidence of a difference in proportions.Note: The conclusion obtained using test is the same as using Z Test.

Caution! Each

expected frequency

should be 5.


2 Test for Two Proportionsin PHStat

PHStat | Two-Sample Tests | Chi-Square Test for Differences in Two Proportions …




2 Test for More Than Two Proportions

Extends the 2 Test to the General Case of c Independent Populations

Tests for Equality (=) of Proportions Only

Uses Contingency Table Assumptions:

Independent random samples “Large” sample sizes All expected frequencies 1


2 Test for c Proportions:Hypotheses and Statistic

Hypotheses H0: p1 = p2 = ... = pc H1: Not all pj are equal

Test statistic

Degrees of freedom: (r - 1)(c - 1)

2

2

all cells

o e

e

f f

f

Observed frequencyObserved frequency

Expected frequencyExpected frequency

# Rows # Columns# Rows # Columns


2 Test for c Proportions:Example

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

Opinion Under Grad Faculty

Favor 63 20 37

Oppose 37 30 13

Totals 100 50 50

Test at the .01 level of significance to determine if there is evidence of a difference in attitude between the groups.



(continued)

1. Set Hypotheses:H0: p1 = p2 = p3 H1: Not all pj are equal

2. Contingency Table:

Opinion Under Grad Faculty Totals

Favor 63 20 37 120

Oppose 37 30 13 80

Totals 100 50 50 200



(continued)

Opinion Under Grad Faculty Totals

Favor 60 30 30 120

Oppose 40 20 20 80

Totals 100 50 50 200

3. Compute Expected Frequencies

(100)(120)/200=60(50)(80)/200=20

All expected frequencies are large.



(continued)

4. Compute Test Statistic:

f0 fe (f0 - fe) (f0 - fe)2 (f0 - fe)2 / fe

63 60 3 9 .15

20 30 -10 100 3.3333

37 30 7 49 1.6333

37 40 -3 9 .225

30 20 10 100 5

13 20 -7 49 2.45Test Statistic 2 = 12.792


2 Test for c Proportions:Example Solution

H0: p1 = p2 = p3

H1: Not all pj are equal

Decision:

Conclusion:

df = (c – 1)(r - 1) = 3 - 1 = 2

Reject

= .01

20 9.210

Do Not Reject H0.

Since 2 =12.792, there is sufficient evidence of a difference in attitude among the groups.


2 Test for c Proportionsin PHStat

PHStat | c-Sample Tests | Chi-Square Test …




Marascuilo Procedure Used when the Test for c Proportions is

Rejected Compares All Pairs of Groups The Marascuilo Multiple Comparison

Procedure:

Compute among all pairs of groups The critical range for a pair is

A pair is considered significantly different if critical range

'j jS Sp p

'j j

' '

'

211

j j j jS SS S

Uj j

p pp p

n n

'j jS Sp p

2


Marascuilo Procedure :Example

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

Opinion Under Grad 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?


Marascuilo Procedure :Solution

Marascuilo Procedure

Sample Sample Absolute Std. Error CriticalGroup Proportion Size Comparison Difference of Difference Range Results

1 0.63 100 1 to 2 0.23 0.08444525 0.256 Means are not different2 0.4 50 1 to 3 0.11 0.07860662 0.239 Means are not different3 0.74 50 2 to 3 0.34 0.09299462 0.282 Means are different

0.01 9.212

3.034856

Chi-sq Critical ValueOther Data

d.fQ Statistic

Level of significance

At 1% level of significance, there is evidence of a difference in attitude between graduate students and faculty.

Excel Output:


2 Test of Independence

Shows If a Relationship Exists between 2 Factors of Interest One sample drawn Each factor has 2 or more levels of responses Does not show nature of relationship Does not show causality

Similar to Testing p1 = p2 = … = pc

Used Widely in Marketing Uses Contingency Table


2 Test of Independence:Example

A survey was conducted to determine whether there is a relationship between architectural style (Split-Level or Ranch) and geographical location (Urban or Rural).

Given the survey data, test at the = .01 level to determine whether there is a relationship between location and architectural style.


House Location

House Style Urban Rural Total

Split-Level 63 49 112Ranch 15 33 48

Total 78 82 160


1. Set Hypotheses:

H0: The 2 categorical variables (Architectural Style and Location) are independent

H1: The 2 categorical variables are related

2. Contingency Table:

Levels of Variable 2

Levels of Variable 1

(continued)



(continued)

3. Computing Expected Frequencies Statistical independence : P(A and B) = P(A)·P(B) Compute marginal (row & column) probabilities & multiply

for joint probability Expected frequency is sample size times joint probability

House LocationUrban Rural

House Style Obs. Exp. Obs. Exp. TotalSplit-Level 63 54.6 49 57.4 112Ranch 15 23.4 33 24.6 48

Total 78 78 82 82 160

78·112160

82·112160


f0 fe (f0 - fe) (f0 - fe)2 (f0 - fe)2 / fe

63 54.6 8.4 70.56 1.292

49 57.4 -8.4 70.56 1.229

15 23.4 -8.4 70.56 3.015

33 24.6 8.4 70.56 2.868

8.404


(continued)

4. Calculate Test Statistic: 2

02

All Cells

e

e

f f

f

2 Test Statistic = 8.404 All expected frequencies are large, i.e. > 1.


2 Test of Independence:Example Solution

H0: The 2 categorical variables (Architectural Style and

Location) are independent H1: The 2 categorical variables are related

Decision:

Conclusion:

df = (r - 1)(c - 1) = 1 Reject

= .01

0 6.635

Reject H0 at = .01.

Since 2 =8.404, there is evidence that the choice of architectural design and location are related.

2


2 Test of Independencein PHStat

PHStat | c-Sample Tests | Chi-Square Test …




Chapter Summary

Performed Z Test for Differences in Two Proportions (Independent Samples)

Discussed Test for Differences in Two Proportions (Independent Samples)

Addressed 2 Test for Differences in More Than Two Proportions (Independent Samples)

Illustrated Marascuilo Procedure Described 2 Test of Independence

Documents

© 2004 Prentice-Hall, Inc.Chap 12-1 Basic Business Statistics (9 th Edition) Chapter 12 Tests for Two or More Samples with Categorical Data