chap 12-1 copyright ©2012 pearson education, inc. publishing as prentice hall chap 12-1 chapter 12...

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 12-1Chap 12-1

Chapter 12

Chi-Square Tests and Nonparametric Tests

Basic Business Statistics12th Edition

Chap 12-2Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 12-2

Learning Objectives

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

contingency tables How to use the Marascuilo procedure for

determining pairwise differences when evaluating more than two proportions

How and when to use the McNemar test How to use the Chi-Square to test for a variance

or standard deviation How and when to use nonparametric tests


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

Chap 12-14Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

Chi-Square Test In Excel

Chap 12-14

Chi-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 Example

Chap 12-21

DCOVAChi-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 Example

Chap 12-22

DCOVATabulated 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

* * * *



The Marascuilo Procedure

Used when the null hypothesis of equal proportions is rejected

Enables you to make comparisons between all pairs

Start with the observed differences, pj – pj’, for all pairs (for j ≠ j’) then compare the absolute difference to a calculated critical range

DCOVA


The Marascuilo Procedure

Critical Range for the Marascuilo Procedure:

(Note: the critical range is different for each pairwise comparison)

A particular pair of proportions is significantly different if

| pj – pj’| > critical range for j and j’

(continued)

'j

'j'j

j

jj2

n

)p(1p

n

)p(1pχrange Critical

α

DCOVA


Marascuilo Procedure Example

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


Marascuilo Procedure: Solution

Marascuilo Procedure

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

1 0.63 100 1 to 2 0.23 0.084445249 0.2563 Means are not different2 0.4 50 1 to 3 0.11 0.078606615 0.2386 Means are not different3 0.74 50 2 to 3 0.34 0.092994624 0.2822 Means are different

0.01 9.21032

3.034854

Chi-sq Critical ValueOther Data

d.fQ Statistic

Level of significance

Calculations In Excel:

compare

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

DCOVA

Minitab does not do the Marascuilo procedure


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 Example

Chap 12-37

DCOVAChi-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 Example

Chap 12-38

DCOVATabulated 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.

* * * *



McNemar Test (Related Samples)

Used to determine if there is a difference between proportions of two related samples

Uses a test statistic the follows the normal distribution

DCOVA



Consider a 2 X 2 contingency table:

(continued)

Condition 2

Condition 1 Yes No Totals

Yes A B A+B

No C D C+D

Totals A+C B+D n

DCOVA



The sample proportions of interest are

Test H0: π1 = π2 (the two population proportions are equal)

H1: π1 ≠ π2 (the two population proportions are not equal)

(continued)

2 condition to yesanswer whosrespondent of proportionn

CAp

1 condition to yesanswer whosrespondent of proportionn

BAp

2

1

DCOVA



The test statistic for the McNemar test:

where the test statistic Z is approximately normally distributed

(continued)

CB

CBZ

STAT

DCOVA


McNemar TestExample

Suppose you survey 300 homeowners and ask them if they are interested in refinancing their home. In an effort to generate business, a mortgage company improved their loan terms and reduced closing costs. The same homeowners were again surveyed. Determine if change in loan terms was effective in generating business for the mortgage company. The data are summarized as follows:


McNemar TestExample

Survey response before change

Survey response after change

Yes No Totals

Yes 118 2 120

No 22 158 180

Totals 140 160 300

Test the hypothesis (at the 0.05 level of significance):H0: π1 ≥ π2: The change in loan terms was ineffective H1: π1 < π2: The change in loan terms increased business

DCOVA


McNemar Test Example

Survey response before change

Survey response after change

Yes No Totals

Yes 118 2 120

No 22 158 180

Totals 140 160 300

The critical value (0.05 significance) is Z0.05 = -1.645

The test statistic is:

08.4222

222

CB

CBZSTAT

Since ZSTAT = -4.08 < -1.645, you reject H0 and conclude that the change in loan terms significantly increase business for the mortgage company.

DCOVA


Excel Output Of McNemar Test

Chap 12-46

Minitab does not have a command for the McNemar test

McNemar Test

Observed Frequencies After Change

Before Change Yes No TotalYes 118 2 120No 22 158 180

Total 140 160 300

DataLevel of Significance 0.05

Intermediate CalculationsNumerator -20=C6-B7Denominator 4.8990=SQRT(C6+B7)Z Test Statistic -4.0825=B14/B15

One-Tail TestLower Critical Value -1.6449=NORMSINV(B11)p-Value 0.0000=NORMSDIST(B16)

Reject the null hypothesis =IF(B21<B11,"Reject the null hypothesis",Do not reject the null hypothesis)

DCOVA


Chi-Square Test for a Variance or Standard Deviation

A χ2 test statistic is used to test whether or not the population variance or standard deviation is equal to a specified value:

Where n = sample size

S2 = sample variance

σ2 = hypothesized population variance

follows a chi-square distribution with d.f. = n - 1

2

22

σ

1)S-(n

STATχ

2STATχ

H0: σ2 = σ02

Ha: σ2 ≠ σ02

Reject H0 if > or if < 2STAT 2

2/ 2STAT 2

2/1

DCOVA

Chap 12-47


DCOVA

Chi-Square Test For A Variance: An Example

Suppose you have gathered a random sample of size 25 andobtained a sample standard deviation of S = 7 and want to do the following hypothesis test:

H0: σ2 = 81

Ha: σ2 ≠ 81

5185.1481

49*24

σ

1)S-(n2

22 STATχ

Since < 14.185 < you fail to reject H0401.122

975.0 364.392025.0

Chap 12-48


Excel Output For Chi-Square Test For A Variance

Chap 12-49

DCOVAChi-Square Test For A Variance

Data Null Hypothesis s^2= 81Level of Significance 0.05Sample Size 25Sample Standard Deviation 7

Intermediate CalculationsDegrees of Freedom 24=B6-1Half Area 0.025=B5/2Chi-Square Statistic 14.5185=B10*B7^2/B4

Two-Tail Test

Lower Critical Value 12.4012=CHIINV(1-B11,B10)

Upper Critical Value 39.3641=CHIINV(B11,B10)

p-Value 0.9341=IF((B12-B15)<0,1-CHIDIST(B12,B10),CHIDIST(B12,B10))

Do not reject the null hypothesis =IF(B17<$B$5/2,"Reject the null hypothesis",Do not reject the null hypothesis)

Minitab does not have a command for the McNemar test


Wilcoxon Rank Sum Test for Differences in 2 Medians

Test two independent population medians

Populations need not be normally distributed

Distribution free procedure

Used when only rank data are available

Must use normal approximation if either of the

sample sizes is larger than 10

DCOVA


Wilcoxon Rank Sum Test: Small Samples

Can use when both n1 , n2 ≤ 10

Assign ranks to the combined n1 + n2 sample observations If unequal sample sizes, let n1 refer to smaller-sized

sample Smallest value rank = 1, largest value rank = n1 + n2 Assign average rank for ties

Sum the ranks for each sample: T1 and T2

Obtain test statistic, T1 (from smaller sample)

DCOVA


Checking the Rankings

The sum of the rankings must satisfy the formula below

Can use this to verify the sums T1 and T2

2

1)n(nTT 21

where n = n1 + n2

DCOVA


Wilcoxon Rank Sum Test:Hypothesis and Decision Rule

H0: M1 = M2

H1: M1 ¹ M2

H0: M1 £ M2

H1: M1 > M2

H0: M1 ³ M2

H1: M1 < M2

Two-Tail Test Left-Tail Test Right-Tail Test

M1 = median of population 1; M2 = median of population 2

Reject

T1L T1U

RejectDo Not Reject

Reject

T1L

Do Not Reject

T1U

RejectDo Not Reject

Test statistic = T1 (Sum of ranks from smaller sample)

Reject H0 if T1 ≤ T1L

or if T1 ≥ T1U

Reject H0 if T1 ≤ T1L Reject H0 if T1 ≥ T1U

DCOVA


Sample data are collected on the capacity rates (% of capacity) for two factories.

Are the median operating rates for two factories the same?

For factory A, the rates are 71, 82, 77, 94, 88

For factory B, the rates are 85, 82, 92, 97

Test for equality of the population medians at the 0.05 significance level

Wilcoxon Rank Sum Test: Small Sample Example DCOVA


Wilcoxon Rank Sum Test: Small Sample Example

Capacity Rank

Factory A Factory B Factory A Factory B

71 1

77 2

82 3.5

82 3.5

85 5

88 6

92 7

94 8

97 9

Rank Sums: 20.5 24.5

Tie in 3rd and 4th places

RankedCapacityvalues:

(continued)DCOVA



(continued)

Factory B has the smaller sample size, so the test statistic is the sum of the Factory B ranks:

T1 = 24.5

The sample sizes are:

n1 = 4 (factory B)

n2 = 5 (factory A)

The level of significance is = .05

DCOVA


n2

a n1

One-Tailed

Two-Tailed 4 5

4

5

.05 .10 12, 28 19, 36

.025 .05 11, 29 17, 38

.01 .02 10, 30 16, 39

.005 .01 --, -- 15, 40

6


Lower and Upper Critical Values for T1 from

Appendix table E.8:

(continued)

T1L = 11 and T1U = 29

DCOVA


H0: M1 = M2

H1: M1 ¹ M2

Two-Tail Test

Reject

T1L=11 T1U=29

RejectDo Not Reject

Reject H0 if T1 ≤ T1L=11

or if T1 ≥ T1U=29

a = .05 n1 = 4 , n2 = 5

Test Statistic (Sum of ranks from smaller sample):

T1 = 24.5

Decision:

Conclusion:

Do not reject at a = 0.05

There is not enough evidence to prove that the medians are not equal.

Wilcoxon Rank Sum Test:Small Sample Solution

(continued)

DCOVA


Wilcoxon Rank Sum Test (Large Sample)

For large samples, the test statistic T1 is

approximately normal with mean and standard deviation :

Must use the normal approximation if either n1 or n2 > 10

Assign n1 to be the smaller of the two sample sizes

Should not use the normal approximation for small samples

1Tμ

1Tσ

2

1μ 1

1

)(nnT

12

)1(σ 21

1

nnnT

DCOVA


Wilcoxon Rank Sum Test (Large Sample)

The Z test statistic is

Where ZSTAT approximately follows a standardized normal distribution

12

1)(nnn2

1)(nnT

σ

μTZ

21

11

T

T1

1

1

STAT

(continued)

DCOVA


Wilcoxon Rank Sum Test: Normal Approximation Example

Use the setting of the prior example:

The sample sizes were:

n1 = 4 (factory B)

n2 = 5 (factory A)

The level of significance was α = .05

The test statistic was T1 = 24.5

DCOVA


Wilcoxon Rank Sum Test: Normal Approximation Example

The test statistic is

202

)19(4

2

)1n(nμ 1

T1

082412

1954

12

1σ 21

1

.)()()n(nn

T

(continued)

10.14.082

205.24

σ

μZ

1

11

T

TSTAT

T

Z = 1.10 is not greater than the critical Z value of 1.96 (for α = .05) so we do not reject H0 – there is not

sufficient evidence that the medians are not equal

DCOVA


Example Excel Output

Chap 12-63

DCOVAWilcoxon Rank Sum Test

Data Level of Significance 0.05

Population 1 Sample Sample Size 4Sum of Ranks 24.5

Population 2 Sample Sample Size 5Sum of Ranks 20.5

Intermediate CalculationsTotal Sample Size n 9=B7+B10T1 Test Statistic 24.5=IF(B7<=B10,B8,B11)T1 Mean 20=IF(B7<=B10,B7*(B14+1)/2,B10*(B14+1)/2)Standard Error of T1 4.0825=SQRT(B7*B10*(B14+1)/12)Z Test Statistic 1.1023=(B15-B16)/B17

Two-Tail Test Lower Critical Value -1.9600=NORMSINV(B4/2)Upper Critical Value 1.9600=NORMSINV(1-B4/2)p-Value 0.2703=2*(1-NORMSDIST(ABS(B18)))

Do not reject the null hypothesis =IF(B23<$B$4,"Reject the null hypothesis",

Do not reject the null hypothesis)


Example Minitab Output

Chap 12-64

DCOVAMann-Whitney Test and CI: Factory B, Factory A

N Median

Factory B 4 88.50

Factory A 5 82.00

Point estimate for ETA1-ETA2 is 6.50

96.3 Percent CI for ETA1-ETA2 is (-9.00,21.00)

W = 24.5

Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.3272

The test is significant at 0.3252 (adjusted for ties)


Kruskal-Wallis Rank Test

Tests the equality of more than 2 population medians

Use when the normality assumption for one-way ANOVA is violated

Assumptions: The samples are random and independent Variables have a continuous distribution The data can be ranked Populations have the same variability Populations have the same shape

DCOVA


Kruskal-Wallis Test Procedure

Obtain rankings for each value In event of tie, each of the tied values gets the

average rank

Sum the rankings for data from each of the c groups Compute the H test statistic

DCOVA



The Kruskal-Wallis H-test statistic: (with c – 1 degrees of freedom)

)1n(3n

T

)1n(n

12H

c

1j j

2j

where:n = sum of sample sizes in all groupsc = Number of groupsTj = Sum of ranks in the jth groupnj = Number of values in the jth group (j = 1, 2, … , c)

(continued)

DCOVA


Decision rule Reject H0 if test statistic H > 2

α

Otherwise do not reject H0

(continued)


Complete the test by comparing the calculated H value to a critical 2 value from the chi-square distribution with c – 1 degrees of freedom

2

2α

0


DCOVA


Do different departments have different class sizes?

Kruskal-Wallis Example

Class size (Math, M)

Class size (English, E)

Class size (Biology, B)

2345547866

5560724570

3040183444

DCOVA


Do different departments have different class sizes?


Class size (Math, M)

Ranking Class size (English, E)

Ranking Class size (Biology, B)

Ranking

2341547866

269

1512

5560724570

101114 813

3040183444

35147

= 44 = 56 = 20

(continued)

DCOVA


The H statistic is

(continued)


6.721)3(155

20

5

56

5

44

1)15(15

12

1)3(nn

T

1)n(n

12H

222

c

1j j

2j

equal are Medians population allNot :H

MedianMedianMedian :H

1

BEM0

DCOVA


Since H = 7.12 > ,

reject H0

(continued)


5.991205.0 χ

Compare H = 7.12 to the critical value from the chi-square distribution for 3 – 1 = 2 degrees of freedom and = 0.05:

5.991χ 20.05

There is sufficient evidence to reject that the population medians are all equal

DCOVA


Example Excel Output

Chap 12-73

DCOVA

Kruskal-Wallis Rank Test

CalculationsGroup Sample Size Sum of Ranks Mean Ranks

Data Supplier 1 5 44 8.8Level of Significance 0.05 Supplier 2 5 56 11.2

Supplier 3 5 20 4Intermediate Calculations

Sum of Squared Ranks/Sample Size 1094.4 =(G5*F5)+(G6*F6)+(G7*F7)Sum of Sample Sizes 15 =SUM(E5:E7)Number of Groups 3

Test Result H Test Statistic 6.7200 =(12/(B10*(B10+1)))*B9-(3*(B10+1))Critical Value 5.9915 =CHIINV(B6,B11-1)p-Value 0.0347 =CHIDIST(B14,B11-1)

Reject the null hypothesis =IF(B16<B6,"Reject the null hypothesis","Do not reject the null hypothesis")


Example Minitab Output

Chap 12-74

DCOVAKruskal-Wallis Test: Class Size versus Class Type

Kruskal-Wallis Test on Class Size

Class Type N Median Ave Rank Z

B 5 34.00 3.8 -2.57

E 5 60.00 11.1 1.90

M 5 54.00 9.1 0.67

Overall 15 8.0

H = 7.11 DF = 2 P = 0.029

H = 7.13 DF = 2 P = 0.028 (adjusted for ties)


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

Applied the Marascuilo procedure for comparing all pairs of proportions after rejecting a 2 test

Examined the 2 test for independence Applied the McNemar test for proportions from two

related samples


Chapter Summary

Examined how to use the Chi-Square to test for a variance or standard deviation

Used the Wilcoxon rank sum test for two population medians

Applied the Kruskal-Wallis rank H-test for multiple population medians

(continued)


Learning Objectives

In this topic, you learn:How to use the Wilcoxon signed ranks test for

comparing paired samples



A nonparametric test for two related populations Steps:

1. For each of n sample items, compute the difference, Di, between two measurements

2. Ignore + and – signs and find the absolute values, |Di|

3. Omit zero differences, so sample size is n’4. Assign ranks Ri from 1 to n’ (give average rank to

ties)

5. Reassign + and – signs to the ranks Ri

6. Compute the Wilcoxon test statistic W as the sum of the positive ranks


Wilcoxon Signed Ranks Test Statistic

The Wilcoxon signed ranks test statistic is the sum of the positive ranks:

For small samples (n’ < 20), use Table E.9 for the critical value of W

n'

1i

)(iRW


Wilcoxon Signed Ranks Test Statistic

For samples of n’ > 20, W is approximately normally distributed with

24

1)1)(2n'(n'n'σ

4

1)(n'n'μ

W

W



The large sample Wilcoxon signed ranks Z test statistic is

To test for no median difference in the paired values:

H0: MD = 0

H1: MD ≠ 0

24

1)1)(2n'(n'n'4

1)(n'n'W

ZSTAT


Topic Summary

Used the Wilcoxon signed ranks test for comparing paired samples


Learning Objectives

In this topic, you learn:How to use the Friedman rank test for comparing

multiple population medians in a randomized block design


Friedman Rank Test

Use the Friedman rank test to determine whether c groups (i.e., treatment levels) have been selected from populations having equal medians

H0: M.1 = M.2 = . . . = M.c

H1: Not all M.j are equal (j = 1, 2, …, c)


Friedman Rank Test

Friedman rank test for differences among c medians:

where = the square of the total ranks for group j

r = the number of blocks

c = the number of groups

(continued)

1)3r(cR1)rc(c

12F

c

1j

2.jR

2.jR


Friedman Rank Test

The Friedman rank test statistic is approximated by a chi-square distribution with c – 1 d.f.

Reject H0 if

Otherwise do not reject H0

(continued)

2R

χFα


Topic Summary

Applied the Friedman rank test for comparing multiple population medians in a randomized block design

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