Basic Business Statistics, 10/e

Basic Business Statistics

10 th Edition

Chapter 12

Chi-Square Tests and

Nonparametric Tests

Basic Business Statistics, 10e © 2006 Prentice-Hall, Inc.

Chap 12-1

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 to use the chi-square test to evaluate the goodness of fit of a set of data to a specific probability distribution



How and when to use nonparametric tests


Chap 12-2

Contingency Tables

Contingency Tables



Useful in situations involving multiple population proportions



Used to classify sample observations according to two or more characteristics



Also called a cross-classification table .


Chap 12-3

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 12-4

Contingency Table Example

(continued)

Sample results organized in a contingency table: sample size = n = 300:

120 Females, 12 were left handed

180 Males, 24 were left handed

Gender

Female

Male

Hand Preference

Left Right

12 108

24

36

156

264

120

180

300


Chap 12-5



2 Test for the Difference

Between Two Proportions

H

0

: π

1

= π

2

H

1

: π

1

≠ π

2

(Proportion of females who are left handed is equal to the proportion of males who are left handed)

(The two proportions are not the same –

Hand preference is not independent of gender)



If H

0 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


Chap 12-6

The Chi-Square Test Statistic

The Chi-square test statistic is:

 2   all cells

( f o

 f e

)

2 f e

 where: f o

= observed frequency in a particular cell f e

= expected frequency in a particular cell if H

0 is true



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

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


Chap 12-7

Decision Rule

The



2 test statistic approximately follows a chisquared distribution with one degree of freedom

Decision Rule:

If



2 >



2

U

, reject H

0

, otherwise, do not reject H

0



0

Do not reject H

0



2

U

Reject H

0


 2

Chap 12-8

Computing the

Average Proportion

The average proportion is: p



X

1 n

1



X

2

 n

2



X n

120 Females, 12 were left handed

180 Males, 24 were left handed

Here: p



12



24

120



180



36

300



0 .

12 i.e., the proportion of left handers overall is 0.12, that is, 12%


Chap 12-9

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


Chap 12-10

Observed vs. Expected

Frequencies

Gender

Female

Male

Hand Preference

Left

Observed = 12

Expected = 14.4

Right

Observed = 108

Expected = 105.6

Observed = 24

Expected = 21.6

Observed = 156

Expected = 158.4

36 264

120

180

300


Chap 12-11


Gender

Female

Male

Hand Preference

Left

Observed = 12

Expected = 14.4

Observed = 24

Expected = 21.6

Right

Observed = 108

Expected = 105.6

Observed = 156

Expected = 158.4

120

180

36 264 300

The test statistic is:

χ 2 



 all cells

(f o

 f e

) 2 f e

(12



14.4) 2

14.4



(108



105.6) 2

105.6



(24



21.6)

21.6

2



(156



158.4) 2

158.4



0.7576


Chap 12-12

Decision Rule

The test statistic is χ 2 

0.7576

, χ

U

2 with 1 d.f.



3.841

0



Do not reject H

0

Reject H

0



2

U

=3.841


Decision Rule:

If



2 > 3.841, reject H

0


0

 2

Here,



2 = 0.7576 <



2

U

= 3.841, so we do not reject H

0 and conclude that there is not sufficient evidence that the two proportions are different at



= 0.05

Chap 12-13



2 Test for Differences Among

More Than Two Proportions



Extend the



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

H

0

: π

1

= π

2

= … = π c

H

1

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


Chap 12-14




( f o

 f e

)

2 f e

 where: f o

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


0 is true



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



Chap 12-15

Computing the

Overall Proportion

The overall proportion is: p



X

1 n

1



X

2

 n

2

  

X c

   n c



X n



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

Decision Rule:

If



2 >



2

U

, reject H

0


0

Where



2

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


Chap 12-16

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, p all pairs (for j ≠ j ’ ) . . .

j

– p j ’

, for



. . .then compare the absolute difference to a calculated critical range


Chap 12-17

The Marascuilo Procedure

(continued)



Critical Range for the Marascuilo Procedure:

Critical range

 χ 2

U p j

(1

 p j

) n j

 p j '

(1

 p j '

) n j '



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



A particular pair of proportions is significantly different if

| p j

– p j ’

| > critical range for j and j ’


Chap 12-18

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

Totals

37

100

30

50

13

50

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


Chap 12-19

Marascuilo Procedure: Solution

Excel Output: compare

Marascuilo Procedure

Sample Sample Absolute Std. Error Critical

Group Proportion Size Comparison Difference of Difference Range Results

1

2

3

0.63

0.4

0.74

100

50

50

1 to 2

1 to 3

2 to 3

0.23

0.11

0.34

0.08444525

0.07860662

0.09299462

0.256 Means are not different

0.239 Means are not different

0.282 Means are different

Level of significance d.f

Q Statistic

0.01

2

Other Data

Chi-sq Critical Value in attitude between students and faculty

9.21


Chap 12-20



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

H

0

: The two categorical variables are independent

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

H

1

: The two categorical variables are dependent

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


Chap 12-21



2 Test of Independence

(continued)



( f o

 f e

)

2 f e

 where: f o

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


0 is true



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



Chap 12-22

Expected Cell Frequencies



Expected cell frequencies: f e

 row total

 column total n

Where: row total = sum of all frequencies in the row column total = sum of all frequencies in the column n = overall sample size


Chap 12-23

Decision Rule



The decision rule is

If



2 >



2

U

, reject H

0


0

Where



2

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


Chap 12-24

Example



The meal plan selected by 200 students is shown below:

Class

Standing

Fresh.

Number of meals per week

20/week 10/week none

24 32 14

Soph.

Junior

22

10

26

14

12

6

Senior

Total

14

70

16

88

10

42

Total

70

60

30

40

200


Chap 12-25

Example

(continued)



The hypothesis to be tested is:

H

0

: Meal plan and class standing are independent

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

H

1

: Meal plan and class standing are dependent

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


Chap 12-26

Example:

Expected Cell Frequencies

Class

Standing

Fresh.

Soph.

Junior

Senior

Total

Observed:


20/wk 10/wk none

24

22

32

26

14

12

10

14

70

14

16

88

6

10

42

Example for one cell: f e

 row total

 column total n



30



70



10 .

5

200


Total

70

60

30

40

200

(continued)

Class

Standing

Fresh.

Soph.

Junior

Senior

Total

Expected cell frequencies if H

0 is true:


20/wk 10/wk none

24.5

30.8

14.7

21.0

26.4

12.6

10.5

13.2

6.3

14.0

17.6

8.4

70 88 42

Total

70

60

30

40

200

Chap 12-27

Example: The Test Statistic

(continued)



The test statistic value is:


( f o

 f e

)

2 f e



( 24



24 .

5 )

2

24 .

5



( 32



30 .

8 )

2

30 .

8

  

( 10



8 .

4 )

2

8 .

4



0 .

709



2

U

= 12.592 for



= 0.05 from the chi-squared distribution with (4 – 1)(3 – 1) = 6 degrees of freedom


Chap 12-28

Example:

Decision and Interpretation

(continued)

The test statistic is

 2 

0 .

709 ,

 2

U

with 6 d.f.



12.592

Decision Rule:

If



2 > 12.592, reject H

0


0



0

Do not reject H

0

Reject H

0



2

U

=12.592


 2

Here,



2 = 0.709 <



2

U

= 12.592, so do not reject H

0

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



= 0.05

Chap 12-29

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


Chap 12-30


(continued)



Consider a 2 X 2 contingency table:

Condition 2

Condition 1 Yes

Yes A

No C

Totals A+C


No

B

D

B+D

Totals

A+B

C+D n

Chap 12-31


(continued)



The sample proportions of interest are p

1



A



B

 proportion of respondent s who answer yes to condition 1 n p

2



A



C

 proportion of respondent s who answer yes to condition 2 n



Test H

0

: π

1

= π

2

(the two population proportions are equal)

H

1

: π

1

≠ π

2

(the two population proportions are not equal)


Chap 12-32


(continued)



The test statistic for the McNemar test:

Z



B



C

B



C where the test statistic Z is approximately normally distributed


Chap 12-33

Chi-Square Test for a Variance or

Standard Deviation



A χ 2 test statistic is used to test whether or not the population variance or standard deviation is equal to a specified value:

χ 2 

(n 1)S

2

σ 2

Where n = sample size

S 2 = sample variance

σ 2 = hypothesized population variance

χ 2 follows a chi-square distribution with n – 1 d.f.


Chap 12-34

Chi-Square Goodness-of-Fit Test



Does sample data conform to a hypothesized distribution?



Examples:



Are technical support calls equal across all days of the week? (i.e., do calls follow a uniform distribution?)



Do measurements from a production process follow a normal distribution?


Chap 12-35

Chi-Square Goodness-of-Fit Test



(continued)

Are technical support calls equal across all days of the week? (i.e., do calls follow a uniform distribution?)



Sample data for 10 days per day of week:

Monday

Tuesday

Wednesday

Thursday

Friday

Saturday

Sunday

Sum of calls for this day:

290

250

238

257

265

230

192



= 1722


Chap 12-36

Logic of Goodness-of-Fit Test

 If calls are uniformly distributed, the 1722 calls would be expected to be equally divided across the 7 days:

1722



246 expected calls per day if uniform

7

 Chi-Square Goodness-of-Fit Test: test to see if the sample results are consistent with the expected results


Chap 12-37

Monday

Tuesday

Wednesday

Thursday

Friday

Saturday

Sunday

TOTAL

Observed vs. Expected

Frequencies

Observed f o

290

250

238

257

265

230

192

1722

Expected f e

246

246

246

246

246

246

246

1722


Chap 12-38

Chi-Square Test Statistic

H

0

: The distribution of calls is uniform over days of the week

H

1

: The distribution of calls is not uniform






2   k

(f o

 f e

)

2

(where df f e

 k

 p



1) where: k = number of categories f o f e

= observed frequency

= expected frequency p = number of parameters estimated from the data


Chap 12-39

The Rejection Region

H

0


H

1




Reject H

0 if



2   k

(f o

 f e

)

2 f e

 2   2

α

 k – 2 degrees of freedom, since p = 1 here (the mean was estimated)

0

Do not reject H

0



2



Reject H

0




2

Chap 12-40

Chi-Square Test Statistic

H

0


H

1


 2 

(290



246)

2

246



(250



246)

2

246



...



(192



246)

2

246



23.05

k – 2 = 5 (k = 7 days of the week) so use 5 degrees of freedom:



2

.05

= 11.0705



= .05

Conclusion:



2 = 23.05 >



2

 reject H

0

= 11.0705 so and conclude that the distribution is not uniform

0

Do not reject H

0

Reject H

0



2

.05

= 11.0705




2

Chap 12-41

Normal Distribution Example



Do measurements from a production process follow a normal distribution with

μ = 50 and σ = 15?



Process:



Get sample data



Group sample results into classes (cells)

(Expected cell frequency must be at least

5 for each cell)



Compare actual cell frequencies with expected cell frequencies


Chap 12-42


(continued)



Sample data and values grouped into classes:

150 Sample

Measurements

80

65

36

66

50

38

57

77

59

…etc…

Class less than 30

30 but < 40

40 but < 50

50 but < 60

60 but < 70

70 but < 80

80 but < 90

90 or over

TOTAL

Frequency

10

21

33

41

26

10

7

2

150


Chap 12-43




(continued)

What are the expected frequencies for these classes for a normal distribution with μ = 50 and σ = 15?

Expected

Frequency Class less than 30

30 but < 40

40 but < 50

50 but < 60

60 but < 70

70 but < 80

80 but < 90

90 or over

TOTAL

Frequency

10

21

33

41

26

10

7

2

150


?

Chap 12-44

Expected Frequencies

Value less than 30

30 but < 40

40 but < 50

50 but < 60

60 but < 70

70 but < 80

80 but < 90

90 or over

TOTAL

P(X < value)

0.09121

0.16128

0.24751

0.24751

0.16128

0.06846

0.01892

0.00383

1.00000

Expected frequency

13.68

24.19

37.13

37.13

24.19

10.27

2.84

0.57

150.00

Combine class groups so no class has expected frequency <1


Expected frequencies in a sample of size n=150 , from a normal distribution with

μ=50, σ=15

Example:

P(x



30)





P

 z



30



50

15



P(z

 

1.3333)



.0912

(.0912)(15 0)



13.68

Chap 12-45

The Test Statistic

Class less than 30

30 but < 40

40 but < 50

50 but < 60

60 but < 70

70 but < 80

80 or over

Frequency

(observed, f o

)

10

21

33

41

26

10

9

Expected

Frequency, f e

13.68

24.19

37.13

37.13

24.19

10.27

3.41

TOTAL 150


150.00




2   k

(f o

 f e

)

2 f e



Reject H

0

 2   2

α if

(with k – p – 1 degrees of freedom)

Chap 12-46

The Rejection Region

H

0

: The distribution of values is normal with μ = 50 and σ = 15

H

1

: The distribution of calls does not have this distribution

χ 2   k

(f o

 f e

)

2 f e



(10



13.68)

2

13.68



...



(9



3.41)

2

3.41



11.580

8 classes so use 8 – 2 – 1 = 5 d.f.:



2

.05

= 11.0705



=.05

Conclusion:



2 = 11.580 >



2



= 11.0705 so reject H

0

:

There is sufficient evidence that the data are not normal with μ = 50 and σ = 15

0

Do not reject H

0

Reject H

0



2

.05

= 11.0705



2


Chap 12-47

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


Chap 12-48

Wilcoxon Rank-Sum Test:

Small Samples



Can use when both n

1

, n

2

≤ 10



Assign ranks to the combined n

1 observations

+ n

2 sample

 If unequal sample sizes, let n

1 sample refer to smaller-sized





Smallest value rank = 1, largest value rank = n

1

+ n

2

Assign average rank for ties



Sum the ranks for each sample: T

1 and T

2



Obtain test statistic, T

1

(from smaller sample)


Chap 12-49

Checking the Rankings



The sum of the rankings must satisfy the formula below



Can use this to verify the sums T

1 and T

2

T

1



T

2

 n(n



1)

2 where n = n

1

+ n

2


Chap 12-50


Hypothesis and Decision Rule

M

1

= median of population 1; M

2

= median of population 2

Test statistic = T

1

(Sum of ranks from smaller sample)

Two-Tail Test

H

0

: M

1

H

1

: M

1

= M

2



M

2

Left-Tail Test

H

H

0

1

: M

: M

1

1



M

2



M

2

Right-Tail Test

H

0

H

1

: M

: M

1

1



M

2



M

2

Reject Do Not

Reject

T

1L

Reject

T

1U

Reject H

0 if T

1

< T

1L or if T

1

> T

1U


Reject

T

1L

Do Not Reject

Reject H

0 if T

1

< T

1L

Do Not Reject Reject

T

1U

Reject H

0 if T

1

> T

1U

Chap 12-51


Small Sample Example

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


Chap 12-52



(continued)

Ranked

Capacity values:

Tie in 3 rd and

4 th places

Capacity Rank

Factory A Factory B Factory A Factory B

71

77

82

1

2

3.5

82

85

3.5

5

88 6

92 7

94 8

97

Rank Sums: 20.5

9

24.5


Chap 12-53



(continued)

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

Factory B ranks:

T

1

= 24.5

The sample sizes are: n

1

= 4 (factory B) n

2

= 5 (factory A)

The level of significance is



= .05


Chap 12-54



Lower and

Upper

Critical

Values for

T

1 from

Appendix table E.8:



(continued)

 n

1 n

2

4

5

One-

Tailed

.05

.025

.01

Two-

Tailed

4 5

.10

12, 28 19, 36

.05

11, 29 17, 38

.02

10, 30 16, 39

.005

.01

--, -15, 40

6

T

1L

= 11 and T

1U

= 29


Chap 12-55


Small Sample Solution

(continued)







= .05

n

1

= 4 , n

2

= 5

Two-Tail Test

H

0

: M

1

H

1

: M

1

= M

2



M

2

Test Statistic (Sum of ranks from smaller sample):

T

1

= 24.5

Decision:

Do not reject at



= 0.05

Reject Do Not

Reject

Reject

T

1L

=11 T

1U

=29

Reject H

0 if T

1

< T

1L

=11 or if T

1

> T

1U

=29


Conclusion:

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

Chap 12-56

Wilcoxon Rank-Sum Test

(Large Sample)



For large samples, the test statistic T

1 is approximately normal with mean and

T

1

T

1

μ

T

1

 n

1

( n



1 )

2

σ

T

1

 n

1 n

2

( n



1 )

12





Must use the normal approximation if either n

1 or n

2

> 10

Assign n

1 to be the smaller of the two sample sizes



Can use the normal approximation for small samples


Chap 12-57

Wilcoxon Rank-Sum Test

(Large Sample)

(continued)



The Z test statistic is

Z



T

1

 μ

T

1

σ

T

1



T

1

 n

1

(n

 n

1 n

2

(n

2



1)

1)

12



Where Z approximately follows a standardized normal distribution


Chap 12-58


Normal Approximation Example

Use the setting of the prior example:

The sample sizes were: n

1

= 4 (factory B) n

2

= 5 (factory A)

The level of significance was α = .05

The test statistic was T

1

= 24.5


Chap 12-59


Normal Approximation Example

(continued)

μ

T

1

 n

1

( n



1 )

2



4 ( 9



1 )



20

2

σ

T

1

 n

1 n

2

( n



1 )

12



4 ( 5 ) ( 9



1 )



2 .

739

12






Z



T

1

μ

T

1

σ

T

1



24 .

5



20

2.739



1 .

64



Z = 1.64 is not greater than the critical Z value of 1.96

(for α = .05) so we do not reject H

0

– there is not sufficient evidence that the medians are not equal


Chap 12-60

Wilcoxon Signed Ranks Test





A nonparametric test for two related populations

2.

3.

4.

Steps:

1.

5.

6.

For each of n sample items, compute the difference,

D i

, between two measurements

Ignore + and – signs and find the absolute values, |D i

|

Omit zero differences, so sample size is n ’

Assign ranks R i ties) from 1 to n ’ (give average rank to

Reassign + and – signs to the ranks R i

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


Chap 12-61

Wilcoxon Signed Ranks

Test Statistic



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

W

 i n' 



1

R ( i



)



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


Chap 12-62

Wilcoxon Signed Ranks

Test Statistic



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

μ

W

 n' (n'



1)

4

σ

W

 n' (n'



1)(2n'



1)

24


Chap 12-63

Wilcoxon Signed Ranks Test



The large sample Wilcoxon signed ranks Z test statistic is

Z



W n' (n'

 n' (n'



4



1)(2n'

1)



1)

24



The appropriate hypothesis test is

H

0

: M

D

≤ 0

H

1

: M

D

> 0


Chap 12-64

Kruskal-Wallis Rank Test







Tests the equality of more than 2 population medians

Use when the normality assumption for oneway 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


Chap 12-65

Kruskal-Wallis Test Procedure



Obtain relative 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


Chap 12-66


(continued)



The Kruskal-Wallis H-test statistic:

(with c – 1 degrees of freedom)

H









12 n ( n



1 ) c  j



1

T j

2 n j









3 ( n



1 ) where: n = sum of sample sizes in all samples c = Number of samples

T n j j

= Sum of ranks in the j th sample

= Number of values in the j th sample (j = 1, 2, … , c)


Chap 12-67


(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

0

Do not reject H

0



2

U



Reject H

0

 2



Decision rule





Reject H

0 if test statistic H >



2

U

Otherwise do not reject H

0


Chap 12-68

Kruskal-Wallis Example



Do different departments have different class sizes?

Class size

(Math, M)

23

45

54

78

66

Class size

(English, E)

55

60

72

45

70

Class size

(Biology, B)

30

40

18

34

44


Chap 12-69


(continued)



Do different departments have different class sizes?

Class size

(Math, M)

Ranking

23

41

54

78

66

2

6

9

15

12



= 44

Class size

(English, E)

55

60

72

45

70

Ranking

10

11

14

8

13



= 56

Class size

(Biology, B)

30

40

18

34

44

Ranking

3

5

1

4

7



= 20


Chap 12-70


(continued)

H

0

: Median

M



Median

E



Median

B

H

1

: Not all population Medians are equal



The H statistic is

H











12 n(n



1) c  j



1

T j

2 n j









3(n



1)







12

15(15



1)





44 2

5



56 2

5



20 2

5













3(15



1)



6.72


Chap 12-71


(continued)



Compare H = 6.72 to the critical value from the chi-square distribution for 3 – 1 = 2 degrees of freedom and



= 0.05:

χ

U

2 

5.991

Since H = 6.72 >



U

2 

5.991

, reject H

0

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


Chap 12-72

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

H

0

: M

.1

= M

.2

= . . . = M

.c

H

1

: Not all M

.j

are equal (j = 1, 2, …, c)


Chap 12-73

Friedman Rank Test

(continued)



Friedman rank test for differences among c medians:

F

R

 rc(c

12



1) c  j



1

R

.j

2 

3r(c



1)

R

.j

r = the number of blocks c = the number of groups


Chap 12-74

Friedman Rank Test

(continued)



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

Reject H

0 if F

R

 

U

2

Otherwise do not reject H

0


Chap 12-75

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

Used the Wilcoxon rank sum test for two population medians





Small Samples

Large sample Z approximation

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


Chap 12-76

Basic Business Statistics, 10/e

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