Nonparametric Statistics. Salad bar, Pizza Hut, China, State of the Art in Structural Engineering.

Nonparametric Statistics

Salad bar, Pizza Hut, China, State of the Art in Structural

Engineering

Learning Objectives

1. Distinguish Parametric & Nonparametric Test Procedures

2. Explain a Variety of Nonparametric Test Procedures

3. Solve Hypothesis Testing Problems Using Nonparametric Tests

Hypothesis Testing Procedures

HypothesisTesting

Procedures

NonparametricParametric

Z Test

Kruskal-W allisH-Test

W ilcoxonRank Sum

Test

t Test One-W ayANOVA

Many More Tests Exist!

Parametric Test Procedures

1. Require Interval Scale or Ratio Scale Whole Numbers or Fractions Example: Height in Inches (72, 60.5, 54.7)

2. Have Stringent Assumptions Example: Normal Distribution

3. Examples: t Test, Anova

Advantages of Nonparametric Tests

Easy to understand and useEasy to understand and use

Usable with nominal data, fewer assumptions

Usable with nominal data, fewer assumptions

Appropriate for non-normal population distributions

Appropriate for non-normal population distributions

Disadvantages of Nonparametric Tests

1. May Waste Information If Data Permit Using Parametric

Procedures Example: Converting Data From

Ratio to Nominal

2. Difficult to Compute by Hand for Large Samples

3. Tables Not Widely Available

© 1984-1994 T/Maker Co.

Recommended Statistical Techniques

Frequently Used Nonparametric Tests

1. Sign Test

2. Wilcoxon Rank Sum Test

3. Wilcoxon Signed Rank Test

4. Kruskal Wallis H-Test

5. Friedman’s Fr-Test

6. Spearman’s Rank Correlation Coefficient

Single-Variable Chi-Square Test


1. Compares the observed frequencies of categories to frequencies that would be expected if the null hypothesis were true.

2. Data are assumed to be a random sample.

3. Each subject can only have one entry in the chi-square table.

4. The expected frequency for each category should be at least 5.


• Open single variable chi.sav in SPSS.

• In SPSS, click Analyze Nonparametric Tests Chi-Square

• Move pbrand into the Test Variable List

• Click Options and select Descriptive under Statistics

• Click OK.


• Ho: There is no preference for the three brands.

• Chi-square p-value = 0.032 < 0.05, reject Ho.

• Brand C is the favored brand.

Test Statistics

Preferred Brand

Chi-Square 6.857a

df 2

Asymp. Sig. .032

a. 0 cells (.0%) have expected frequencies less than 5. The minimum expected cell frequency is 14.0.

Preferred Brand

Observed N Expected N Residual

Brand A 10 14.0 -4.0

Brand B 10 14.0 -4.0

Brand C 22 14.0 8.0

Total42

Sign Test

Sign Test

1. Tests Population Median, (eta)

2. Corresponds to t-Test for 1 Mean

Sign Test Uses P-Value

to Make Decision

.031.109

.219.273

.219

.109.031.004 .004

0%

10%

20%

30%

0 1 2 3 4 5 6 7 8 X

P(X) Binomial: n = 8 p = 0.5

P-Value Is the Probability of Getting an Observation At Least as Extreme as We Got. If 7 of 8 Observations ‘Favor’ Ha, Then P-Value = P(x 7) = .031 + .004 = .035. If = .05, Then Reject H0 Since P-Value .

Sign Test Example

7 people rated a new product on a 5-point Likert scale (1 = poor to 5 = excellent). The ratings are: 4 5 1 4 4 4 5. Is there evidence that the product has good market potential?

R Sign Test

If there is no difference in preferences, then we would expect a rating > 3 to be equally likely as a rating < 3.

Hence, the probability that a rating > 3 (and < 3) is binomially distributed with p = 0.5 and n = 7.

Test statistic = 6

Sign Test Solution

H0: = 3

Ha: > 3

= .05

Test Statistic:

• Sign = 1 -pbinom(6,size=7,prob=.5)

• [1] 0.0078125 • Since p-value = 0.0078125

< 0.05, reject Ho.• Median is significantly

larger than 3• abs(qnorm(sign))/sqrt(7)

gives the approximate effect size.

S = 6 (Only the third rating is less than 3:4, 5, 1, 4, 4, 4, 5)

R Sign Test

Twenty patients are given two treatments each (blindly and in randomized order) and then asked whether treatment A or B worked better. It turned out that 16 patients liked A better.

If there was no difference between the two treatments, then we would expect the number of people favouring treatment A to be binomially distributed with p = 0.5 and n = 20. How (im)probable would it then be to obtain what we have observed?

What we need is the probability of the observed 16 or more extreme, so we need to use “15 or less”:

> 1-pbinom(15,size=20,prob=.5)[1] 0.005908966If you want a two-tailed test because you have no prior idea about

which treatment is better, then you will have to add the probability of obtaining equally extreme results in the opposite direction.

> 1-pbinom(15,20,.5)+pbinom(4,20,.5)[1] 0.01181793

Cumulative Binomial Distribution

SPSS Sign Test

• Open rating.sav in SPSS, which contains 7 people;s rating of the new product on a 5-point Likert scale (1 = terrible to 5 = excellent).

• Ho: <= 3 vs. Ha: > 3

• In SPSS, click Analyze Nonparametric Tests Legacy Dialogs Binomial

• Move “ratings” into the Test Variable List, and set Cut point = 3.

• Click OK.

SPSS Sign Test

• Ours is a one-tail test, so the Sig. value has to be halved. • Since p-value = 0.125/2 = 0.0625 > 0.05, do not reject Ho.• The new product does not have sufficient good market

potential to justify investment.

Mann-Whitney Test


1. Sign Test






Mann-Whitney Test

Assumptions: Independent Random Samples The assumption of normality has been

violated in a t-test (especially if the sample size is small.)

The assumption of homogeneity of variance has been violated in a t-test

Mann-Whitney Test

Example• Corresponds to the independent samples t-test.

• You want to see if the buying intentions for 2 two product designs are the same. For design 1, the ratings are 71, 82, 77, 92, 88. For design 2, the ratings are 86, 82, 94 & 97.

Ho: Identical buying intention vs. Ha: Different buying intention

Mann-Whitney Test Computation Table

Design 1 Design 2Rate Rank Rate Rank

71 1 85 582 3 3.5 82 4 3.577 2 94 892 7 97 988 6 ... ...

Rank Sum 19.5 25.5

Mann-Whitney Test

1. Corresponds to independent samples t-Test.

2. Assign Ranks, Ri, to the n1 + n2 Sample Observations

3. Average Ties

4. Sum the Ranks, Ti, for Each Sample

5. The rank sum of the smaller-sized sample is used to test hypotheses.

Mann-Whitney Test

Table (Portion)

n1

3 4 5 ..

TL TU TL TU TL TU ..

3 6 15 7 17 7 20 ..n2 4 7 17 12 24 13 27 ..

5 7 20 13 27 19 36 ..: : : : : : : :

= .05 one-tailed; = .10 two-tailed

Mann-Whitney Test

SolutionH0: Identical Distrib.

Ha: Shifted Left or Right

= .10

n1 = 4 n2 = 5

Critical Value(s):

Test Statistic:

Decision:

Conclusion:Do Not Reject at = .10

There Is No Evidence Distrib. Are Not Equal

Reject RejectDo Not Reject

13 27 Ranks

T2 = 5 + 3.5 + 8+ 9 = 25.5 (Smallest Sample)

R Wilcoxon Rank Sum Test

Ho: The two product designs do not give rise to different buying intentions

> ratings1 <- c(71,82,77,92,88)> ratings2 <- c(86,82, 94,97)> mann=wilcox.test(ratings1, ratings2, paired = F)> mann

Wilcoxon rank sum test with continuity correction

data: ratings1 and ratings2W = 4.5, p-value = 0.2187alternative hypothesis: true location shift is not equal to 0

> abs(qnorm(mann$p.value))/sqrt(5 + 4)[1] 0.2588163>

SPSS Mann-Whitney Test

• Read “product design.sav” into SPSS, which contains data of 9 respondents’ intention to buy a product of two different designs.

• Under Analyze Nonparametric Tests Legacy Dialogs Two-Independent-Samples Tests

• Move “Probability of Buying” into Test Variable List, and design into Grouping Variable box.

• Click Define Groups, enter “1” for Group 1, and “2” for Group 2.

• Click Continue and then OK to get your output.

SPSS Mann-Whitney Test

• Since Sig. = 0.190 > 0.05, do not reject Ho.

• The two product designs do not give rise to different buying intentions.

Wilcoxon Signed Rank Test


1. Sign Test






Wilcoxon Signed Rank Test

1. For repeated measurements taken from the same subject.

2. Corresponds to paired samples t-test.

Signed Rank TestExample

Is the new loyalty program better in boosting consumption (.05 level)?

Buyer Old NewDonna 9.98 9.88Santosha 9.88 9.86Sam 9.90 9.83Tamika 9.99 9.80Brian9.94 9.87Jorge 9.84 9.84

Signed Rank Test Computation Table

Old New Di |Di| Ri Sign Sign Ri

9.98 9.88 +0.10 0.10 4 + +4

9.88 9.86 +0.02 0.02 1 + +1

9.90 9.83 +0.07 0.07 2 2.5 + +2.5

9.99 9.80 +0.19 0.19 5 + +5

9.94 9.87 +0.07 0.07 3 2.5 + +2.5

9.84 9.84 0.00 0.00 ... ... Discard

Total T+ = 15, T- = 0

Signed Rank Test Procedure

1. Obtain Difference Scores, Di = Old - New

2. Take Absolute Value of Differences, Di

3. Delete Differences With 0 Value

4. Assign Ranks, Ri, starting with 1

5. Assign Ranks Same Signs as Di

6. Sum ‘+’ Ranks (T+) & ‘-’ Ranks (T-)

7. Use T- for One-Tailed Test, and the Smaller of T- or T+ for 2-Tail Test

Signed Rank Test Computation Table

X1i X2i Di = X1i - X2i |Di| Ri Sign Sign Ri

X11 X21 D1 = X11 - X21 |D1| R1 ± ? R1

X12 X22 D2 = X12 - X22 |D2| R2 ± ? R2

X13 X23 D3 = X13 - X23 |D3| R3 ± ? R3

: : : : : : :

X1n X2n Dn = X1n - X2n |Dn| Rn ± ? Rn

Total T+ & T-

Wilcoxon Signed Rank

Table (Portion)

One-Tailed Two-Tailed n = 5 n = 6 n = 7 .. = .05 = .10 1 2 4 .. = .025 = .05 1 2 .. = .01 = .02 0 .. = .005 = .01 ..

n = 11 n = 12 n = 13

: : : :

Signed Rank Test Solution

H0: Identical Distrib.

Ha: Current Shifted Right

= .05

n’ = 5 (not 6; 1 elim.)

Critical Value(s):

Test Statistic:

Decision:

Conclusion:Reject at = .05

RejectDo Not Reject

1 T0

Since One-Tailed Test & Current Shifted Right, Use T-: T- = 0

R Signed Rank Test

> old <- c(9.98, 9.88, 9.9, 9.99, 9.94, 9.84)> new <- c(9.88, 9.86, 9.83, 9.8, 9.87, 9.84)> wil = wilcox.test(old, new, paired = TRUE, alternative = "greater")> wil

Wilcoxon signed rank test with continuity correction

data: old and newV = 15, p-value = 0.02895alternative hypothesis: true location shift is greater than 0

> abs(qnorm(wil$p.value))/sqrt(6 + 6) # Gives the approximate effect size.

[1] 0.5474433>

SPSS Signed Rank Test

• Read “loyalty program.sav” into SPSS.

• Under Analyze Nonparametric Tests Legacy Dialogs Two-Related-Samples Tests

• Set “old consumption level” as Variable 1, and “consumption level under new loyalty program” as Variable 2.

• Click Options, and select Descriptive.


SPSS Signed Rank Test

• Since Sig. = 0.042 < 0.05, reject Ho.

• Consumption level under new loyalty program is significantly worse than the old level.

Wilcoxon test

The procedure is to subtract the theoretical mu and rank the differences according to their numerical value, ignoring the sign, and then calculate the sum of the positive or negative ranks.

Assuming only that the distribution is symmetric around mu, the test statistic corresponds to selecting each number from 1 to n with probability 1/2 and calculating the sum.

The distribution of the test statistic can be calculated exactly. It becomes computationally excessive in large samples, but the distribution is then very well approximated by a normal distribution.

> wilcox.test(daily.intake, mu=7725)The test statistic V is the sum of the positive ranks.

Kruskal-Wallis H-Test


1. Sign Test







1. Tests the Equality of More Than 2 (p) Population Probability Distributions

2. Used to Analyze Completely Randomized Experimental Designs

4. Uses 2 Distribution with p - 1 df If At Least 1 Sample Size nj > 5

Kruskal-Wallis H-Test Assumptions

1. Corresponds to ANOVA for More Than 2 Populations

2. Independent, Random Samples

3. At Least 5 Observations Per Sample

Kruskal-Wallis H-Test Procedure

1. Assign Ranks, Ri , to the n Combined Observations Smallest Value = 1; Largest Value = n Average Ties

2. Sum Ranks for Each Group

3. Compute Test Statistic

131

12

1

2

nn

R

nnH

p

j j

j

Squared total of each group

Kruskal-Wallis H-Test Example

As a marketing manager, you want to see how 3 different price levels affect sales. You assign 15 branches, 5 per price level, to the three price levels. At the .05 level, is there a difference in sales under the three price levels?

Price1 Price2 Price325.40 23.40 20.0026.31 21.80 22.2024.10 23.50 19.7523.74 22.75 20.6025.10 21.60 20.40


Solution

Raw Data

Price1 Price2 Price325.40 23.40 20.0026.31 21.80 22.2024.10 23.50 19.7523.74 22.75 20.6025.10 21.60 20.40

Ranks

Price1 Price2 Price314 9 215 6 712 10 111 8 413 5 365 38 17Total


Solution

58.11

486.191240

12

1635

17

5

38

5

65

1615

12

131

12

222

1

2

nn

R

nnH

p

j j

j

20 5.991


SolutionH0: Identical Distrib.

Ha: At Least 2 Differ

= .05

df = p - 1 = 3 - 1 = 2

Critical Value(s):

Test Statistic:

Decision:


There Is Evidence Pop. Distrib. Are Different

= .05

H = 11.58


> mydata <- read.table('sales.csv', header = T, sep=',')> krus = kruskal.test(sales ~ price, data = mydata)> krus

Kruskal-Wallis rank sum test

data: sales by priceKruskal-Wallis chi-squared = 11.58, df = 2, p-value = 0.003058

> abs(qnorm(krus$p.value))/sqrt(nrow(mydata)) # Gives the approximate effect size.[1] 0.7078519>

Multiple Comparison

> library(pgirmess) # The package is for doing multiple comparison, install.packages("pgirmess") if necessary.> kruskalmc(mydata$sales ~ mydata$price)Multiple comparison test after Kruskal-Wallis p.value: 0.05 Comparisons obs.dif critical.dif differencePrice1-Price2 5.4 6.771197 FALSEPrice1-Price3 9.6 6.771197 TRUEPrice2-Price3 4.2 6.771197 FALSE

SPSS Kruskal-Wallis H-Test

• Read “sales.sav” into SPSS.

• Under Analyze Nonparametric Tests Legacy Dialogs K Independent Samples

• Move “sales” under Test Variable List, and “price” in Grouping Variable Box.

• Click Define Range, enter ‘1’ as Minimum and ‘3’ as Maximum.

• Click Options, and select Descriptive.


SPSS Kruskal-Wallis H-Test


• There is a significant difference in sales under the three price levels.

• Price1 gives rise to the highest sales.

Friedman Fr-Test


1. Sign Test






Friedman Fr-Test

1. Tests the Equality of More Than 2 (p) Population Probability Distributions

2. Corresponds to ANOVA for More Than 2 Means

3. Used to Analyze Randomized Block Experimental Designs

4. Uses 2 Distribution with p - 1 df If either p, the number of treatments, or b,

the number of blocks, exceeds 5

Friedman Fr-Test Assumptions

1. The p treatments are randomly assigned to experimental units within the b blocks Samples

2. The measurements can be ranked within the blocks

3. Continuous population probability distributions

Friedman Fr-Test Example

• For dependent samples.• Three price levels were

tested one after another in each branch.

• At the .05 level, is there a difference in the sales under the three price levels?

Price1 Price2 Price33 5 0

23 17 1511 5 78 4 2

19 11 5

Friedman Fr-Test Procedure

1. Assign Ranks, Ri = 1 – p, to the p treatments in each of the b blocks Smallest Value = 1; Largest Value = p Average Ties

2. Sum Ranks for Each Treatment

3. Compute Test Statistic

131

12

1

2

pbRpbp

Fp

jjr

Squared total of each treatment

Friedman Fr-Test Solution

Raw Data

Price1 Price2 Price33 5 0

23 17 1511 5 78 4 2

19 11 5

Ranks

Price1 Price2 Price32 3 13 2 13 1 23 2 1

3 2 1

14 10 6Total


64.660)332(60

12

)13()5(3)61014()13()3)(5(

12

131

12

222

1

2

r

r

p

jjr

F

F

pbRpbp

F

20 5.991


H0: Identical Distrib.

Ha: At Least 2 Differ

= .05

df = p - 1 = 3 - 1 = 2

Critical Value(s):

Test Statistic:

Decision:


There Is Evidence Pop. Distrib. Are Different

= .05

Fr = 6.64

Friedman Fr-Test

> mydata <- read.table('related sample sales.csv', header = T, sep=',')

> temp = as.matrix(mydata)

> fri = friedman.test(temp)

> fri

Friedman rank sum test

data: temp

Friedman chi-squared = 8.4, df = 2, p-value = 0.015

> abs(qnorm(fri$p.value))/sqrt(nrow(temp)*ncol(temp))

[1] 0.5603451

Friedman Fr-Test

> friedmanmc(temp)

Multiple comparisons between groups after Friedman test

p.value: 0.05

Comparisons

obs.dif critical.dif difference

1-2 6 7.570429 FALSE

1-3 9 7.570429 TRUE

2-3 3 7.570429 FALSE

>

SPSS Friedman Fr-Test

• Read “related samples sales.sav” into SPSS.

• Under Analyze Nonparametric Tests Legacy Dialogs K Related Samples

• Move price1 to price3 into the Test Variables Box.

• Click Statistics, and select Descriptive.


SPSS Friedman Fr-Test


• There is a significant difference in sales under the three price levels.

• Price1 gives rise to the highest sales.


1. Sign Test






Nonparametric Correlation

• Spearman’s Rho– Pearson’s correlation on the ranked data

• Kendall’s Tau– Better than Spearman’s for small samples

Spearman’s Rank Correlation Coefficient

1. Measures Correlation Between Ranks

2. Corresponds to Pearson Product Moment Correlation Coefficient

3. Values Range from -1 to +1

4. Equation (Shortcut)

1

61

2

2

nn

drs

Spearman’s Rank Correlation Procedure

1. Assign Ranks, Ri , to the Observations of Each Variable Separately

2. Calculate Differences, di , Between Each Pair of Ranks

3. Square Differences, di 2, Between Ranks

4. Sum Squared Differences for Each Variable

5. Use Shortcut Approximation Formula

Spearman’s Rank Correlation Example

You’re a research assistant for the FBI. You’re investigating the relationship between a person’s attempts at deception & % changes in their pupil size. You ask subjects a series of questions, some of which they must answer dishonestly. At the .05 level, what is the correlation coefficient?

Subj. DeceptionPupil

1 87 102 63 63 95 114 50 75 43 0

Spearman’s Rank Correlation Table

Subj. Decep. R1i Pupil R2i di di2

Total



1 87 10

2 63 6

3 95 11

4 50 7

5 43 0

Total



1 87 4 10

2 63 3 6

3 95 5 11

4 50 2 7

5 43 1 0

Total



1 87 4 10 4

2 63 3 6 2

3 95 5 11 5

4 50 2 7 3

5 43 1 0 1

Total



1 87 4 10 4 0

2 63 3 6 2 1

3 95 5 11 5 0

4 50 2 7 3 -1

5 43 1 0 1 0

Total



1 87 4 10 4 0 0

2 63 3 6 2 1 1

3 95 5 11 5 0 0

4 50 2 7 3 -1 1

5 43 1 0 1 0 0

Total 2

Spearman’s Rank Correlation Solution

90.0

10.01

155

261

1

61

2

21

2

nn

dr

n

ii

s

Doing a Rank Correlation

Correlation Output

Correlations

bring shopping

bags use scrap paper

Kendall's tau_b bring shopping bags Correlation Coefficient 1.000 .356*

Sig. (2-tailed) . .040

N 25 25

use scrap paper Correlation Coefficient .356* 1.000

Sig. (2-tailed) .040 .

N 25 25

Spearman's rho bring shopping bags Correlation Coefficient 1.000 .552**

Sig. (2-tailed) . .004

N 25 25

use scrap paper Correlation Coefficient .552** 1.000

Sig. (2-tailed) .004 .

N 25 25

*. Correlation is significant at the 0.05 level (2-tailed).

**. Correlation is significant at the 0.01 level (2-tailed).

20-88

Related-Samples Nonparametric Tests:

McNemar Test

BeforeAfter

Do Not FavorAfterFavor

Favor A B

Do Not Favor C D

20-89

An Example of the McNemar Test

BeforeAfter

Do Not FavorAfterFavor

Favor A=10 B=90

Do Not Favor C=60 D=40

Chi-square Test

The Chi-Square Test procedure tabulates a variable into categories and tests the hypothesis that the observed frequencies do not differ from their expected values.

From the menus choose:

Analyze Nonparametric Tests Chi-Square...

Chi-square Test

Select Day of the Week as the test variable.

Click OK.

Chi-square Test

Average Daily Sales

Observed N Expected N Residual

44 44 84.1 -40.1

78 78 84.1 -6.1

84 84 84.1 -.1

89 89 84.1 4.9

90 90 84.1 5.9

94 94 84.1 9.9

110 110 84.1 25.9

Total589

Test Statistics

Average Daily Sales

Chi-Square 29.389a

df 6

Asymp. Sig. .000

a. 0 cells (.0%) have expected frequencies less than 5. The minimum expected cell frequency is 84.1.

Chi-square Test

Chi-square Test

End of Chapter

95

Nonparametric Statistics. Salad bar, Pizza Hut, China, State of the Art in Structural Engineering.

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