Slides by John Loucks St. Edward’s University

1 Slide© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied

or duplicated, or posted to a publicly accessible website, in whole or in part.

Slides by

JohnLoucks

St. Edward’sUniversity



Chapter 19Nonparametric Methods

Sign Test Wilcoxon Signed Rank Test Mann-Whitney-Wilcoxon Test Kruskal-Wallis Test Rank Correlation



Nonparametric Methods

Most of the statistical methods referred to as parametric require the use of interval- or ratio-scaled data.

Nonparametric methods are often the only way to analyze categorical ( nominal or ordinal) data and draw statistical conclusions.

Nonparametric methods require no assumptions about the population probability distributions. Nonparametric methods are often called distribution-free methods.



Nonparametric Methods

Whenever the data are quantitative, we will transform the data into categorical data in order to conduct the nonparametric test.



Sign Test

The sign test is a versatile method for hypothesis testing that uses the binomial distribution with p = .50 as the sampling distribution. We present two applications of the sign test: A hypothesis test about a population

median A matched-sample test about the difference between two populations



Hypothesis Test about a Population Median

We can apply the sign test by:• Using a plus sign whenever the data in the

sample are above the hypothesized value of the median• Using a minus sign whenever the data in the sample are below the hypothesized value of the median

• Discarding any data exactly equal to the hypothesized median



Hypothesis Test about a Population Median

The assigning of the plus and minus signs makes the situation into a binomial distribution application.• The sample size is the number of trials.• There are two outcomes possible per trial, a

plus sign or a minus sign.

• We let p denote the probability of a plus sign.• If the population median is in fact a particular value, p should equal .5.

• The trials are independent.



Hypothesis Test about a Population Median:

Small-Sample Case The small-sample case for this sign test should

be used whenever n < 20. The hypotheses are

The population median is differentthan the value assumed.

The population median equals thevalue assumed.

The number of plus signs is our test statistic. Assuming H0 is true, the sampling distribution

for the test statistic is a binomial distribution with p = .5. H0 is rejected if the p-value < level of significance, a.

0 : .50H p

0 : .50H p



Example: Potato Chip Sales


Smaller Sample Size Lawler’s Grocery Store made the decision to carry Cape May Potato Chips based on the manufacturer’s estimate that the median sales should be $450 per week on a per-store basis.

Lawler’s has been carrying the potato chips for three months. Data showing one-week sales at 10 randomly selected Lawler’s stores are shown on the next slide.




Smaller Sample Size

StoreNumber

56193612812

WeeklySales$485 562 415 860 426

Sign++-+-

StoreNumber

633984

10244

WeeklySales$474 662 380 515 721

Sign++-++

Example: Potato Chip Sales




Smaller Sample Size Example: Potato Chip Sales Lawler’s management requested the following hypothesis test about the population median weekly sales of Cape May Potato Chips (using a = .10).

H0: Median Sales = $450Ha: Median Sales ≠ $450

H0: p = .50Ha: p ≠ .50

In terms of the binomial probability p:




Smaller Sample Size Example: Potato Chip Sales

Number ofPlus Signs

012345

Probability.0010 .0098 .0439 .1172 .2051 .2461

Number ofPlus Signs

678910

Probability.2051 .1172 .0439 .0098 .0010

Binomial Probabilities with n = 10 and p = .50




Smaller Sample Size Example: Potato Chip Sales Because the observed number of plus signs is 7, we begin by computing the probability of obtaining 7 or more plus signs. The probability of 7, 8, 9, or 10 plus signs is: .1172 + .0439 + .0098 + .0010 = .1719. We are using a two-tailed hypothesis test, so: p-value = 2(.1719) = .3438. With p-value > a, (.3438 > .10), we cannot reject H0.




Smaller Sample Size ConclusionBecause the p-value > a, we cannot reject H0.

There is insufficient evidence in the sample to reject

the assumption that the median weekly sales is $450.




Larger Sample Size With larger sample sizes, we rely on the normal distribution approximation of the binomial distribution to compute the p-value, which makes the computations quicker and easier.

Normal Approximation ofthe Number of Plus Signs when H0: p = .50

Mean: = .50n Standard Deviation: .25n

Distribution Form: Approximately normal for n > 20




Larger Sample Size

H0: Median Age = 34 yearsHa: Median Age ≠ 34 years

Example: Trim Fitness CenterA hypothesis test is being conducted about the

median age of female members of the Trim Fitness

Center.

In a sample of 40 female members, 25 are older

than 34, 14 are younger than 34, and 1 is 34. Is there

sufficient evidence to reject H0? Use a = .05.




Larger Sample Size Example: Trim Fitness Center• Letting x denote the number of plus signs,

we will use the normal distribution to approximate the binomial probability P(x < 25).• Remember that the binomial distribution is discrete and the normal distribution is continuous.• To account for this, the binomial probability of 25 is computed by the normal probability interval 24.5 to 25.5.



p-Value = 2(1.0000 - .9726) = .0548

= .5(n) = .5(39) = 19.5


Larger Sample Size

p-Valuez = (x – )/ = (25.5 – 19.5)/3.1225 = 1.92

Test Statistic

Mean and Standard Deviation

.25 .25(39) 3.1225n




Larger Sample Size Rejection Rule

ConclusionDo not reject H0. The p-value for this

two-tail test is .0548. There is insufficient evidence in the sample to conclude that the median age is not 34 for female members of Trim Fitness Center.

Using .05 level of significance: Reject H0 if p-value < .05



Hypothesis Test with Matched Samples

A common application of the sign test involves using a sample of n potential customers to identify a preference for one of two brands of a product. The objective is to determine whether there is a difference in preference between the two items being compared.

To record the preference data, we use a plus sign if the individual prefers one brand and a minus sign if the individual prefers the other brand. Because the data are recorded as plus and minus signs, this test is called the sign test.



Hypothesis Test with Matched Samples:Small-Sample Case

The small-sample case for the sign test should be used whenever n < 20.

The hypotheses are

a : .50H p

0 : .50H p

A preference for one brandover the other exists.

No preference for one brandover the other exists.

The number of plus signs is our test statistic. Assuming H0 is true, the sampling distribution

for the test statistic is a binomial distribution with p = .5. H0 is rejected if the p-value < level of significance, a.




Maria Gonzales is the supervisor responsible forscheduling telephone operators at a major call center. She is interested in determining whether her operators’ preferences between the day shift (7 a.m.to 3 p.m.) and evening shift (3 p.m. to 11 p.m.) are different.

Example: Major Call Center

Maria randomly selected a sample of 16 operators

who were asked to state a preference for the one of

the two work shifts. The data collected from the

sample are shown on the next slide.





Worker12345678

ShiftPreference

DayEveningEveningEvening

DayEvening

Day(none)

Sign+---+-+

Worker 910111213141516

ShiftPreference

Evening EveningEvening(none)

EveningDay

EveningEvening

Sign---

-+--



4 plus signs10 negative

signs (n = 14)



Can Maria conclude, using a level of significance of a = .10, that operator preferences are different for the two shifts?

a : .50H p

0 : .50H p

A preference for one shiftover the other does not exist.

A preference for one shiftover the other does exist.



Number ofPlus Signs

01234567

Probability

.00006 .00085.00555 .02222 .06110 .12219 .18329.20947

Number ofPlus Signs

8 91011121314

Probability

.18329 .12219 .06110 .02222 .00555.00085.00006

Binomial Probabilities with n = 14 and p = .50





Because the observed number of plus signs is 4, we begin by computing the probability of obtaining 4 or less plus signs. The probability of 0, 1, 2, 3, or 4 plus signs is: .00006 + .00085 + .00555 + .02222 + .06110 = .08978. We are using a two-tailed hypothesis test, so: p-value = 2(.08978) = .17956. With p-value > a, (.17956 > .10), we cannot reject H0.





ConclusionBecause the p-value > a, we cannot reject H0. There

is insufficient evidence in the sample to conclude that

a difference in preference exists for the two work shifts.




Using H0: p = .5 and n > 20, the sampling distribution for the number of plus signs can be approximated by a normal distribution.

When no preference is stated (H0: p = .5), the sampling distribution will have:

The test statistic is:

H0 is rejected if the p-value < level of significance, a.

Mean: = .50nStandard Deviation: .25n

xz -

(x is the number of plus signs)

Hypothesis Test with Matched Samples:Large-Sample Case



Example: Ketchup Taste TestAs part of a market research study, a sample of

80 consumers were asked to taste two brands of

ketchup and indicate a preference. Do the data

shown on the next slide indicate a significantdifference in the consumer preferences for

the twobrands?




45 preferred Brand A Ketchup (+ sign recorded)

27 preferred Brand B Ketchup (_ sign recorded)

8 had no preference

Example: Ketchup Taste Test

The analysis will be based on

a sample size of 45 + 27 = 72.




Hypotheses

a : .50H p

0 : .50H p

A preference for one brand over the other exists

No preference for one brand over the other exists




Sampling Distribution for Number of Plus Signs

= .5(72) = 36

.25 .25(72) 4.243n




p-Value = 2(1.0000 - .9875) = .025

Rejection Rule

p-Value

z = (x – )/ = (45.5 - 36)/4.243 = 2.24 Test Statistic

Using .05 level of significance:Reject H0 if p-value < .05




ConclusionBecause the p-value < a, we can reject H0. There

is sufficient evidence in the sample to conclude that

a difference in preference exists for the two brands of

ketchup.




Wilcoxon Signed-Rank Test

The Wilcoxon signed-rank test is a procedure for analyzing data from a matched samples experiment. The test uses quantitative data but does not require the assumption that the differences between the paired observations are normally distributed.

It only requires the assumption that the differences have a symmetric distribution.

This occurs whenever the shapes of the two populations are the same and the focus is on determining if there is a difference between the two populations’ medians.




Let T - denote the sum of the negative signed ranks.

If the medians of the two populations are equal, we would expect the sum of the negative signed ranks and the sum of the positive signed ranks to be approximately the same.

Let T + denote the sum of the positive signed ranks.

We use T + as the test statistic.



Sampling Distribution of T +

for the Wilcoxon Signed-Rank Test

Mean:

( 1)(2 1)Standard Deviation: 24T

n n n

Distribution Form: Approximately normal for n > 10

( 1)4T

n n




Example: Express Deliveries


A firm has decided to select one of two expressdelivery services to provide next-day deliveries to itsdistrict offices. To test the delivery times of the two services, thefirm sends two reports to a sample of 10 district offices, with one report carried by one service and theother report carried by the second service. Do the dataon the next slide indicate a difference in the twoservices?




SeattleLos Angeles

BostonClevelandNew YorkHoustonAtlantaSt. LouisMilwaukeeDenver

32 hrs.30191615181410 716

25 hrs.241515131515 8 911

District Office OverNight NiteFlite




HypothesesH0: The difference in the median delivery times of

the two services equals 0.Ha: The difference in the median delivery times of

the two services does not equal 0.



Wilcoxon Signed-Rank Test Preliminary Steps of the Test

• Compute the differences between the paired observations.

• Discard any differences of zero.• Rank the absolute value of the differences

from lowest to highest. Tied differences are assigned the average ranking of their positions.• Give the ranks the sign of the original difference in the data.

• Sum the signed ranks.. . . next we will determine whether the

sum is significantly different from zero.




SeattleLos Angeles

BostonClevelandNew YorkHoustonAtlantaSt. LouisMilwaukeeDenver

7 6 4 1 2 3-1 2-2 5

District Office Differ. |Diff.| Rank Sign. Rank1097

1.546

1.5448

+10+9+7

+1.5+4+6-1.5+4-4

+8T+ = 44.0




Test Statistic

p-Valuep-Value = 2(1.0000 - .9535) = .093

44 27.5( 44) ( 1.68)9.81

P T P z p z -

( 1)(2 1) 10(11)(21) 9.8124 24T

n n n

( 1) 10(10 1) 27.54 4T

n n



ConclusionReject H0. The p-value for this two-tail

test is .093. There is insufficient evidence in the sample to conclude that a difference exists in the median delivery times provided by the two services.


Rejection RuleUsing .05 level of significance,

Reject H0 if p-value < .05



Mann-Whitney-Wilcoxon Test

This test is another nonparametric method for determining whether there is a difference between two populations.

This test is based on two independent samples. Advantages of this procedure are; It can be used with either ordinal data or

quantitative data. It does not require the assumption that the

populations have a normal distribution.




Ha: The two populations are not identicalH0: The two populations are identical

Instead of testing for the difference between the medians of two populations, this method tests to determine whether the two populations are identical. The hypotheses are:




Example: Westin FreezersManufacturer labels indicate the annual energy cost

associated with operating home appliances such as

freezers.The energy costs for a sample of 10 Westin freezers

and a sample of 10 Easton Freezers are shown on the

next slide. Do the data indicate, using a = .05, that a

difference exists in the annual energy costs for the two

brands of freezers?




$55.10

54.50

53.20

53.00

55.50

54.90

55.80

54.00

54.20

55.20

$56.10

54.70 54.40 55.40 54.10 56.00 55.50 55.00 54.30 57.00

Westin FreezersEaston Freezers



Hypotheses


Ha: Annual energy costs differ for the two brands of freezers.

H0: Annual energy costs for Westin freezers and Easton freezers are the same.



Mann-Whitney-Wilcoxon Test:Large-Sample Case

First, rank the combined data from the lowest to the highest values, with tied values being assigned the average of the tied rankings. Then, compute W, the sum of the ranks for the first sample.

Then, compare the observed value of W to the sampling distribution of W for identical populations. The value of the standardized test statistic z will provide the basis for deciding whether to reject H0.



Mann-Whitney-Wilcoxon Test:Large-Sample Case

1 2 1 21 ( 1)12W n n n n

Approximately normal, providedn1 > 7 and n2 > 7

Sampling Distribution of W with Identical Populations• Mean

• Standard Deviation

• Distribution Form

W = 1/2 n1(n1 + n2 + 1)




$55.10

54.50

53.20

53.00

55.50

54.90

55.80

54.00

54.20

55.20

$56.10

54.70 54.40 55.40 54.10 56.00 55.50 55.00 54.30 57.00

Westin Freezers Easton Freezers

Sum of Ranks Sum of Ranks

Rank Rank

86.5 123.5

12

128

15.510173513

1997

144

1815.5116

20



Sampling Distribution of W with Identical Populations

W = ½(10)(21) = 105


1 2 1 21 ( 1)12

1 (10)(10)(21)12 13.23

W n n n n

W



Rejection RuleUsing .05 level of significance,

Reject H0 if p-value < .05 Test Statistic

p-Valuep-Value = 2(.0808) = .1616


86.5 105( 44) ( 1.40)13.23

P W P z p z- -




ConclusionDo not reject H0. The p-value > a. There

is insufficient evidence in the sample data to conclude that there is a difference in the annual energy cost associated with the two brands of freezers.



Kruskal-Wallis Test

The Mann-Whitney-Wilcoxon test has been extended by Kruskal and Wallis for cases of three or more populations.

The Kruskal-Wallis test can be used with ordinal data as well as with interval or ratio data. Also, the Kruskal-Wallis test does not require the assumption of normally distributed populations.

Ha: Not all populations are identicalH0: All populations are identical



Test Statistic

Kruskal-Wallis Test

-

2

1

12 3( 1)( 1)k

iT

iT T i

RH nn n n

where: k = number of populations ni = number of observations in sample I nT = Sni = total number of observations in all samples Ri = sum of the ranks for sample i



Kruskal-Wallis Test

When the populations are identical, the sampling distribution of the test statistic H can be approximated by a chi-square distribution with k – 1 degrees of freedom.

This approximation is acceptable if each of the sample sizes ni is > 5.

The rejection rule is: Reject H0 if p-value < a This test is always expressed as an upper-tailed

test.



Kruskal-Wallis Test

John Norr, Director of Athletics at Lakewood High School, is curious about whether a student’s total number of absences in four years of high school is the same for students participating in no varsity sport, one varsity sport, and two varsity sports.

Example: Lakewood High School

Number of absences data were available for 20 recent graduates and are listed on the next slide. Test whether the three populations are identical in terms of number of absences. Use a = .10.



Kruskal-Wallis Test

Example: Lakewood High SchoolNo Sport 1 Sport 2 Sports

13 18 1216 12 226 19 927 7 1120 15 1514 20 21

17 10



Kruskal-Wallis Test

No Sport Rank 1 Sport Rank 2 Sports Rank

13 8 18 14 12 6.516 12 12 6.5 22 196 1 19 15 9 327 20 7 2 11 520 16.5 15 10.5 15 10.514 9 20 16.5 21 18

17 13 10 4Total 66.5 77.5 66

Example: Lakewood High School



Kruskal-Wallis Test

Using test statistic: Reject H0 if c2 > 4.60517 (2 d.f.)Using p-value: Reject H0 if p-value < .10

Rejection Rule

-

2

1

12 3( 1)( 1)k

iT

iT T i

RH nn n n

-

2 2 2(66.5) (77.5) (66.0)12 3(20 1) .353220(20 1) 6 7 7

k = 3 populations, n1 = 6, n2 = 7, n3 = 7, nT = 20 Kruskal-Wallis Test Statistic



Kruskal-Wallis Test

Do no reject H0. There is insufficient evidence to conclude that the populations are not identical. (H = .3532 < 4.60517)

Conclusion



Rank Correlation

The Pearson correlation coefficient, r, is a measure of the linear association between two variables for which interval or ratio data are available. The Spearman rank-correlation coefficient, rs , is a measure of association between two variables when only ordinal data are available.

Values of rs can range from –1.0 to +1.0, where• values near 1.0 indicate a strong positive

association between the rankings, and• values near -1.0 indicate a strong negative

association between the rankings.



Rank Correlation

Spearman Rank-Correlation Coefficient, rs

2

261 ( 1)

is

dr

n n -

-

where: n = number of observations being ranked xi = rank of observation i with respect to the first variable yi = rank of observation i with respect to the second variable di = xi - yi



Test for Significant Rank Correlation

0 : 0sH p

a : 0sH p

We may want to use sample results to make an inference about the population rank correlation ps. To do so, we must test the hypotheses:

(No rank correlation exists)(Rank correlation exists)



Rank Correlation

0sr

11sr n

-

Approximately normal, provided n > 10

Sampling Distribution of rs when ps = 0• Mean

• Standard Deviation

• Distribution Form



Rank Correlation

Example: Crennor InvestorsCrennor Investors provides a portfolio management

service for its clients. Two of Crennor’s analysts ranked

ten investments as shown on the next slide. Use rank

correlation, with a = .10, to comment on the agreement

of the two analysts’ rankings.



Rank Correlation

Analyst #2 1 5 6 2 9 7 3 10 4 8Analyst #1 1 4 9 8 6 3 5 7 2 10Investment A B C D E F G H I J

Example: Crennor Investors

0 : 0sH p

a : 0sH p (No rank correlation exists)(Rank correlation exists)

• Analysts’ Rankings

• Hypotheses



Rank Correlation

ABCDEFGHIJ

14986357210

1562973

1048

0-136-3-42-3-22

019369164944

Sum = 92

InvestmentAnalyst #1

RankingAnalyst #2

Ranking Differ. (Differ.)2



Sampling Distribution of rs

Assuming No Rank Correlation

Rank Correlation

1 .33310 1sr -

r = 0rs



Test Statistic2

26 6(92)1 1 0.4424( 1) 10(100 1)

is

dr

n n - -

- -

Rank Correlation

z = (rs - r )/r = (.4424 - 0)/.3333 = 1.33

Rejection RuleWith .10 level of significance:

Reject H0 if p-value < .10

p-Valuep-Value = 2(1.0000 - .9082) = .1836



Do no reject H0. The p-value > a. There is not a significant rank correlation. The two analysts are not showing agreement in their ranking of the risk associated with the different investments.

Rank Correlation

Conclusion



End of Chapter 19

Slides by John Loucks St. Edward’s University

Documents

Transcript of Slides by John Loucks St. Edward’s University