chapter

© 2010 Pearson Prentice Hall. All rights reserved

Chapter

Nonparametric Statistics

15


Section

An Overview of Nonparametric Statistics

15.1

© 2010 Pearson Prentice Hall. All rights reserved 15-3

Objective

1. Understand the difference between parametric statistical procedures and nonparametric statistical procedures


Parametric statistical procedures are inferential procedures conducted under the assumption that the underlying distribution of the data belongs to some parametric family of distributions (such as the normal distribution).


Nonparametric statistical procedures are inferential procedures that do not make any assumptions about the underlying distribution of the data. They do not require that the population belong to any particular parametric family of distributions (such as the normal distribution) and, therefore, are often referred to as distribution-free procedures.


Advantages of Nonparametric Statistical Procedures

• Most of the tests have very few requirements, so it is unlikely that these tests will be used improperly.



• For some nonparametric procedures, the computations are fairly easy.




• For some nonparametric procedures, the computations are fairly easy.

• The procedures can be used for count data or rank data, so nonparametric methods can be used on data, such as the rankings of a movie as excellent, good, fair, or poor.



Disadvantages of Nonparametric Statistical Procedures

• Nonparametric procedures are less efficient than parametric procedures. This means that a larger sample size is required when conducting a nonparametric procedure to have the same probability of a Type I error as the equivalent parametric procedure.


• Nonparametric procedures often discard useful information. For example, the sign test uses only the sign of the data and rank tests merely preserve order-the magnitude of the actual data values is lost. As a result, nonparametric procedures are typically less powerful. Recall that the power of a test refers to the probability of making a Type II error. A Type II error occurs when a researcher does not reject the null hypothesis when the alternative hypothesis is true.



• Because fewer requirements must be satisfied to conduct these tests, researchers sometimes use these procedures when parametric procedures can be used.



Nonparametric Test Parametric Test Efficiency

Runs test for randomness

No corresponding test --

Sign test Single sample z-test or t-test

0.955 (for small samples that come from a normal population)0.75 (for large samples if data are normal)

Wilcoxon matched-pairs test

Inference about the difference of two means –dependent samples

0.955 (if the differences are normal)

Mann-Whitney test Inference about the difference of two means – independent samples

0.955 (if data are normal)

Spearman rank-correlation coefficient

Linear correlation 0.912 (if the data are bivariate normal)

Kruskal-Wallis Test One-way ANOVA 0.955 (if the data are normal)


“In Other Words”

The lower the efficiency is, the larger the sample size must be for a nonparametric test to have the probability of a Type I error the same as it would be for its equivalent parametric test.


Section

Runs Test for Randomness

15.2


Objective

1. Perform a runs test for randomness


A runs test for randomness is used to test whether data have been obtained or occur randomly. A run is a sequence of similar events, items, or symbols that is followed by an event, item, or symbol that is mutually exclusive from the first event, item, or symbol. The number of events, items, or symbols in a run is called its length.


CAUTION!

Runs tests are used to test whether it is reasonable to conclude that data occur randomly, not whether the data are collected randomly. For example, we might wonder whether defective parts come off an assembly line randomly or systematically. If broken parts occur systematically (such as every fourth part), we might be led to believe that we have a broken machine. We don’t collect the data randomly; instead, we select 100 consecutive parts. We want to know whether the defective parts in the 100 selected occur randomly.


Notation Used in Conducting a Runs Test for Randomness

• Let n represent the sample size of which there are two mutually exclusive types.

• Let n1 represent the number of observations of the first type.

• Let n2 represent the number of observations of the second type.

• Let r represent the number of runs.


Parallel Example 1: Notation in a Runs Test for Randomness

The following data represent the league that won the World Series for the years 1996-2007. Let “AL” represent the American League and “NL” represent the National League.

AL NL AL AL AL NL AL NL AL AL NL AL

Identify the values of n, n1, n2 and r.


Solution

Let n represent the number of World Series in the sample. Let n1 represent the number of World Series won by the American League and n2 the number of World Series won by the National League. Lastly, let r represent the number of runs.

Then, there are n =12 World Series in the sample, n1 = 8 World Series won by the American League, n2 =4 World Series won by the National League and r =9 runs.


Test Statistic for a Runs Test for Randomness

Small-Sample Case: If n1≤20 and n2≤20, the test statistic in the runs test for randomness is r, the number of runs.

Large-Sample Case: n1>20 or n2>20, the test statistic in the runs test for randomness is

z r r

r

where

r 2n1n2

n1 and r

2n1n2 2n1n2 n n2 n 1


Critical Values for a Runs Test for Randomness

Small-Sample Case: To find the critical value at the = 0.05 level of significance for a runs test, we use Table X if n1≤20 and n2≤20.

Large-Sample Case: If n1>20 or n2>20, the critical value is found from Table V, the standard normal table.


Parallel Example 2: Obtaining Critical Values from Table X

Find the upper and lower critical values if n1=8 and n2=4.


Solution

From Table X, the lower critical value is 3 and the upper critical value is 10.


Runs Test for Randomness

To test the randomness of data, we can use the following steps, provided that

1. the sample is a sequence of observations recorded in the order of their occurrence, and

2. the observations can be categorized into two mutually exclusive categories.


Step 1: Assume the data are random. This formsthe basis of the null and alternative hypotheses, which are structured as follows:H0: The sequence of data is randomH1: The sequence of data is not random


Step 2: Determine a level of significance, ,based on the seriousness of making a Type I error. The level of significance is used to determine the critical value.

Note: For the small-sample case, we must use the level of significance =0.05.


Step 3: Use the number of runs, r, to compute the test statistic.

Small-Sample Case Large-Sample Case

r

z0 r r

r


Step 4: Compare the critical value to the teststatistic.


If r ≤ lower critical value of r ≥ upper

critical value, reject the null hypothesis

If or

, reject the null hypothesis

z0 z 2

z0 z 2


Step 5: State the conclusion.


Parallel Example 3: Testing for Randomness (Small-Sample Case)

The following data represent the league that won the World Series for the years 1996-2007. Let “AL” represent the American League and “NL” represent the National League.

AL NL AL AL AL NL AL NL AL AL NL AL

Test the claim that leagues win the World Series in a non-random way at the = 0.05 level of significance.


Solution

The sample is a sequence of observations (which league won the World Series in a particular year) recorded in the order of occurrence. The observations are in two mutually exclusive categories, American League or National League. The requirements for the test are satisfied.


Solution

Step 1: We are testing the hypothesis that the sequence of observations is random. Thus, H0: The sequence of data is randomH1: The sequence of data is not random

Step 2: The level of significance is = 0.05. The lower critical value is 3 and the upper critical value is 10 (Parallel Example 2).


Solution

Step 3: The test statistic is r = 9 (Parallel Example 1).

Step 4: Since the test statistic is between the lower and upper critical values, we do not reject the null hypothesis.

Step 5: There is insufficient evidence to conclude that the World Series were won by the two leagues in a nonrandom way during the years 1996-2007.


Section

Inferences About Measures of Central Tendency

15.3


Objective

1. Conduct a one-sample sign test


A one-sample sign test is a nonparametric test that uses data, converted to plus and minus signs, to test a hypothesis regarding the median of a population. Data values equal to the assumed value of the median are ignored during the test.


Test Statistic for a One-Sample Sign TestThe test statistic will depend on the structure of the hypothesis test and on the sample size.

Small-Sample Case: (n ≤ 25)

Two-Tailed Left-Tailed Right-Tailed

H0: M =M0 H0: M =M0 H0: M =M0

H1: M≠ M0 H1: M < M0 H1: M > M0

The test statistic, k, will be the smaller of the number of minus signs or plus signs

The test statistic, k, will be the number of plus signs.

The test statistic, k, will be the number of minus signs.


Large-Sample Case: (n > 25)The test statistic, z, is

where n is the number of minus and plus signs and k is obtained as described in the small-sample case.

z0 k 0.5 n

2n

2


Critical Values for a One-Sample Sign Test

Small-Sample Case: To find the critical value for a one-sample sign test, we use Table XI if n ≤ 25.

Large-Sample Case: If n >25, the critical value is found from Table V, the standard normal table. The critical value is always located in the left tail of the standard normal distribution. For a two-tailed test, the critical value is . For a left-tailed or right-tailed test, the critical value is .

z 2

z


One-Sample Sign TestTo test hypotheses regarding the median of apopulation, we use the following steps, providedthat the sample is a random sample.Step 1: Determine the null and alternative hypotheses.

The hypotheses can be structured in one of three ways:

Note: M0 is the assumed value of the median.


H0: M =M0 H0: M =M0 H0: M =M0

H1: M≠ M0 H1: M < M0 H1: M > M0


Step 2: Count the number of observations below M0, and assign them minus (-) signs. Count the number of observations above M0, and assign them plus (+) signs.


Step 3: Select a level of significance, ,based on the seriousness of making a Type I error. The level of significance is used to determine the critical value. The critical value for small samples (n ≤ 25) is found from Table XI. Thecritical value for large samples (n > 25) is found from Table V.


Step 4: Obtain the test statistic, k.

Note that k is the smaller of the number of minus signs and plus signs in the two-tailed test, that k is the number of plus signs in the left-tailed test, and that k is the number of minus signs in the right tailed test. In addition, n is the total number of plus and minus signs.


k

z0 k 0.5 n

2n

2


Step 5: Compare the critical value to the teststatistic.


If k ≤ critical value, reject the null hypothesis

Two-tailed: If , reject the null hypothesis.

Left-tailed or right-tailed:

If , reject the null hypothesis.

z0 z 2

z0 z


Parallel Example 1: Conducting a One-Sample Sign Test (Small-Sample Case)

According to the United States Bureau of Labor Statistics, in 2000, the median tenure of employees with their current employer is 3.5 years. An economist believes that the median has increased since then. To test this claim, he randomly selects 16 employed individuals, determines their length of employment and obtains the following data.

0.3 0.8 0.7 3.2 10.3 1.4 0.2 0.93.6 6.3 11.2 12.8 7.3 13.0 3.8 23.6

Test the claim at the =0.05 level of significance.


Solution

The data were obtained from a random sample so the conditions of the test are met.

Step 1: We want to know if the median tenure of employees with their current employer is greater than 3.5 years. This is a right-tailed test.

H0: M=3.5 versus H1: M > 3.5


Solution

Step 2: There are 7 observations less than 3.5 and 9 observations greater than 3.5. Thus, we have 7 minus signs and 9 plus signs with n=16.

Step 3: Because this is a right-tailed test and n ≤ 25, we find the critical value at the = 0.05 level of significance with n=16 to be 4 (see Table XI).

Step 4: The test statistic is the number of minus signs. Thus, k =7.


Solution

Step 5: Since the test statistic is greater than the critical value, 4, we do not reject the null hypothesis.

Step 6: There is insufficient evidence to support the hypothesis that the median tenure of employees with their employer is greater than 3.5 years.


Section

Inferences About The Difference Between Two Medians: Dependent Samples

15.4


Objective

1. Test a hypothesis about the difference between the medians of two dependent samples


The Wilcoxon Matched-Pairs Signed-Ranks Test is a nonparametric procedure used to test the equality of two population medians by dependent sampling.


Test Statistic for the Wilcoxon Matched-Pairs Signed-Ranks Test

The test statistic will depend on the size of the sample and on the alternative hypothesis. Let n represent the number of nonzero differences.

Small-Sample Case: (n ≤ 30)


H0: MD =0 H0: MD =0 H0: MD =0

H1: MD≠ 0 H1: MD < 0 H1: MD > 0

Test Statistic: T is the smaller of T+ or |T-|

Test Statistic: T = T+

Test Statistic: T = |T-|


Large-Sample Case: (n > 30)The test statistic is given by

where T is the test statistic from the small-sample case.

z0 T

n n 1 4

n n 1 2n 1 24


Critical Value for Wilcoxon Matched-Pairs Signed-Ranks Test

Small-Sample Case: (n ≤ 30)Using as the level of significance, the critical value(s) is (are) obtained from Table XII in Appendix A.


T 2

T

T


Large-Sample Case: (n > 30)Using as the level of significance, the critical value(s) is obtained from Table V in Appendix A. The critical value is always in the left tail of the standard normal distribution.

z 2

z


z


Wilcoxon Matched-Pairs Signed-Ranks Test

If a hypothesis is made regarding the medians of twopopulations, we can use the following steps to test thehypothesis, provided that

1. the samples are dependent random samples and2. the distribution of the differences is symmetric.

Although tests for verifying the symmetry of data exist,we do not present them in this text. All the data givensatisfy the second requirement.


Step 1: Determine the null and alternative hypotheses. The hypotheses can be structured in one of three ways:

Note: MD is the median of the differences of matched pairs.


H0: MD = 0 H0: MD = 0 H0: MD = 0

H1: MD≠ 0 H1: MD < 0 H1: MD > 0


Step 2: Compute the differences in the matched-pairs observations. Rank the absolute value of all sample differences from smallest to largest after discarding those differences that equal 0. Handle ties by finding the mean of the ranks for tied values. Assign negative values to the ranks where the differences are negative and positive values to the ranks where the differences are positive. Find the sum of the positive ranks, T+, and the sum of the negative ranks T-.


Step 3: Draw a boxplot of the differences to compare the sample data from the two populations. This helps to visualize the difference in the medians.


Step 4: Choose a level of significance, ,based on the seriousness of making a Type I error. The level of significance is used to determine the critical value. The critical value is found from Table XII for small samples (n ≤ 30). The criticalvalue is found from Table V for large samples (n > 30).


Step 5: Compute the test statistic.

Small-Sample Case (n ≤ 30)


H0: MD =0 H0: MD =0 H0: MD =0

H1: MD≠ 0 H1: MD < 0 H1: MD > 0

Test Statistic: T is the smaller of T+ or |T-|

Test Statistic: T = T+

Test Statistic: T = |T-|


Large-Sample Case (n > 30)


z0 T

n n 1 4

n n 1 2n 1 24


Step 6: Compare the critical value with the teststatistic.


Two-tailed: If T < , reject H0.

Two-tailed: If , reject H0.

Left-tailed: If T < , reject H0.

Left-tailed: If , reject H0.

Right-tailed: If T < , reject H0.

Right-tailed: If , reject H0.

z0 z 2

z0 z

z0 z

T 2

T

T


Parallel Example 1: Wilcoxon Matched-Pairs Signed-Ranks Test (Small-Sample Case)

The data on the following slide represent the cost of a one night stay in Hampton Inn Hotels and La Quinta Inn Hotels for a random sample of 10 cities. Test the claim that Hampton Inn Hotels are priced differently than La Quinta Hotels at the =0.05 level of significance.


City Hampton Inn La Quinta

Dallas 129 105

Tampa Bay 149 96

St. Louis 149 49

Seattle 189 149

San Diego 109 119

Chicago 160 89

New Orleans 149 72

Phoenix 129 59

Atlanta 129 90

Orlando 119 69


Solution

The data were obtained randomly. We assume that the symmetry requirement is satisfied.

Step 1: We want to know if the hotels are priced differently. This is a two-tailed test.

H0: MD =0 versus H1: MD ≠ 0


Solution

Step 2: In order to calculate T+ and T-, we must find the differences, rank them, and then attach the sign of the difference to the ranks. The differences and their signed ranks are given in the next slide.


City

Hampton Inn

La Quinta

D=HI-LQ |D|

Signed

RanksDallas 129 105 24 24 +2Tampa Bay 149 96 53 53 +6

St. Louis 149 49 100 100 +10Seattle 189 149 40 40 +4San Diego 109 119 -10 10 -1Chicago 160 89 71 71 +8New Orleans 149 72 77 77 +9Phoenix 129 59 70 70 +7Atlanta 129 90 39 39 +3Orlando 119 69 50 50 +5


Solution

Step 2: From the previous slide, we find that T+=54 and |T-| = 1.


Solution

Step 3: The figure below shows a boxplot of the differences. The boxplot indicates that the sample-median difference is about 51.


Solution

Step 4: We are testing the hypothesis at the =0.05 level of significance. Since this is a two-tailed test and the sample size is less than 30, we find the critical value with n=10 by using Table XII and obtain T0.025=8.

Step 5: The test statistic is the smaller of T+ and |T-| which is 1.


Solution

Step 6: The test statistic is less than the critical value (1< 8), so we reject the null hypothesis.

Step 7: There is sufficient evidence at the =0.05 level of significance to conclude that the median room price at Hampton Inns is different than the median room price at La Quinta Inns.


Section

Inferences About The Difference Between Two Medians: Independent Samples

15.5


Objective

1. Test a hypothesis about the difference between the medians of two independent samples


The Mann-Whitney Test is a nonparametric procedure that is used to test the equality of two population medians from independent samples.


Test Statistic for the Mann-Whitney Test

The test statistic will depend on the size of the samples from each population. Let n1 represent the sample size for population X and n2 represent the sample size for population Y.

Small-Sample Case: (n1 ≤ 20 and n2 ≤ 20 ) If S is the sum of the ranks corresponding to the sample from population X, then the test statistic, T, is given by

Note: The value of S is always obtained by summing the ranks of the sample data that correspond to Mx, the median of population X, in the hypothesis.

T S n1 n1 1

2


Large-Sample Case: (n1 > 20 or n2 > 20 )From the Central Limit Theorem, the test statistic is given by


z0 T

n1n2

2n1n2 n1 n2 1

12


Critical Value for Mann-Whitney Test

Small-Sample Case: (n1 ≤ 20 and n2 ≤ 20 )Using as the level of significance, the critical value(s) is(are) obtained from Table XIII in Appendix A.


w 2

w

w1 n1n2 w

w1 2 n1n2 w 2


Large-Sample Case: (n1 > 20 or n2 > 20 )Using as the level of significance, the critical value(s) is(are) obtained from Table V in Appendix A.


z 2 and z 2

z

z


Mann-Whitney Test

To test hypotheses regarding the medians of twopopulations, we can use the following steps provided that

1. the samples are independent random samples and2. the shape of the distributions are the same.

Throughout this section, we will assume that thecondition that the shape of the distributions be the sameis satisfied.


Step 1: Draw a side-by-side boxplot to compare the sample data from the two populations. This helps to visualize the difference in the medians.


Step 2: Determine the null and alternative hypotheses. The hypotheses are structured as follows:

Note: Mx is the median of population X and My is the median of population Y.


H0: Mx = My H0: Mx = My H0: Mx = My

H1: Mx ≠ My H1: Mx < My H1: Mx > My


Step 3: Rank all sample observations from smallest to largest. Handle ties by finding the mean of the ranks for tied values. Find the sum of the ranks for the sample from population X.


Step 4: Choose a level of significance, ,to match the seriousness of making a Type I error. The level of significance is used to determine the critical value. The critical value is found from Table XIII for small samples (n1 ≤ 20 and n2 ≤ 20) and from Table V for large samples (n1 > 20 or n2 > 20).


Step 5: Compute the test statistic. Note that S is the sum of the ranks obtained from the sample observations from population X. In addition, n1 is the size of the sample from population X, and n2 is the size of the sample from population Y.


T S n1 n1 1

2

z0 T

n1n2

2n1n2 n1 n2 1

12


Step 6: Compare the critical value with the teststatistic.


Two-tailed: If T < , or T > , reject H0.

Two-tailed: If or , reject H0.

Left-tailed: If T < , reject H0.

Left-tailed: If , reject H0.

Right-tailed: If T > , reject H0.

Right-tailed: If , reject H0.

z0 z 2

z0 z

z0 z

w 2

w

w1

w1 2

z0 z 2


Parallel Example 1: Mann-Whitney Test (Small-Sample Case)

A researcher wanted to know whether “state” quarters had a weight that is more than “traditional” quarters. He randomly selected 18 “state” quarters and 16 “traditional” quarters, weighed each of them and obtained the following data.


Parallel Example 1: Mann-Whitney Test (Small-Sample Case)

Test the claim that state quarters have a higher median weight than traditional quarters at the =0.05 level of significance.


Solution

Step 1:


Solution

Step 1: Based on the boxplots, the median weight for the state quarters is higher. We want to estimate whether this difference is due to differences in the population medians or to sampling error.


Solution

Step 2: We want to know if the median weight for the state quarters is higher than the median weight of the traditional quarters. This is a right-tailed test.

H0: MState = MTraditional versus

H1: MState > MTraditional


Solution

Step 3: In order to calculate the test statistic, we combine the two data sets into one data set and arrange the data in ascending order. Ranks are shown on the following slide.


5.7 (20) 5.67 (13.5) 5.67 (13.5) 5.55 (2.5)5.73 (27.5) 5.61 (6) 5.7 (20) 5.61 (6)5.7 (20) 5.67 (13.5) 5.72 (24.5) 5.58 (4)5.65 (9.5) 5.62 (8) 5.66 (11) 5.74 (30.5)5.73 (27.5) 5.65 (9.5) 5.7 (20) 5.68 (16.5)5.79 (34) 5.73 (27.5) 5.68 (16.5) 5.53 (1)5.77 (33) 5.71 (23) 5.67 (13.5) 5.55 (2.5)5.7 (20) 5.76 (32) 5.61 (6) 5.74 (30.5)5.73 (27.5) 5.72 (24.5)

TraditionalState


Solution

Step 3: After ranking the observations, we add up the ranks corresponding to the state quarters to obtain

S = 20+27.5+20+9.5+27.5+34+33+20+ 27.5+13.5+6+13.5+8+9.5+27.5+23+32+24.5

=376.5


15-100

Solution

Step 4: Since this is a right-tailed test and both sample sizes are less than 20, we determine the right critical value with n1=18 and n2=16 at the =0.05 level of significance from Table XIII and obtain w0.95 = n1n2-w0.05 = 18(16)-96 = 192.


15-101

Solution

Step 5: The test statistic is

T S n1 n1 1

2376.5

18(19)

2205.5


15-102

Solution

Step 6: Since the test statistic is greater than the critical value (205.5 > 192), we reject the null hypothesis.

Step 7: There is sufficient evidence at the = 0.05 level of significance to conclude that the median weight of “state” quarters is greater than that of “traditional” quarters.


Section

Spearman’s Rank-Correlation Test

15.6


15-104

Objective

1. Perform Spearman’s rank-correlation test


15-105

The Spearman’s rank-correlation test is a nonparametric procedure that is used to test hypotheses regarding the association between two variables.


15-106

Test Statistic for Spearman’s Rank-Correlation Test

The test statistic will depend on the size of the sample, n, and on the sum of the squared differences

where di = the difference in the ranks of the two observations in the ith ordered pair.

The test statistic, rs, is also called Spearman’s rank-correlation coefficient.

rs 16 di

2n n2 1


15-107

CAUTION!

means to square the differences first

and then add up the squared differences.

di2


15-108

Critical Value for Spearman’s Rank-Correlation Test

Using as the level of significance, the critical value(s) is(are) obtained from Table XIV in Appendix A. For a two-tailed test, be sure to divide the level of significance, , by 2.


15-109

Spearman’s Rank-Correlation Test

To test hypotheses regarding the association between twovariables X and Y, we can use the following steps,provided that

1. the data are a random sample of n ordered pairs and2. each pair of observations is two measurements taken

on the same individual.

Notice that there is no requirement about the form of thedistribution of the data.


15-110

Step 1: Determine the null and alternative hypotheses which are structured as follows:

Two-Tailed One-Tailed One-Tailed

H0: X and Y are

not associated

H0: X and Y are

not associated

H0: X and Y are

not associated

H1: X and Y are

associated

H1: X and Y are

positively associated

H1: X and Y are

negatively associated


15-111

Step 2: Rank the X-values and rank the Y-values.Compute the differences between ranks and then square these differences. Compute the sum of the squared differences.


15-112

Step 3: Choose a level of significance, ,based on the seriousness of making a Type I error. The level of significance is used to determine the critical value. The critical value is found in Table XIV.


15-113


where n is the sample size and di is the difference in the ranks of the two observations in the ith ordered pair.

rs 16 di

2n n2 1


15-114

Step 5: Compare the critical value with the test statistic.

Hypothesis Decision RuleH0: X and Y are not associated

H1: X and Y are associated

Reject H0 if rs is greater than the critical value or if rs is less than the negative of the critical value in Table XIV

H0: X and Y are not associated

H1: X and Y are positively associated

Reject H0 if rs is greater than the critical value in Table XIV

H0: X and Y are not associated

H1: X and Y are negatively associated

Reject H0 if rs is less than the negative of the critical value in Table XIV


15-115



15-116

Parallel Example 1: Spearman’s Rank-Correlation Test

Is the price of a sport’s car associated with its performance? The following data represent the ranks of the price and performance of 8 sport’s cars. Using Spearman’s Rank Correlation, determine if the two variables are associated at the = 0.05 level of significance.


15-117

Car Rank of Price

Rank of Performance

BMW M3 Coupe 5 8

Chevy Corvette Z06 4 4

Ferrari 360 Modena 1 1

Lotus Elise 7 2

Mazda MP3 8 7

Mitsubishi Lancer Evolution VII

6 3

Porsche Boxster S 3 6

Porsche 911 Turbo 2 5


15-118

Solution

Step 1: We are looking for evidence that price and performance of sport’s cars are associated. Let X represent the price of the sport’s car and Y represent performance. The null and alternative hypotheses are as follows:H0: X and Y are not associatedH1: X and Y are associated


15-119

Solution

Step 2: Rank the X-values and rank the Y-values. Compute the differences in ranks and then square the differences. Calculate the sum of the squared differences to obtain . Details are on the following slide.

di2


15-120

Rank of X

Rank of Y d = X-Y d2

5 8 -3 9

4 4 0 0

1 1 0 0

7 2 5 25

8 7 1 1

6 3 3 9

3 6 -3 9

2 5 -3 9

di2 62


15-121

Solution

Step 3: This is a two-tailed test with n=8 and = 0.05. From Table XIV we determine the critical value to be 0.738.


15-122

Solution

Step 4: The test statistic is

rs 16 di

2n n2 1

16(62)

8(64 1)0.262


15-123

Solution

Step 5: Since the test statistic is less than the critical value and greater than the negative of the critical value (-0.738 < 0.262 < 0.738), we fail to reject the null hypothesis.

Step 6: There is insufficient evidence at the = 0.05 level of significance to conclude that the price and performance of sport’s cars are associated.


15-124

Large-Sample (n > 100) Approximation

If n > 100, the test statistic for Spearman’s Rank-Correlation Test is

Compare this test statistic with the critical value obtained from the standard normal table, Table V. For a two-tailed test, the critical values are . When testing for positive association, the critical value is . When testing for negative association, the critical value is .

z0 rs n 1

z 2

z

z


Section

Kruskal-Wallis Test

15.7


15-126

Objective

1. Test a hypothesis using the Kruskal-Wallis test


15-127

The Kruskal-Wallis Test is a nonparametric procedure that is used to test whether k independent samples come from populations with the same distribution.


15-128

Test Statistic for the Kruskal-Wallis Test

The test statistic for the Kruskal-Wallis test is

H 12

N(N 1)

1

ni

Ri ni N 1

2

2


15-129

A computational formula for the test statistic is

where

• Ri is the sum of the ranks of the ith sample

• is the sum of the ranks squared for the first sample• is the sum of the ranks squared for the second sample, and so on

• n1 is the number of observations in the first sample

• n2 is the number of observations in the second sample, and

so on

• N is the total number of observations (N=n1+n2+···+nk)

• k is the number of populations being compared

H 12

N(N 1)

R12

n1

R2

2

n2

Rk

2

nk

3 N 1

R12

R22


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Critical Value for Kruskal-Wallis Test

Small-Sample CaseWhen three populations are being compared and when the sample size from each population is 5 or less, the critical value is obtained from Table XV in Appendix A.

Large-Sample CaseWhen four or more populations are being compared or the sample size from one population is more than 5, the critical value is with k-1 degrees of freedom, where k is the number of populations and is the level of significance.

2


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Kruskal-Wallis Test

To test hypotheses regarding the distribution of threeor more populations, we can use the following steps,provided that two requirements are satisfied:

1. The samples are independent random samples2. The data can be ranked

Step 1: Draw side-by-side boxplots to compare the sample data from the populations. Doing so helps to visualize the differences, if any, between the medians.


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Step 2: State the null and alternative hypotheses, which are structured as follows:H0: the distributions of the populations are the sameH1: the distributions of the populations are not the same

Step 3: Rank all sample observations from smallest to largest. Handle ties by finding the mean of the ranks for tied values. Find the sum of the ranks for each sample.


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Step 4: Choose a level of significance, ,to match the seriousness of making a Type I error. The level of significance is used to determine the critical value. The critical value is found from Table XV for small samples. The critical value is

with k-1 degrees of freedom (found in Table VII) for large samples.

2


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H 12

N(N 1)

R12

n1

R2

2

n2

Rk

2

nk

3 N 1


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Step 6: Compare the critical value to the teststatistic. We reject the null hypothesis if the test statistic is greater than the critical value.


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Parallel Example 1: Kruskal-Wallis Test

The following data represent the weight (in grams) of pennies minted at the Denver mint in 1990, 1995, and 2000. Test the claim that the distribution of penny weights differs for the three years at the = 0.05 level of significance.


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1990 1995 20002.50 2.52 2.502.50 2.54 2.482.49 2.50 2.492.53 2.48 2.502.46 2.52 2.482.50 2.50 2.522.47 2.49 2.512.53 2.53 2.492.51 2.48 2.512.49 2.55 2.502.48 2.49 2.52


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Solution

The samples are independent random samples that can be ranked. Therefore, the conditions for the Kruskal-Wallis test are met.


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Solution

Step 1: Based on boxplots of the data, the medians do not appear to differ significantly.


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Solution

Step 2: We are interested in determining whether the distribution of penny weights differs for the three years.The null and alternative hypotheses are as follows:

H0: the distribution of penny weights are the same for the three years

H1: the distribution of penny weights are not the same for the three years


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Solution

Step 3: The ranks of the pennies are given in parentheses.

1990 1995 20002.50 (17.5) 2.52 (26.5) 2.50 (17.5)2.50 (17.5) 2.54 (32) 2.48 (5)2.49 (10.5) 2.50 (17.5) 2.49 (10.5)2.53 (30) 2.48 (5) 2.50 (17.5)2.46 (1) 2.52 (26.5) 2.48 (5)2.50 (17.5) 2.50 (17.5) 2.52 (26.5)2.47 (2) 2.49 (10.5) 2.51 (23)2.53 (30) 2.53 (30) 2.49 (10.5)2.51 (23) 2.48 (5) 2.51 (23)2.49 (10.5) 2.55 (33) 2.50 (17.5)2.48 (5) 2.49 (10.5) 2.52 (26.5)


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Solution

Step 3: We sum the ranks for each of the three years to obtain the following:

Year

1990 1995 2000

Sample size n1=11 n2=11 n3=11

Sum of ranks

R1=164.5 R2=214 R3=182.5


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Solution

Step 4: Since the sample sizes for each population are greater than 5, we find the critical value from the chi-square distribution with k-1=3-1=2 degrees of freedom with = 0.05. Thus, the critical value is

0.052 5.991.


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Solution

Step 5: Note that N=11+11+11=33. The test statistic is

H 12N N 1

R12

n1

R22

n2

R32

n3

3 N 1

12

33(331)

164.52

11

214 2

11

182.52

11

3(331)

1.221


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Solution

Step 6: Since the test statistic is less than the critical value, we fail to reject the null hypothesis.

Step 7: There is insufficient evidence at the = 0.05 level of significance to conclude that the distribution of penny weights differs for the years 1990, 1995 and 2000.

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