Chapter 21

21.1Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.

Chapter 21

Nonparametric Statistics


Nonparametric Statistics…

This chapter deals with statistical techniques that deal with ordinal data.

Recall: when the data are ordinal, the mean is not an appropriate measure of central location. Instead, we will test characteristics of populations without referring to specific parameters, hence the term nonparametric.

Rather than testing to determine whether the population means differ, we will test to determine whether the population locations differ…


Population Locations…

These two populations have the same location…

population 1

population 2


Population Locations…The location of pop’n 1 is to the left of the location of pop’n 2…

The location of pop’n 1 is to the right of the location of pop’n 2…

population 1 population 2

population 2 population 1


Problem Objectives…When the problem objective is to compare two populations the null hypothesis will state:

H0: The two population locations are the same.

The alternative hypothesis can take on any one of the following three forms:

H1: The location of population 1 is different from the location of population 2

H1: The location of population 1 is to the right of the location of population 2

H1: The location of population 1 is to the left of the location of population 2


The Alternative Hypotheses…

H1: The location of population 1 is different from the location of population 2

Used when we want to know whether there is sufficient evidence to infer that there is a difference between the two populations.




Used when we want to know whether we can conclude that the random variable in population 1 is larger in general than the random variable in population 2,

and, not surprisingly…



H1: The location of population 1 is to the left of the location of population 2

Used when we want to know whether we can conclude that the random variable in population 1 is smaller in general than the random variable in population 2.

NOTE: all of our hypotheses are phrased in terms of “1 then 2”. This is for consistency. Rather than state:

H1: The location of population 2 is to the left of the location of population 2, we would want to phrase this as:



Wilcoxon Rank Sum Test…

We’ll use the Wilcoxon Rank Sum Test for problems where:

— We’re asked to compare two populations,

— The data are ordinal or interval (where the normality requirement is unsatisfied), and

— The samples are independent.


Example 21.1

From these samples:

: 22, 23, 20

: 18, 27, 26 Can we conclude (at 5% confidence level of course) that the location of population 1 is to the left (i.e. “smaller”) that the location of population 2?

That is, we want to test:


H1: The location of population 1 is to the left of the location of population 2.

We can test this, we just need a test statistic…


Test Statistic…

Step #1… rank the observations from smallest to largest, assign a rank number, and add up the “rank sum”…

rank rank

22 3 18 1

23 4 27 6

20 2 26 5

T1=9 T2=12

*in the case of “ties” we average the ranks

of the tied observations.

We arbitrarily select T1 as the test statistic and label it “T”


Sampling Distribution of the Test StatisticA small value of T indicates most of the smaller observations are in sample 1 which was drawn from population 1 — but how small is “small”? Is 9 “small” enough?

We have our test statistic, T=9. We need to compare it to some critical value of “T” to know if we’re in the rejection region for H0 (or not).

So, what then, does the sampling distribution of “ranks” look like?


Sampling Distribution of the Test StatisticWe can build up the sampling distribution of the test statistic in much the same way we we built histograms for the outcomes of rolls of 2 and 3 dice…

Enumerate all possible combinations of ranks

Calculate ranks sums for the combinations

The probability of any rank sum is the number of occurrences divided by the total number of combinations…


Sampling Distribution of the Test Statistic Enumerate & Calculate & Probabilities…

Total of20 combinations

1 combination

3 combinations


X

Sampling Distribution of the Test Statistic

5%

P(T≤6) = 1/20 = .05Thus our critical value of T is 6

Since T=9 < TCritical=6, we cannotreject H0…


Example 21.1…

We cannot reject the null hypothesis, that is, there is not enough evidence to conclude that the location of population 1 is located to the left of population 2 (at 5% significance).

INTERPRET


Critical Values: Wilcoxon Rank Sum Test…For sample sizes smaller than 10 observations (in each sample), refer to the Critical Values in Table 8 (Appendix B)

For sample sizes larger than 10, the test statistic is approximately normally distributed with:

Mean: Hence:

Standard Deviation:

ni=size of sample i, i=1,2


Example 21.2…

A drug company is trialing a new painkiller. 30 people were selected at random, half were given the new drug, half given aspirin, and all were told to rate the effectiveness on a five point scale (hence ordinal data):

5 = The drug was extremely effective.

4 = The drug was quite effective.

3 = The drug was somewhat effective.

2 = The drug was slightly effective.

1 = The drug was not at all effective.


Example 21.2…

The data were recorded. Can we conclude (at 5% significance) that the new painkiller is perceived to be more effective?

Its important to note here that “5” is a “good” score, so if the drug is effective, we’d likely see its location “greater than” the location of aspirin users, hence:

H1: The location of population 1 is to the right of the location of population 2, and so:


IDENTIFY


Example 21.2…

The data looks like:

IDENTIFY

These three ones would occupy ranks 1, 2, & 3 — we average them (2) and each is assigned

that rank…These five twos would occupy

ranks 4,5,6,7, & 8 — again, average them to

(4+5+6+7+8)/5 = 6

and so on and so forth…


Example 21.2…

(though not shown here) The rank sum for the new painkiller is T1=276.5, and the rank sum for aspirin: T2=188.5

Set T= T1=276.5, and begin calculating…

COMPUTE


Example 21.2…

The p-value of the test is:

p-value = P(Z > 1.83) = .5 - .4664 = .0336

(or Z=1.83 > ZCritical=1.645), hence:

“There is sufficient evidence to infer that the new painkiller is perceived to be more effective than aspirin”

COMPUTE


Example 21.2…

We can use the Wilcoxon Rank Sum Test in the Data Analysis Plus set of tools to come to the same conclusion…

COMPUTE

compare…p-value


Required Conditions…The Wilcoxon rank sum test actually tests to determine whether the population distributions are identical. This means that it tests not only for identical locations, but for identical spreads (variances) and shapes (distributions) as well.

The rejection of the null hypothesis may be due instead to a difference in distribution shapes and/or spreads.

To avoid this problem, we will require that the two probability distributions be identical except with respect to location.


Identifying Factors…

Factors that identify the Wilcoxon Rank Sum…


Tests for Matched Pairs Experiments…We will now look at two nonparametric techniques (Sign Test and Wilcoxon Signed Rank Sum Test) that test hypotheses in problems with the following characteristics:

— We want to compare two populations,

— The data are either ordinal or interval (nonnormal),

— and the samples are matched pairs.

As before, we’ll compute matched pair differences and work from there…


The Sign Test…

We can use the Sign Test when we’re dealing with two populations of ordinal data in a matched pairs experiment.

For each matched pair, take the differences and count up the number of positive differences and negative differences.

If population locations are the same (say), we’d expect the number of positives and negatives to net out to zero. If we have more positives than negatives (or vice versa) what can we learn? Again, how many is enough to make a difference?


Sign Test…

We can think of the sign test in terms of a binomial experiment, getting a positive sign is like flipping heads on a coin. We use this notion along with previously developed statistics to come up with our standardized test statistic (assuming the null hypothesis is true):

Our null hypothesis:

H0: the two population locations are the same

is equivalent to:

H0: p = .5 (i.e. equal proportions of +’s & –’s)

n≥10


Sign Test Hypotheses…

Since our null hypothesis is:

H0: the two population locations are the same(i.e. p=.5)

Our research hypothesis must be:

H1: the two population locations are different

which is the same as:

H1: p ≠ .5


Example 21.3…

25 people were asked to ride in a European car (and rate the ride) then ride in a North American car (and again, rate the ride). The ratings were ordinal, from 1 – very uncomfortable to 5 – very comfortable, and it’s a matched pairs experiment since the same rider tried both cars. [Xm21-03.xls]

Can we conclude (at 5% significance) that the European car is perceived to be more comfortable than the North American car?


Example 21.3…

The data was analyzed…

We had 25 pairs of data initially, two pairs gave

identical ratings (i.e. delta = zero) so these data points are

dropped, hence n=23

We had 18 positive responses, thus x=18

We had 5 negative responses.

COMPUTE


Example 21.3…

The p-value is P(Z > 2.71) = .0034, hence we reject H0 in favor of H1, and conclude:


Or, in the context of this problem…

“There is relatively strong evidence to indicate that people perceive the European car to provide a more comfortable

ride than the North American car.”

INTERPRET


Example 21.3…

Again, we can leverage Excel to reduce the amount of work that we have to do to perform the Sign Test*

COMPUTE

*Data Analysis Plus

compare…p-value


Checking the Required Conditions…

The sign test requires:

The populations be similar in shape and spread:

The sample size exceeds 10 (n=23).


Wilcoxon Signed Rank Sum Test…

We’ll use Wilcoxon Signed Rank Sum test when we want to compare two populations of interval (but not normally distributed) date in a matched pairs type experiment.

Compute paired differences, discard zeros.

Rank absolute values of differences smallest (1) to largest (n), averaging ranks of tied observations.

Sum the ranks of positive differences (T+) and of negative differences (T–).

Use T=T+ as our test statistic…


Wilcoxon Signed Rank Sum Test…

Now we have a test statistic, but what to compare it against?

For small sample sizes, i.e. n ≤ 30, critical values of T can be read from Table 9 in Appendix B.

For large sample sizes, i.e. n > 30, T is approximately normally distributed, so we have:


Example 21.4…

Do travel times to the office vary between an 8:00 am start and a “flextime” start? 32 workers recorded their travel times

We want to research this hypothesis:


Thus we require:

H0: the two population locations are the same.

IDENTIFY


Example 21.4…

The data are interval (i.e. times) and were produced by a matched pairs experiment (same drivers, same day of the week – Wednesday). Why aren’t we using a t-test for ?

A histogram of the paired differences reveals a non-normal distribution, hence we must use a non-parametric technique.

IDENTIFY


Example 21.4… COMPUTE

The Original Data

ranks of +ve differences…ranks of -ve

differences…

Rank Sums

Sorted ascending by |difference|


Example 21.4…

We compute our test statistic as follows…

Our rejection region is…

COMPUTE


Example 21.4…

The Wilcoxon Signed Rank Sum Test tool in Data Analysis Plus yields the same result: there is not enough evidence to infer that flextime commute times differ from 8:00 am start commute times.

INTERPRET

compare…

p-value


Identifying Factors I…

Factors that Identify the Sign Test…


Identifying Factors II…

Factors that Identify the Wilcoxon Signed Rank Sum Test…


Kruskal-Wallis Test…

So far we’ve been comparing locations of two populations, now we’ll look at comparing two or more populations.

The Kruskal-Wallis test is applied to problems where we want to compare two or more populations or ordinal or interval (but nonnormal) data from independent samples.

Our hypotheses will be:

H0: The locations of all k populations are the same.

H1: At least two population locations differ.


Test Statistic…

In order to calculate the Kruskal-Wallis test statistic, we need to:

Rank all the observations from smallest (1) to largest (n), and average the ranks in the case of ties.

We calculate rank sums for each sample: T1, T2, …, Tk

Lastly, we calculate the test statistic (denoted H):


Sampling Distribution of the Test Statistic:For sample sizes greater than or equal to 5, the test statistic H is approximately Chi-squared distributed with k–1 degrees of freedom.

Our rejection region is:

And our p-value is:


Example 21.5…

Can we compare customer ratings (4=good … 1=poor) for “speed of service” across three shifts in a fast food restaurant? Our hypotheses will be:

H0: The locations of all 3 populations are the same.

(that is, there is no difference in service between shifts), and


Customer ratings for service were recorded…

IDENTIFY


Example 21.5…

One way to solve the problem is to take the original data,

“stack” it, and then

sort by customer response

& rank bottom to top…

COMPUTEsorte

d b

y re

sp

on

se


Example 21.5…

Once its in “stacked” format, put in straight rankings from 1 to 30, average the rankings for the same response, then parse them out by shift to come up with rank sum totals…

COMPUTE


Example 21.5…

Our critical value of Chi-squared (5% significance and k–1=2 degrees of freedom) is 5.99147, hence there is not enough evidence to reject H0.

COMPUTE


Example 21.5…

From Data Analysis Plus, a similar finding…

“There is not enough evidence to infer that a difference in speed of service exists between the three shifts, i.e. all three

of the shifts are equally rated, and any action to improve service should be applied to all three shifts”

COMPUTE

compare…

p-value



Factors that Identify the Kruskal-Wallis Test…


Friedman Test…

The Friedman Test is a technique used compare two or more populations of ordinal or interval (nonnormal) data that are generated from a matched pairs experiment.

The hypotheses are the same as before:

H0: The locations of all k populations are the same.



Friedman Test – Test Statistic…

Since this is a matched pairs experiment, we first rank each observation within each of b blocks from smallest to largest (i.e. from 1 to k), averaging any ties. We then compute the rank sums: T1, T2, …, Tk. The we calculate our test statistic:

This test statistic is approximate Chi-squared with k–1 degrees of freedom (provided either k or b ≥ 5). Our rejection region and p-value are:


Example 21.6…

Four managers evaluate and score job applicants on a scale from 1 (good) to 5 (not so good). There have been complaints that the process isn’t fair. Is it the case that all managers score the candidates equally or not? That is:

H0: The locations of all 4 populations are the same.

(i.e. all managers score like candidates alike)


(i.e. there is some disagreement between managers on scores)

IDENTIFY


Example 21.6…

The data looks like this:

Applicant #1 for example, received a top score from manager and next-to-top scores from the other three.

Applicant #7 received a top score from manager as well, but the other three scored this candidate very low…

COMPUTE

There are k=4 populations (managers) and b=8 blocks (applicants) in this set-up.


Example 21.6…

“rank each observation within block from smallest to largest (i.e. from 1 to k), averaging any ties”… For example, consider the case of candidate #2:

COMPUTE

Manager

Manager

Manager

Manager

Original Scores 4 2 3 2 checksum

“straight” ranking 4 1 3 2 10

averaged ranking 4

(1+2)/2=

1.5 3(1+2)/2=

1.5 10

checksum = 1 + 2 + 3 + … + k


Example 21.6…

Compute the rank sums: T1, T2, …, Tk and our test statistic…

COMPUTE


Example 21.6…

The value of our Friedman test statistic is 10.61 compared to a critical value of Chi-squared (at 5% significance and 3 d.f.) which is: 7.81473

Thus, there is sufficient evidence to reject H0 in favor of H1

INTERPRET

It appears that the managers’ evaluations of applicants do indeed differ



Factors that Identify the Friedman Test…


Spearman Rank Correlation Coefficient…Previously we looked at the t-test of the coefficient of correlation ( ). In many situations, one or both variables may be ordinal; or if both variables are interval, the normality requirement may not be satisfied.

In such cases, we measure and test to determine whether a relationship exists by employing a nonparametric technique, the Spearman rank correlation coefficient.


Spearman Rank Correlation Coefficient…We are interested whether a relationship exists between the two variables, hence the hypotheses to be tested are:

H0: = 0 (no linear pattern, hence no correlation)

H1: ≠ 0 (correlation; we can also do one-tail tests)

Since is a population parameter, our sample statistic is rs,

and is calculated as:

(where a and b are the ranks of x and y respectively)

[ is referred to as the Spearman correlation coefficient]


Spearman Rank Correlation Coefficient…For values of n between 5 and 30, critical values of rs are available in Table 10 of Appendix B.

When n is greater than 30, rs is approximately normally distributed with

— a mean of zero, and

— a standard deviation of

Hence our standardized test statistic is:


Example 21.7…

Is there a relationship between aptitude test scores before being hired and performance ratings 3 months into the job?

Aptitude test scores range: [0…100] (i.e. interval data)

Performance ratings scale: 1 – below average

:

5 – above average (ordinal)

The problem is we’re trying to correlate interval & ordinal data. We’ll treat the aptitude scores as ordinal, and apply the Spearman rank correlation coefficient…


Example 21.7…

We specify our hypotheses as:

H0: = 0

H1: ≠ 0

At a 5% significance level and n=20 observations, the rejection region (from Table 10) is:

rs < –.450 -or- rs > .450

IDENTIFY


Example 21.7…

As before, we rank each of the variables separately and average any ties…

Now we use the ranks columns, compute their standard deviations (sa, sb) and covariance (sab)…

COMPUTE


Example 21.7…

Thus…

Compare this to our critical value of rs=.450, and…

COMPUTE


Example 21.7…

“There is not enough evidence to believe that the aptitude test scores and performance ratings are related.”

INTERPRET

compare…

p-value



Factors that Identify the Spearman Rank Correlation Coefficient Test…

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Chapter 21