King M.R., Mody N.A. Numerical and Statistical Methods for Bioengineering: Applications in MATLAB

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Example 4.8

Employees who work night-shifts or varying shifts may suffer from sleepiness at night on the job because

the body’s circadian r hythm routinely falls out of sync with the changing daily sleeping and waking

patterns. For some people, chronic misalignment of the circadian cycle with the sleep cycle can adversely

affect productivity (efficiency) at work and personal health. A fraction of shift workers are unable to cope

well with disrupted sleep cycles and are diagnosed with chronically excessive sleepiness. Patients with

shift-work sleep disorders are more prone to accidents, depression, and ulcers. A clinical study (Czeisler

et al., 2005) was performed to examine if administering the drug modafinil benefited patients with this

disorder. Modafinil is a stimulant drug used to treat daytime sleepiness, a key symptom of narcolepsy and

obstructive sleep apnea. The researchers wished to evaluate the efficacy of the drug for treating night-time

sleepiness suffered by shift-work sleep disorder patients.

The experimental study consisted of a double-blinded trial conducted over a 3 month period. The

treatment group consisted of 96 patients, who received 200 mg of modafinil tablets to be ingested before

commencing night-shift work. The placebo group had 108 patients who received identical-looking tablets.

The severity of sleepiness at baseline (prior to administering drug or placebo) in individuals in both groups

was determined by measuring sleepiness level on the Karolinska Sleepiness scale. Table 4.14 presents

sleepiness severity data for both groups.

A χ

test is used to ensure that the null hypothesis is true at baseline, i.e. that the distribution of severity

of sleepiness is the same in both treatment and control groups. The expected frequencies under the null

hypothesis are tabulated in Table 4.15.

The test statistic is calculated as

42

i¼1

 e

ðÞ

49  48ðÞ

53  54ðÞ

þþ

4  3:2ðÞ

3:2

¼ 0:529:

Table 4.14. A4× 2 contingency table for sleepiness severity at baseline for modaﬁnil

and placebo groups

Modaﬁnil Placebo Total

Moderately sleepy 49 53 102

Markedly sleepy 29 34 63

Severely sleepy 16 17 33

Among the most extremely sleepy 2 4 6

Total 96 108 204

Table 4.15 Expected frequencies for sleepiness severity at baseline for modaﬁnil and

placebo groups

Modaﬁnil Placebo Total

Moderately sleepy 48.0 54.0 102

Markedly sleepy 29.7 33.3 63

Severely sleepy 15.5 17.5 33

Among the most extremely sleepy 2.8 3.2 6

Total 96 108 204

287

4.9 Chi-square tests for nominal scale data

The associated degrees of freedom is given by f ¼ 4  1ðÞ2  1ðÞ¼3. The p value is calculated

using the chi2cdf function and is found to be 0.91. The test result is insignificant and close to unity. This

tells us that if H

is true, the chances of observing differences of this nature randomly within the two

samples are very high. We conclude that at baseline the two groups have an equal distribution of the

nominal variable “severity of sleepiness.”

The conditions for use of the χ

test of homogeneity can be contrasted with that for

the χ

test of independence. For the test of independence, we require only one

random sample from a population that is subjected to classiﬁcation ba sed on two

criteria. The only known or ﬁxed quantity within the contingency table is the grand

total. The column and row totals are depen dent on the distribution of the two

variables within the population and are thus a property of the system being inves-

tigated. For the test of homogeneity, k independent random samples are chosen

from k populations that differ from each other in at least one aspect. Thus, the k

column totals will be equal to the k sample sizes, and are ﬁxed by the experimenter.

Table 4.16 contrasts the two χ

tests.

When the sample size is too small to meet the validity criteria of the χ

tests, you can

still calculate the probabilities corresponding to the test statistic using the theory of

combinatorics. Fisher’s Exact test rigorously calculates all possible outcomes of the

samples and the total number of ways in which one obtains outcomes as extreme as or

more extreme than the observed outcome, when H

is true. The exact p value can be

calculated using this method. The method can be found in most statistics textbooks

(see, for instance, Statistics for the Life Sciences by Samuels and Witmer (2003)).

4.10 More on non-parametric (distribution-free) tests

Non-parametric (distribution-free) tests are routinely used for drawing statistical

inferences when the population distribution of one or more random variables is

either unknown or deviates greatly from normality. Also, non-parametric

Table 4.16 Distinct characteristics of the χ

test of independence and the χ

test of

homogeneity

Property χ

test of independence χ

test of homogeneity

Sample data single random sample from a

population

k independent random samples

from k populations

Classiﬁcation the sample is categorized on the

basis of two nominal variables

the k samples are categorized on

the basis of one nominal variable

Null

hypothesis

the two nominal variables are

independent with respect to

each other

the k populations are

homogeneous with respect to a

nominal variable

Expected

frequencies

calculation is based on

probability of independent

events

calculation is based on combining

sample data to obtain a global

estimate of population proportions

of the different categories of the

nominal variable

288

Hypothesis testing

procedures must be used to test hypotheses that are based on data of ordinal nature,

i.e. the variable of interest is measured in terms of rankings. Use of non-parametric

tests is especially common when the sample size is small and the central-limit

theorem does not apply.

Some non-parametric tests make no assumption of the form or nature of distri-

bution of the random variable within the population and are called distribution-free

tests. In these tests, the observations within one or more samples are ranked, and a

test statistic is calculated from the sum total of all ranks allotted to each sample. In

this way, cardinal data are converted to ordinal type resulting in loss of infor mation.

Since distribution-free tests do not use all the information available in the data (e.g.

shape of population distribution of the random variable), they are less powerful than

the equivalent parametric test s. When data are suitable for analysis using a para-

metric procedure, those methods are thus preferred over a non-parametric analysis.

Non-parametric tests such as the χ

goodness-of-ﬁt test and the χ

test of inde-

pendence do not test hypotheses concerning population parameters. Such tests use

inferential statistics to address questions of a different nature such as “Do tw o

nominal variables demonstrate an association within a population?”. Examples of

such tests were discussed in the previous section. In this section, we introduce three

distribution-free statistical tests: the sign test and the Wilcoxon signed-rank test for

analyzing paired data, and the Wilcoxon rank-sum test for determining if two

independent random samples belong to the same population.

4.10.1 Sign test

The sign test is a non-parametric procedure used to analyze either a single sample or

paired data. The appeal of the sign test lies in its simplicity and ease of use. A

hypothesis tested by a sign test concerns either (1) the median of a single population,

or (2) the median of the differences between paired observations taken from two

different populations. No assumptions are made regarding the shape of the po-

pulation distribution. For this reason we cannot draw inferences regarding the

population mean, unless the distribution is symmetric about the median, in which

case the median and mean will be equal regardless of the shape of the distribution. A

population median is deﬁned as that value that is numerically greater than exactly

50% of all observed values in the population. If we were to choose only one

observation from the population, the probability that its value is less than the

population median is 0.5. Equivalently, the probability that the observed value is

greater than the population median is 0.5.

The null hypothesis of a sign test asserts that the median of a population m is

equal to some value, i.e.

: m = m

If the population consists of differences d

= x

– y

calculated from paired data,

then under H

the hypothesized median of the differences m

is zero, i.e.

: m

=0.

The sign test does not use the magnitude of the differences to generate the test

statistic. Therefore, the sign test is appropriate for use in studies where the difference

between two observations is not accurately known, the data points are ordinal, or

289

4.10 More on non-parametric tests

the experiment/data collection is incomplete. The only information needed is the

direction of the differences, which is recorded either as a “+” sign or a “–” sign. If

the null hypothesis is true, we would expect 50% of the differences to be positive and

50% of them to be negative.

When performing the sign test, the deviation d of each observation in the sample

from the hypothesized media n is tabulated as either (+) or (–). When paired data are

being analyzed, the difference d between paired values is either greater than zero or

less than zero and is marked by a (+) or (–) sign accordingly. If a data pair yields a

difference of exactly zero, or if the deviation of an observation from the hypothe-

sized median is equal to zero, the data point is ignored in the analysis and n is reduced

by 1. The number of (+) signs and (–) signs are calculated. The test statistic for the

sign test is the number of (+) signs, S

, or the number of (–) signs, S

−

. The sidednes s

of the hypothesis dictates whether S

or S

−

is used as the test statistic.

A non-directional null hypothesi s for the sign test is expressed in mathematical

terms as

: P(d >0)=P(d < 0) = 0.5.

The complementary alternate hypothesis is

: P(d >0)≠ P(d < 0).

When the nature of the hypothesis is non-directional, the test statistic is chosen as

either S

or S

−

, whichever is least. If there are n non-zero d’s, and if H

is true, the

expected value for both S

and S

−

is n/2. If H

is true, the probabil ity that S

and S

−

will be close in value is high. On the other hand, if a majority of the observations are

greater (or less) than the hypothesized median, we are less inclined to believe the null

hypothesis since the outcome is less easily explained by sampling variation than by

the alternate hypothesis. The objective criterion for rejection of H

is the same as that

for all other statistical tests: the p value of the test statistic should be less than the

chosen signiﬁcance level α.

The p value for the test is calculated using the binomial theorem. The null

hypothesis states that an observed difference d has an equal probability of being

positive or negative. For a sample size n, we have n independent outcomes, each of

which can assume two values, either (+) or ( –) with a probability of 0.5, when H

true. Thus, we have a binomial process with parameters n and p, where p = 0.5. Let S

be either S

or S

−

, whichever is less. Therefore, S is a binomial random variable and

is either the number of (+) signs or the number of (−) signs. The probability that one

will observe S successes (or failures) out of n trials is given by

0:5

nS

The probability that one will observe S successes (or failures) or fewer out of n trials

is given by

P ¼

x¼0

0:5

nx

: (4:34)

Equation (4.34) is the probability of observing S or fewer plus or minus signs,

whichever sign is in the minority. The p value for a non-directional hypo thesis is

twice the probability calculated using Equation (4.34).

290

Hypothesis testing

For a directional alternate hypothesis such as

: P(d > 0) > 0.5,

the corresponding null hypothesis is

: P(d >0)≤ P(d <0)orP(d >0)≤ 0.5.

If H

is true, the observed differences should contain many more positive differences

than negative differences. If H

is true, then we expect the number of (−) signs to be

either roughly equal to or larger than the number of (+) signs. The appropriate test

statistic S is equal to the number of (−) signs, S

−

. The null distribution of S

−

is a

binomial distribution with μ = n/2. If S

−

> n/2, the null hypothesis is automa tically

retained since the corresponding p value is > 0.5. If S

−

< n/2, then the p value

associated with S

−

is calculated using Equation (4.34) and compared with the

signiﬁcance level α. The rejection region is located only in the left tail of the

distribution.

Similarly, for the directional alternate hypothesis

: P(d < 0) > 0.5,

the test statistic is S

. The null distribution of S

is a binomial distribution with μ =

n/2. The p value associated with S

is calculated. The rejection region is located only

in the left tail of the distribution. If p is less than the signiﬁcance level α, the null

hypothesis H

: P(d <0)≤ 0.5 is rejected in favor of H

Example 4.9

Marfan’s syndrome is an inherited autosomal dominant disease that often ends in premature death of the

patient. A genetic defect in the formation of fibrillin-1 causes abnormalities in the structure of the

extracellular matrix. As a result, these patients tend to have abnormally long limbs, eyesight problems, and

numerous heart conditions. Fibrillin-1 is known to suppress transforming growth factor β (TGF-β) signaling

activity by binding these molecules, thereby limiting their activity in the body. Defective fibrillin-1 is

associated with enhanced signaling of TGF-β, and the latter is found to cause progressive dilation of the

aortic root in Marfan’s syndrome patients.

In Box 4.2A, a study was presented that tested angiotensin II receptor blockers (ARB) for controlling or

preventing upregulation of TGF-β (Bedair et al., 2008). A small clinical trial was performed in recent years

(Brooke et al., 2008) to evaluate the effect of ARB on aortic root enlargement.

In a fictitious study, the annual rate of aortic root enlargement is measured in 12 Marfan syndrome

patients prior to administration of losartan, an ARB drug. At one year following commencement of ARB

therapy, the annual rate of aortic root enlargement is measured. It is found that the annual rate of aortic root

enlargement decreased for 11 patients and increased for only one patient. Does losartan have a beneficial

effect on aortic root enlargement?

The alternate hypothesis is

: “Treatment with ARB is effective, i.e. ARB acts to reduce the rate of aortic root expansion.”

Mathematically, this is written as P(d < 0) > 0.5.

The null hypothesis is

: “Treatment with ARB is not effective,” i.e. P(d <0)≤ 0.5.

Note that we have posed a directional hypothesis in this case. The experiment yields 12 data pairs. Each

pair contains two values of the annual rate of aortic root enlargement, one measured at baseline and one at

291

4.10 More on non-parametric tests

follow-up. The difference d for each pair is already calculated and the direction of the difference is supplied

in the problem statement. We have S

−

= 11 and S

= 1. Since the direction of the alternate hypothesis is

negative, the test statistic is S

. A small number of (+) signs will lead to rejection of the null hypothesis.

The data appear to support the alternate hypothesis, but we can confirm this only after we calculate the

p value.

According to Equation (4.34),

P ¼ C

0:5

n0

þ C

0:5

n1

¼ 0:0032:

If the null hypothesis were non-directional, an equally likely extreme outcome under H

would be S

−

and S

= 11, and the probability calculated from Equation (4.34) must be doubled to obtain the p value.

Here, because the hypothesis is one-sided, the probability calculated above is the p value itself and no

doubling is required. Note that for the directional hypothesis considered here, the reverse outcome of

= 11 and S

−

= 1 is not an extreme outcome when H

is true, but is, in fact, a highly likely observation.

If α = 0.05, then p = 0.0032 < α, and we conclude that the result supports the alternate hypothesis, i.e.

ARB produces the desired effect.

If n is large, calculating the p value using Equation (4.34) can become cumbersome.

Instead, the normal approximation of the binomial distribution of S for large n

(see Section 4.7) can be used to calculate the p value. The test statistic S is normalized

to obtain the z statistic and is calculated as follows:

z ¼

S  n=2

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

n  0:5  0:5

; (4:35)

where z is the number of standard deviations of the binomial variable S from its

mean, when H

is true and n is large. The p value corresponding to z is obtained using

the cumulative distribution function for the normal probability curve.

Using MATLAB

You can perform a sign test in MATLAB using Statistics Toolbox function

signtest. There are several variations in the syntax as shown below.

The following statement carries out a sign test for the non-directional null

hypothesis, H

: m = m

p = signtest(x, m0)

where x is the vector of values and m

is the hypothesized median.

The following statement performs a sign test for paired data to test the null

hypothesis, H

: m

=0:

p = signtest(x, y)

where x and y are vectors that represent two samples with matched data pairs, and

whose sizes must be equal. You can also specify the method to perform the sign test,

i.e. the exact method using combinatorics or the approximate method using the

normal approximation. See MATLAB help for more details.

4.10.2 Wilcoxon signed-rank test

The Wilcoxon signed-rank test is a non-param etric test also used for analyzing paired

data, and is essentially an extension of the sign test. This test procedure is the method

292

Hypothesis testing

of choice when experimental data are on an ordinal scale, or when the data values are

cardinal but the population distribution is considerably non-normal. The Wilcoxon

signed-rank test is a slightly more involved procedure than the sign test, and it is

more powerful because it not only uses information about the sign of the difference

between the pa ired data values, but also considers the magnitude of the absolute

difference. The sign test is preferred over the Wilcoxon signed-rank test only when a

study produces incomplete data and ranking of observations is not possible, or when

requirements of the signed-rank test such as symmetric distribution of the random

variable about the median are not met.

In this method, the difference between each of the paired values is calculated.

Observations that yield differences exactly equal to zero are removed from the

analysis, and the number of pairs n is reduced accordingly. The non-zero differences

are ranked in ascending order of their absolute magnitude. The ranks are preﬁxed

with the positive or negative signs of the differences to which the ranks correspond.

All positive ranks are summed to obtain a positive rank sum R

. Similarly, the

negative ranks are summ ed to obtain a negative rank sum R

−

. The appropriate test

statistic R

or R

−

is chosen based on the directionality of the hypothesis.

Box 4.11 Neurostimulation as a treatment option for Parkinson’s disease

Parkinson’s disease is a chronic and progressive neurodegenerative disorder of impaired motor

functioning. This disease is characterized by involuntary tremors, muscle rigidity or stiffness, slowness

of voluntary movements, and poor posture and balance. Currently, there is no cure for this disease;

however, treatments to alleviate symptoms are available. Prescription medications such as levodopa

are accompanied by side-effects such as dyskinesia (or involuntary movements) and hallucinations.

Recently, deep-brain stimulation was proposed as an alternate treatment method for patients suffering

from drug-related complications. This is a surgical procedure in which a pacemaker-type device is

implanted within the body to stimulate a particular region of the brain continuously with electricity.

An unblinded clinical study (Deuschl et al., 2006) was conducted in Germany and Austria to

determine if deep-brain stimulation was more effective in attenuating the motor problems of Parkinson’s

disease patients compared to medication alone. Eligible patients for the study were paired up, and one

member of a pair was randomly assigned to electrostimulation treatment and the other was provided

with conventional medical treatment. The efficacy of the treatment was measured in terms of the change

in quality of life over 6 months from baseline to follow-up as assessed using the Parkinson’s Disease

Questionnaire summary index. Where follow-up values were missing, the worst possible outcome score

was used for the patients allotted to neurostimulation and the best possible outcome score was used for

patients allotted to medical treatment to yield a more conservative test.

For 78 pairs of patients, the index favored neurostimulation in 50 pairs and medical treatment in 28

pairs. There were no ties in the scores within any pair. Based on the study data, we wish to determine

whether neurostimulation is effective as a treatment method for Parkinson’s disease.

Since n = 78 is large, we use the normal approximation given by Equation (4.35) to calculate the

test statistic. The test is two-sided, so S is chosen to be the smaller of either of the two totals (S

= 50,

−

= 28). We obtain

z ¼

28  78=2

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

78  0:5  0:5

¼2:49:

The p value associated with the test statistic value is 0.0128 (verify this yourself). We conclude that

deep-brain stimulation is more effective than standard medical treatment in treating the symptoms of

Parkinson’s disease.

293

4.10 More on non-parametric tests

Example 4.10

A crossover trial is perfor med to determine the effectiveness of two statin drugs in lowering LDL

cholesterol levels in patients with high cardiovascular risk. Eight patients receive both drugs one at a time in

randomized order with a wash-out period between the two treatments. The reduction in LDL cholesterol

level from baseline is measured in each patient following treatment with each drug. The results are

presented in Table 4.17. We wish to determine if one drug is more effective than the other.

The reduction of LDL cholesterol levels quantifies the effect of each drug on the body. The difference

between the effects of Statin A and Statin B is calculated for each individual. The differences are ranked

based on their absolute values. If some of the differences are tied, the rank assigned to each of the tied

differences is equal to the average of the ranks that would have been assigned had the difference values

been slightly unequal. For example, for patients 5 and 7, the absolute difference in reduced LDL cholesterol

values is 4. Since the next two ranks that can be allotted are 3 and 4, the average rank is calculated as 3.5.

The signs are added to their respective ranks, and the positive and negative rank totals, R

and R

−

respectively, are calculated. For this problem, the hypothesis is non-directional and we choose the lesser

value of either R

or R

−

as the test statistic.

The Wilcoxon signed-rank test requires that the following conditions are met.

(1) The distribution of d

’s about zero (or the median) is symmetrical. When this is true,

the population mean difference and populationmediandifferenceareoneandthesame.

Note that we make no assumption about the shape of the distribution of the differences d

A symmetrical distribution about the median is needed to use the Wilcoxon signed-rank

method to test a hypothesis concerning the population mean difference, μ

. This method is

distribution-free as we do not use the actual data values but convert the data to an ordinal

scalebeforewecalculatethemagnitudeoftheteststatistic.

(2) The data should be on a continuous scale. In other words, no ties should occur

between the d

’s. If ties between diff erence values d

occur, a correction term must be

included in the formula; otherwise the calculated p value is approximate.

The null hypothesis tested by this non-parametric procedure is

: μ

¼ 0.

When H

is true, the population mean difference μ

and population median differ-

ence m

is 0, and we expect the positive rank total R

to be close in value to the

Table 4.17. Calculation of the test statistic for the Wilcoxon signed-rank method

Patient Statin A (mg/dl) Statin B (mg/dl) Difference (mg/dl) Rank Signed rank

135 23 128+8

2 41 35 6 5.5 + 5.5

329 22 77+7

438 41 −32− 2

532 36 −4 3.5 − 3.5

6 40 34 6 5.5 + 5.5

7 37 33 4 3.5 + 3.5

833 31 21+1

= 30.5, R− = 5.5.

294

Hypothesis testing

negative rank total R

−

. The distribution of the positive rank total R

or negative

rank total R

−

under the null hypothesis is calculated using combinatorics. If R



−

, H

is retained. If either R

or R

−

is very small, then it is unlikely that the mean

(or median) of the differences is zero.

The non-directional alternate hypothesis is

: μ

6¼ 0.

Here, we illustrate how the p value is calcul ated for this test, assuming that no ties

among ranks occur. Suppose the number of pairs n = 8, then we have a set of eight

ranks,

12345678:

If H

is true, each rank can be assigned a (+) sign or a (–) sign with 0.5 probability,

since the population median difference is zero and the distribution of differences

about the median is symmetrical. The probability of obtaining the combination

þþþþþþ

12345678

is 0:5

. For this combination, R

= 29 and R− = 7. Other combinations that yield

the same values for R

and R

−

are

þþþþþþþ þþþþþþ

12345678 12345678

þþþþþþ þþþþþ

12345678 12345678

Thus, the probability of obtaining the result R

= 29 and R

−

= 7 is 5  0:5

The p value is equal to the probability of obtaining the result R

=29 and

−

= 7, and even more extreme results. Try to determine the probability of obtain-

ing the following outcomes:

= 30 and R

−

=6,

= 31 and R

−

=5,

= 32 and R

−

=4,

= 33 and R

−

=3,

= 34 and R

−

=2,

= 35 and R

−

=1,

= 36 and R

−

=0.

When you sum all the probabilities you get a one-sided p value of

5 þ 4 þ 3 þ 2 þ 2 þ 1 þ 1 þ 1ðÞ0:5

¼ 0:0742:

The two-sided probability or two-sided p value is 0:0742  2 ¼ 0: 1484.

Since this process of calculating the p value is cumbersome and we may encounter

ties in our difference values (such as in Example 4.10), it is best to use the computer

to determine the p value.

Using MATLAB

A two-sided Wilcoxon signed-rank test can be performed using the signrank func-

tion of Statistics Toolbox. The syntax for using signrank to analyze paired data is

295

4.10 More on non-parametric tests

p = signrank(x,y)

where x and y are tw o vectors of equal length. It is assumed that the differences x – y

represent a random sample derived from a continuous distribution symmetric about

the median. This function corrects for tied ranks.

Example 4.10 (continued)

Let’s now calculate the p value associated with the calculated test statistic. We can do this in one of

two ways. We first round the test statistic values to R

= 30 and R

−

= 6. Using the method illustrated

above, the two-sided p value associated with this outcome is

2  4 þ 3 þ 2 þ 2 þ 1 þ 1 þ 1ðÞ0:5

¼ 0:1094.

Now we round the test statistic values to R

= 31 and R− = 5. The p value associated with this

outcome is 2  3 þ 2 þ 2 þ 1 þ 1 þ 1ðÞ0:5

¼ 0:0781.

Using the MATLAB function signrank to calculate the p value, we obtain p = 0.0938, a value

intermediate to the two values calculated above.

When n is large, the sampling distribution of the test statistic R

or R

−

assumes a

normal distribution and the z statistic can be calculated for this test. The Wilcoxon

signed-rank test can also be used to test a hypothesis regarding the mean (or median)

of a single sample. When applying the signed-rank method to test the null hypothesis

: μ ¼ μ

, μ

is subtracted from each data value and ranks are allotted to the

differences. The test then proceeds as usual.

The parametric equivalent of the Wilcoxon signed-rank test is the paired t test.

When data are normally distributed, or when the central-limit theorem applies, the

paired t test should be preferred over the signed-rank test.

4.10.3 Wilcoxon rank-sum test

The Wilcoxon rank-sum test is a non-parametric procedure used to test the hypoth-

esis of the equality of two populatio n medians. Its parametric equivalent is the

independent two sample t test. This procedure does not require that the variable of

interest have any particular distribution within the two populations. The conditions

that must be met in order to use this test procedure are

(1) the random variable is continuous , and

(2) the shape of the variable distribution is the same in both populations.

To perform the rank-sum test we start with two independent representative samples

drawn from two populations. If the median of a population is represented by m, the

(non-directional) null hypothesis is

: m

¼ m

and the corresponding alternate hypothesis is

: m

6¼ m

The alternate hypothesis can also be directional such as H

: m

or H

: m

in which case the respective null hypotheses are H

: m

 m

and H

: m

 m

. The

test procedure is illustrated in Box 4.1B.

How is the p value calculated for the Wilcoxon rank-sum test? Again the method

of calculation is based on the theory of combinatorics. If there are no ties in the

ranks, i.e. the distribution of the variable is continuous, then the calculated p value is

exact. If some of the ranks are tied, the calculated p value is only approximate. If

296

Hypothesis testing