Wilcoxon Test and Calculating Sample Sizes Test and Calculating Sample Sizes Dan Spencer UC Santa Cruz Dan Spencer (UC Santa Cruz) Wilcoxon Test and Calculating Sample Sizes 1 / 33

Wilcoxon Test and Calculating SampleSizes

Dan Spencer

UC Santa Cruz

Dan Spencer (UC Santa Cruz) Wilcoxon Test and Calculating Sample Sizes 1 / 33

Differences in the Means of TwoIndependent Groups

When using the t, t ′ or tp test statistics, we assumethat the responses in both groups are normallydistributedWhat if they are not normally distributed?

I If n1 and n2 are large enough, it is still okay to use thet-distribution

I However, if n1 and n2 are small, this is a problem

This non-normality sometimes occurs in animalstudies


Wilcoxon Rank-Sum Test

Sometimes called the Mann-Whitney-Wilcoxon test,the Mann-Whitney U test, or theWilcoxon-Mann-Whitney test

Test to see if the location of the responses betweenthe groups is different

Interpreted as a test for a difference in medians

An example of a nonparametric test, as it does nottest about parameters in an assumed distribution


Wilcoxon Rank-Sum: Assumptions

Responses are either continuous or ordinal

Observations from both groups are independent

The shape and spread of the response in the twodifferent populations is the same, but not necessarilynormal


t-Test Group Density Assumption

0.0

0.1

0.2

0.3

0.4

−5.0 −2.5 0.0 2.5 5.0Values

Den

sity Group

1

2

Density Assumption for t−Tests


Wilcoxon Group Density Assumption

0.00

0.05

0.10

0.15

0 5 10Values

Den

sity

Group1

Group2

Wilcoxon Density Assumption


Wilcoxon Rank-Sum: Hypotheses

Null Hypothesis (H0): The probability of arandomly-selected response from the first populationexceeding that of a randomly-selected response fromthe second population is equal to 0.5

I A slightly stronger hypothesis is that the distributions areequal in terms of location

I This hypothesis implies the above null hypothesis

Alternative Hypothesis (H1): The probability ofa randomly-selected response from the firstpopulation exceeding that of a randomly-selectedresponse from the second population is

I Not equal to 0.5I Greater than 0.5I Less than 0.5


Case Study: Chick Weights

Newly hatched chicks were separated into twogroups

I Sunflower seed dietI Horsebean seed diet

After six weeks, the weights of the chicks weremeasured in grams



●

265.0

267.5

270.0

horsebean sunflowerFeed Type

Wei

ght (

gram

s)

feed

horsebean

sunflower

Boxplots of Chick Weights by Feed Type



Both distributions look to be somewhat skewed tothe right because they either have a long tail or anoutlier (shown as a solitary point)

Sample sizes are small (8 and 10, respectively), so tand t ′ are not appropriate hereHypotheses:

I H0 : The distribution of chick weights in the two groupsis equal

I H1 : The distribution of chick weights is lower for thehorsebean group


Wilcoxon Rank-Sum Test Statistic

Combine groups, and rank all responses fromsmallest to largest

I The ranks number from 1 to nI n = n1 + n2

If there are ties, the ranks should be averagedI Values 7, 5, 6, 6I Their ranks would be 4, 1, 2.5, 2.5

The test statistic T is the sum of the ranks for thegroup with the smallest sample size

I If n1 = n2, T falls between the two rank sums


Rank Sums

HorsebeanWeights Ranks

266.84 14264.07 6263.82 4263.47 2264.33 8264.25 7263.22 1263.92 5Sum = 47

SunflowerWeights Ranks

267.75 15266.02 12266.29 13264.89 10269.24 17271.63 18264.74 9268.36 16264.99 11263.69 3Sum = 124



T = 47

Wilcoxon Rank-Sum rejection region values can befound in a table athttps://metxstats.soe.ucsc.edu/node/5

Since the research hypothesis is that the horsebeangroup has a lower-shifted distribution than thesunflower group, reject H0 if T is less than thevalues in the table when n1 = 8 and n2 = 10

I T is larger than the critical value for α = 0.025, 0.05,and 0.10

I Fail to reject H0 and conclude that distributions are notsignificantly shifted from one another


https://metxstats.soe.ucsc.edu/node/5

Normal Approximation

When both treatment groups are larger than 10, thenormal distribution approximates the distribution ofthe Wilcoxon Rank-Sum test statistic rather well

z =T − µTσT

µT =n1(n1 + n2 + 1)

2

σT =

√n1n2(n1 + n2 + 1)

12


Normal Approximation: Our Example

µT =n1(n1 + n2 + 1)

2

=8(8 + 10 + 1)

2= 76

σT =

√n1n2(n1 + n2 + 1)

12

=

√(8)(10)(8 + 10 + 1)

12= 11.25463


Normal Approximation: Our Example

z =47− 76

11.25463= −2.576717

This z-score certainly does fall in the rejection region

P-value ≈ 0.00499

This is a contradictory conclusion!

Use this approximation only when samples are largeenough!


Wilcoxon Rank-Sum Test in JMP

Analyze → Fit Y by X

Drag your variables to the appropriate Response andFactor boxes and click OK

Click the → Nonparametric → Wilcoxon Test


Wilcoxon Rank-Sum Test in JMP

JMP calls the test statistic S instead of TOnly the two-sided p-value for the normalaproximation is given

I For the one-sided p-value, divide by 2


Sample Size

Researchers aim to present evidence to support theirhypotheses about how the world worksMost of the time, this hypothesis aims to show thattreatments are significantly different from oneanother

I Usually, the aim is to reject H0

Ideally, sample sizes would be as big as possibleI However, time and money often limit sample sizes


Power

We want to minimize the chance of failing to rejecta false H0

I This chance is often represented by β

An experiment’s power is the chance that a falseH0 is correctly rejected

I 1− βWhen the chance of incorrectly rejecting H0 is fixedat some value α, the power of a test can beestimated for different sample sizes


Power: t Distributions

When H0 is true, the test statistic is centeredaround 0

When H1 is true, the test statistic is proportionallycentered at

∆∗ =µ1 − µ2 − D0

σ√

1n1

+ 1n2

I For simplicity, the quantity µ1 − µ2 − D0 is representedas ∆


Calculating Power

An experiment where n1 = n2 = 5, σ = 10, and∆ = 25α is fixed at 0.05 for the hypotheses

I H0 : µ1 − µ2 = 0I H1 : µ1 − µ2 6= 0


Power Illustrated

t*0.0

0.1

0.2

0.3

0.4

−5 0 5t

Den

sity Hypothesis

H0

H1

β, α, and t


Changing σ

t*0.0

0.1

0.2

0.3

0.4

−5 0 5t

Den

sity Hypothesis

H0

H1

σ = 10

t*0.0

0.1

0.2

0.3

0.4

−5 0 5t

Den

sity Hypothesis

H0

H1

σ = 8


Changing n

t*0.0

0.1

0.2

0.3

0.4

−5 0 5t

Den

sity Hypothesis

H0

H1

n1 = n2 = 5

t*0.0

0.1

0.2

0.3

0.4

−5 0 5t

Den

sity Hypothesis

H0

H1

n1 = n2 = 10


Maximizing Power

Increase n1 and n2 and decrease experimental erroras much as possible

We have previously discussed reducing experimentalerror by standardizing measurement practices

How do we choose the smallest possible sample sizewhile achieving a fixed α and β?


Calculating n

Fix or estimateI α - Chance of incorrectly rejecting H0

I β - Chance of incorrectly failing to reject H0

I σ - Estimated population standard deviationI ∆ - The size of difference that is desirable to detect

One-sided tests for µ1 − µ2:

n1 = n2 = 2σ2(zα + zβ)2

∆2

Two-sided tests for µ1 − µ2:

n1 = n2 = 2σ2(zα/2 + zβ)2

∆2


Calculating n

If |µ1 − µ2 − D0| ≥ ∆, type II error probability ≤ β

Typically, β is chosen to be ≤ 0.2

σ is estimated as s calculated from previousexperiments∆ is set as the minimum difference that is desirableto detect

I A treatment is only preferable if it increases CD4 cellcount by 100 or more, so ∆ ≥ 100


Calculating n: Tooth Growth

In a previous lesson, we examined the effects of thesource of vitamin C on tooth growth in guinea pigsLet’s say we want to conduct another study, butthis time, we want to be able to detect a truedifference of 3 millimeters in tooth length

I We’ll estimate that σ = 7.5, which was our estimate spI Fix α = 0.05I Fix β = 0.20

We’ll assume a two-sided test


Calculating n: Tooth Growth

n1 = n2 = 2(7.52)(z0.05/2 + z0.20)2

32

= 2(7.52)(1.959964 + 0.8416212)2

32

= 98.111

In order to have power = 1 - .2 = .8, the minimumsample size for each group is 99 guinea pigs

I In the case where a non-integer sample size is found,round up to the nearest whole number


Calculating Sample Size in JMP

DOE → Sample Size and PowerTwo Sample Means

I Enter αI σ (Std Dev)I Difference to detect (∆)I Power (1− β)I Continue

Note, small differences may exist due to roundingerrors


JMP Output


Notes on JMP

Note that this tool can also be used to evaluate thepower of a proposed study

A plot of power versus sample size can also beuseful in determining sample size


Documents

Wilcoxon Test and Calculating Sample Sizes Test and Calculating Sample Sizes Dan Spencer UC Santa Cruz Dan Spencer (UC Santa Cruz) Wilcoxon Test and Calculating Sample Sizes 1 / 33