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HYPOTHESIS TESTING:HYPOTHESIS TESTING:ABOUT ABOUT MORE THAN MORE THAN TWOTWO (K) (K) INDEPENDENT POPULATIONSINDEPENDENT POPULATIONS

2

ONE-WAY ANALYSIS OF VARIANCE (ANOVA)Analysis of variance is used for two different purposes:

1. To estimate and test hypotheses about population variances

2. To estimate and test hypotheses about population means

We are concerned here with the latter use.

3

H0: 1= 2= 3=...= k

Ha: Not all the i are equal.

Assumptions:

•We have K independent samples, one from each of K populations.

•Each population has a normal distribution with unknown mean i

•All of the populations have the same standard deviation (unknown)

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MeanT..T.kT.3T.2T.1Total

x3kx33x32x31

x2kx23x22x21

x1kx13x12x11

k321Treatment

1.x 2.x 3.x kx. ..x

11nx

22nx 33n

x knkx

jn

iijj xT

1. columnjth theof total

columnjth theofmean ..

j

jj nT

x

k

j

k

j

n

iijj

j

xTT1 1 1

... nsobservatio all theof total

NT

x ....

k

jjnN

1

5

k

j

n

i

k

j

n

iijij

j j

NT

xxxSST1 1 1 1

2..22

.. )(

The Total Sum of Squares

The Within Groups Sum of Squares

k

j

n

i

k

j

n

i

k

j j

jijjij

j j

nT

xxxSSW1 1 1 1 1

2.22

. )(

The Among Groups Sum of Squares

k

j

k

j j

jjj NT

nT

xxnSSA1 1

2..

.2... /)(

SST=SSA+SSW

6

)1/( kSSAMSA

)/( kNSSWMSW

MSWMSAVR /

Within groups mean square

Among groups mean square

Variance Ratio (F)

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Source SS df MS F(VR)

Among samples SSA k-1 MSA MSA/MSW

Within samples SSW N-k MSW

Total SST N-1

ANOVA TABLE

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Testing for Significant Differences Between Individual Pairs of Means

Whenever the analysis of variance leads to a rejection of the null hypothesis of no difference among population means, the question naturally arise regarding just which pairs of means are different. Over the years several procedures for making individual comparisons have been suggested.

The oldest procedure, and perhaps the one most widely used in the past, is the Least Significant Difference (LSD) procedure.

LSD (LSD (Least Significant Difference ))

TukeyTukey

BonferroniBonferroni

SidakSidak

Dunnett’s CDunnett’s C

Dunnett’s T3Dunnett’s T3

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When sample sizes are equal (n1=n2=n3=...=nk=n)Least Significant Difference (LSD)

nMSWtxx ji

)(2 p<0.05

When sample sizes are not equal (n1n2 n3 ... nk)

)11(ji

ji nnMSWtxx p<0.05

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Example In a study of the effect of glucose on insulin release, specimens of pancreatictissue from experimental animals were randomly assigned to be treated with one of five different stimulants. Later, a determination was made on the amount of insulin released. The experimenters wished to know if they could conclude that there is a difference among the five treatments with respect to the mean amount of insulin released. The resulting measurements of amount of insulin released following treatment are displayed in the table.The five sets of observed data constitute five independent samples from the respective populations.

Each of the populations from which he samples come is normally distributed with mean,i, and variances i

2.

Each population has the same variance.

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Stimulant1 2 3 4 5

1.53 3.15 3.89 8.18 5.861.61 3.96 3.68 5.64 5.463.75 3.59 5.70 7.36 5.692.89 1.89 5.62 5.33 6.493.26 1.45 5.79 8.82 7.81

1.56 5.33 5.26 9.037.10 7.49

8.98Total 13.04 15.60 30.01 47.69 56.81 163.15Mean 2.61 2.60 5.00 6.81 7.10 5.10

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H0: 1= 2= 3= 4 = 5

Ha: Not all the i are equal.

54282.162 32

923.2661173529.994

3215.16398.861.153.1

2222

1 1

2..2

k

j

n

iij

j

NT

xSST

35739.41881.56

769.47

601.30

660.15

504.1398.861.153.1

22222222

1 1 1

2.2

k

j

n

i

k

j j

jij

j

nT

xSSW

SSA=SST-SSW=162.54282-41.35739=121.18543

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ANOVA TABLE

121,185 4 30,296 19,779 ,00041,357 27 1,532

162,543 31

Between GroupsWithin GroupsTotal

Sum ofSquares df Mean Square F Sig.

MSW=SSW/27=41.357/27=1.532

MSA=SSA/(5-1)=121.185/4=30.296

F=MSA/MSW=30.296/1.532=19.779

We conclude that not all population means are equal.

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Since n1n2 n3 n4 n5), reject H0 if )11(21 nn

MSWtxx ji

538.1)61

51(532.105.2

Hypothesis LSD Statistical Decision

H0: 1= 2

0.01<1.538, accept H0.

H0: 1= 3

2.391.538, reject H0.

H0: 4= 5

0.29<1.314, accept H0.

538.1)61

51(532.105.2

314.1)81

71(532.105.2

60.261.2

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Multiple Comparisons LSD

8,000E-03 ,7494 ,992 -1,5297 1,5457-2,3937 * ,7494 ,004 -3,9314 -,8560-4,2049 * ,7247 ,000 -5,6918 -2,7179-4,4933 * ,7056 ,000 -5,9409 -3,0456

-8,0000E-03 ,7494 ,992 -1,5457 1,5297-2,4017 * ,7146 ,002 -3,8678 -,9355-4,2129 * ,6886 ,000 -5,6257 -2,8000-4,5013 * ,6684 ,000 -5,8727 -3,12982,3937 * ,7494 ,004 ,8560 3,93142,4017 * ,7146 ,002 ,9355 3,8678

-1,8112 * ,6886 ,014 -3,2240 -,3984-2,0996 * ,6684 ,004 -3,4710 -,72814,2049 * ,7247 ,000 2,7179 5,69184,2129 * ,6886 ,000 2,8000 5,62571,8112 * ,6886 ,014 ,3984 3,2240-,2884 ,6405 ,656 -1,6027 1,02594,4933 * ,7056 ,000 3,0456 5,94094,5013 * ,6684 ,000 3,1298 5,87272,0996 * ,6684 ,004 ,7281 3,4710,2884 ,6405 ,656 -1,0259 1,6027

(J) Stimulant2,003,004,005,001,003,004,005,001,002,004,005,001,002,003,005,001,002,003,004,00

(I) Stimulant1,00

2,00

3,00

4,00

5,00

MeanDifference

(I-J) Std. Error Sig. Lower Bound Upper Bound95% Confidence Interval

The mean difference is significant at the .05 level.*.

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KRUSKAL- WALLIS ONE-WAY ANOVA

When the assumptions underlying One-way ANOVA are not met, that is, when the populations from which the samples are drawn are not normally distributed with equal variances, or when the data for analysis consist only of ranks, a nonparametric alternative to the one-way analysis of variance may be used to test the hypothesis of equal location parameters.

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The application of the test involves the following steps:

1. The n1, n2, ..., nk observations from the k groups are combined into a single series of size n and arranged in order of magnitude from smallest to largest. The observations are then replaced by ranks.

2. The ranks assigned to observations in each of the k groups are added separately to give k rank sums.

3. The test statistic

is computed.

k

j j

j nnR

nnKW

1

2

)1(3)1(

12# of groups

# of obs. in jth group

Sum of ranks in jth group

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4. When there are three groups and five and fewer observations in each group, the significance of the computed KW is determined by using special tables. When there are more than five observations in one or more of the groups, KW is compared with the tabulated values of 2 with k-1 df.

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ji

ji nnknKWnnntRR 111

12)1(

Determing which groups are significantly different

Like the one-way ANOVA, the Kruskal-Wallis test is an overall test of significant result, the test does not indicate where the differences are among the groups. To determine which groups are significantly different from one another, it is necessary to undertake multiple comparisons.

p<0.05

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Example The effect of two drugs on reaction time to a certain stimulus were studied in three groups of experimental animals. Group III served as a control while the animals in group I treated with drug A and those in group II were treated with drug B prior to the application of the stimulus. Table shows the reaction times in seconds of 13 animals. Can we conclude that the three populations represented by the three samples differ with respect to reaction time?

H0: The population distributions are all identical.

Ha: At least one of the populations tends to exhibit larger values than at least one of the other populations.

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Group

I II III

17 8 2

20 7 5

40 9 4

31 8 3

35

Rank Rank Rank

9 6.5 1

10 5 4

13 8 3

11 6.5 2

12

Ri 55 26 10

68.10)113(3

410

426

555

)113(1312

)1(3)1(

12

222

1

2

k

j j

j nnR

nnKW

KW(5,4,4;0.05)=5.617<KWcal

p<0.05, reject H0.

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Groups Statistical Decision

1-2 4.5 2.115 p<0.05

1-3 8.5 2.115 p<0.05

2-3 4 2.229 p<0.05

ji RR

ji nnknKWnnnt 111

12)1(

Multiple Comparisons Table

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We can use the chi-square test to compare frequencies or proportions in two or more groups. The classification according to two criteria, of a set of entities, can be shown by a table in which the r rows represents the various levels of one criterion of classification and c columns represent the various levels of the second criterion. Such a table is generally called a contingency table.

We will be interested in testing the null hypothesis that in the population the two criteria of classification are independent or associated.

rxc Chi Square Test

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Second Criteria

First Criteria

1 2 c Total

1 O11 O12 O1c O1.

2 O21 O22 O2c O2.

r Or1 Or2 Orc Or.

Total O.1 O.2 O.c N

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r

1i

c

1j ij

2ijij2

E)E(O

χNOO

E .ji.ij

No more than 20% of the cells should have expected frequencies of less than 5.

df = (r-1)(c-1)

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Example A research team studying the relationship between blood type and severity of a certain condition in a population collected data on 1500 subjects as displayed in the below contingency table. The researchers wished to know if these data were compatible with the hypothesis that severity of condition and blood type are independent.

Severity of Condition

Blood Type

A B AB 0 Total

Absent 543 211 90 476 1320

Mild 44 22 8 31 105

Severe 28 9 7 31 75

Total 615 242 105 538 1500

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543 211 90 476 1320

41,1% 16,0% 6,8% 36,1% 100,0%44 22 8 31 105

41,9% 21,0% 7,6% 29,5% 100,0%28 9 7 31 75

75,037,3% 12,0% 9,3% 41,3% 100,0%615 242 105 538 1500

41,0% 16,1% 7,0% 35,9% 100,0%

Severity of conditionCount

% within severityCount

% within severityCount

% within severityCount

% within severity

Absent

Mild

Severe

Total

A B AB OBlood Type

Total

541,2 213,0 92,4 473,4 1320,0

43,1 16,9 7,4 37,7 105,0

30,8 12,1 5,3 26,9

615,0 242,0 105,0 538,0 1500,0

Expected Count

Expected Count

Expected Count

Expected Count

0 cells (,0%) have expected count less than 5. The minimum expected count is 5,25.

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12.5 9.26

)9.2631(96.212

)96.212211(2.541

)2.541543(

E)E(O

χ

222

r

1i

c

1j ij

2ijij2

2(6,0.05)=12.592> 2

(calculated), accept H0, p>0.05

We conclude that these data are compatible with the hypothesis that severity of the condition and blood type are independent.

H0: severity of condition and blood type are independent.

Ha: severity of condition and blood type are not independent.

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Chi-Square Tests

1,998a 6 ,920171

Pearson Chi-SquareN of Valid Cases

Value dfAsymp. Sig.

(2-sided)

5 cells (41,7%) have expected count less than 5. Theminimum expected count is ,84.

a.

Assumption is violated

50 20 9 45 12450,0 22,5 9,4 42,1 124,0

15 8 3 10 3614,5 6,5 2,7 12,2 36,0

4 3 1 3 114,4 2,0 ,8 3,7 11,069 31 13 58 171

69,0 31,0 13,0 58,0 171,0

CountExpected CountCountExpected CountCountExpected CountCountExpected Count

Total

A B AB O Total

Absent

Mild

Severe

Severity of condition Blood TypeWe decide to merge two conditions

When the sample size is small and assumption about expected frequencies is not met;

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After combining mild and severe groups in one group, no more than 20% of the cells have expected frequencies less than 5.

50 20 9 45 12450,0 22,5 9,4 42,1 124,0

19 11 4 13 4719,0 8,5 3,6 15,9 47,0

69 31 13 58 17169,0 31,0 13,0 58,0 171,0

CountExpected CountCountExpected CountCountExpected Count

Present

Total

TotalA B AB O

Absent

Severity of conditionBlood Type

Chi-Square Tests

1,814a 3 ,612Pearson Chi-SquareValue df

Asymp. Sig.(2-sided)

1 cells (12,5%) have expected count less than 5. Theminimum expected count is 3,57.

a.

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32 51 8322,8 60,2 83,0

58,2% 35,2% 41,5%4 8 12

3,3 8,7 12,07,3% 5,5% 6,0%

8 19 277,4 19,6 27,0

14,5% 13,1% 13,5%11 67 78

21,5 56,6 78,020,0% 46,2% 39,0%

55 145 20055,0 145,0 200,0

100,0% 100,0% 100,0%

Blood TypeCountExpected Count% within TromboembolismCountExpected Count% within TromboembolismCountExpected Count% within TromboembolismCountExpected Count% within TromboembolismCountExpected Count% within Tromboembolism

A

AB

B

O

Total

+ -Tromboembolism

Total

12,375a 3 ,006200

Chi-SquareN of Valid Cases

Value dfAsymp. Sig.

(2-sided)

1 cells (12,5%) have expected count less than 5. Theminimum expected count is 3,30.

a.

2=5,118261

2=0,204807

2=0,067016

2=7,038861

Reject H0. Which type of blood group(s) is/are different from the others ?

Exclude Type O from the analysis

If null hypothesis is rejected, how can we find the group which is different?

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Count

32 51 834 8 128 19 27

44 78 122

AABB

Total

+ -Tromboembosim

Total

,747a 2 ,688Pearson Chi-SquareValue df

Asymp. Sig.(2-sided)

1 cells (16,7%) have expected count less than 5. Theminimum expected count is 4,33.

a. p>0.05

Except for blood type O, distribution of tromboembolism is similar within the others.