lecture 8 anova

8/4/2019 Lecture 8 ANOVA

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Analysis of Variance (ANOVA)


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ANOVA

Analysis of variance (ANOVA) is a method of testing

the equality of three or more population means by

analyzing sample variances.

One way ANOVA

Two way ANOVA

Multi-factorial ANOVA


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Type Dependent Variable

(Numerical)

Independent variables

(Categorical)

One way Exam marks One factor ( 2 categories) Race

Malay

Chinese

Indian

Two way Exam marks Two factors Gender

Race

Multi-

factorial

Exam marks > Two factors

Gender Race

Teaching method

..

..


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Assumption

The population under study have normal

distributions

The samples are drawn randomly Each sample is independent of the other

sample

Equal population variances


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Experimental design:

For the effective use of ANOVA, experiment has tobe standardized in terms of the randomness andreplications of all variables, including samples,

surrounding and trial protocol.

The experimental design has to be standardized toreduce the error of all variables mentioned above.


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Completely Randomized Designed

(CRD):(e.g.: Six treatments (A, B,..., F) x

Four replications)

A

D

F

D

B

B

E

A

F

C

E

F

C

D

F

C

E

B

A

C

E

A

B

D


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Randomized Complete Block Designed(RCBD):

(e.g.: Six treatments(A, B,..., F) x four

block replications)

Block I

Block II

Block III

Block IV

C

A

D

E

F

B

B

C

D

A

F

E

E

C

D

A

F

B

F

A

B

C

D

E


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** Each block contains all treatments arranged randomly.

STEPS IN ANOVA

1. Arrange data. Draft ANOVA table.

ANOVA Standard Table:

DF = degree of freedom

SS = sum of squares

MS = mean square; MS treatment=SStreatment/DFtreatment..

Fobserved= MStreatment/MSerror

Source of

Variation

*

DF SS MS Fobserved

Ftabulated

5% 1%

Treatment

Error/residual

Total


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2. Define and HA

Ho: All means are the same value

HA: There are means which are not the same

3. Determine Source of variation

4. Determine D of F for each source

5. Determine CF

6. Determine SS7. Determine MS (=SS/D of F)

8. Calculate Fcalculated9. Obtain Ftabulated from Table-F

10. Compare Ftabulated with FcalculatedIf Fcalculated > Ftabulated reject HoIf Fcalculated < Ftabulated accept Ho


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One way ANOVA

To study the effect of one (independent) factor.

Eg: To access whether the mean BMI of patients are

significantly different among races.

Malay

Chinese

Indian

Unpaired t test

I

Unpaired t test II

Unpaired t test III

Increase

type I

error!!!!!

Mean

BMI

ANOVA


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Example 1 (CRD 1 factor)

One trial was conducted to determine the effectiveness of six types of pesticide, RS1-RS6 in fourreplications, 1-4, with an experimental design CRD. Does the pesticides increase the grainyield by preventing the pest deseases?

Hypothesis:

Ho: All means are the same value

i.e. 1=2=3=4=5=6=controlHA: There are means which are not the same

Treatment Grain yield (kg ha-1)

1 2 3 4

RS1 2537 2069 2104 1797

RS2 3366 2591 2211 2544

RS3 2536 2459 2827 2385

RS4 2387 2453 1556 2116

RS5 1997 1679 1649 1859

RS6 1796 1704 1904 1320

Control 1401 1516 1270 1077


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ANOVA Standard Table:

* for CRD, 1 factor, SV = T, R and TotalDFtreatment = 7-1 = 6Dftotal = (total no. data -1) = 28-1=27

Therefore, DFerror = 27- 6 = 21

........next, calculate total and total means for each treatment, grand mean andgrand total. All these are required to calculate SS.....

Source of

Variation*

DF SS MS Fcalculated

Ftabulated

5% 1%

Treatment 6

Error 21

Total 27


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Treatment Grain yield (kg ha-1) Total Mean

1 2 3 4

RS1 2537 2069 2104 1797 8507 2127

RS2 3366 2591 2211 2544 10712 2678

RS3 2536 2459 2827 2385 10207 2552

RS4 2387 2453 1556 2116 8512 2128

RS5 1997 1679 1649 1859 7184 1796

RS6 1796 1704 1904 1320 6724 1681

Control 1401 1516 1270 1077 5264 1316

Grand total 57110

Grand Mean 2040

Next, calculate correction factor, CF, which is also a required valueto calculate SS


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CF = (Grand Total)2/no. observation

= (57110)2/28

= 116,484,004

SStotal=[(2537)2 + (2069)2 + (2104)2 +...+ (1077)2] - CF

= 7,577,421

SStreatment =1/4 (85072+107122+.....+52642) - CF

= 5,587,175

SSerror = SStotal - SStreatment

= 1,990,246


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ANOVA Standard Table

Ftotal = MStreatment/MSerrorFtable = F (dftreatment, dferror) at a significance level of

= F (x , y) at a significance level of fromFischers Table

Ho is rejected. There are differences among the means........

Source of

VariationDF SS MS Fcalculated Ftabulated

5% 1%

Treatment 6 5,587,175 931,196 9.82 2.57 3.81

Error 21 1,990,246 94,773

Total 27 7,577,421


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Example 2 (CRD 1 factor)In one experiment using CRD, three types of fertilizer (B1,

B2, and B3) were tested on its effectiveness on peanuts

yield, in five replications (R1, R2,.....R5). Based on theresults below, determine whether the fertilizers givedifferent yield.

Ho: B1=B2=B3H1: There are mean which are not the same

Fertilizer

type

Yield (g m-2)

R1 R2 R3 R4 R5

B1 86 79 81 70 84

B2 90 76 88 82 89

B3 82 68 73 71 81

Fertilizer type Yield (g m-2) Total Mean


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CF = (1200)2/15= 96,000

SStotal = [(86)2 + (79)2 +........+(71)2 +( 812) ]- CF= 698

SStreatment = (400)2 + (425)2 + (375)2)CF = 250

5SSerror = SStotal - SStreatment = 698-250=448

Jadual Piawai ANOVA:

Accept Ho. All means are the same, the fertilizers do not give any effect onthe yield of peanut.

Fertilizer type Yield (g m-2) Total Mean

R1 R2 R3 R4 R5

B1 86 79 81 70 84 400 80

B2 90 76 88 82 89 425 85

B3 82 68 73 71 81 375 75

Total 1200

Source of variation DF SS MS Fcalculated F(2,12)

5% 1%

treatment

error

total


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In certain one factorial experiment, anexperimental design, RCBD, is used whenwe use block.

Block is used in certain cases as follows:

1. The experimental site is non-homogenous (non

uniform). For example, a field experiment involvedplanting, there might be uneven amount of soil wateravailable for the plant, or the site is having granularedsoil or shaded (from light) etc..

2. The experiment was conducted bydifferent people fromday to day with differentaccuracy in measurement. In this case, each person canbe considered as one block.


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Example 3 (RCBD 1 factor)

One experiment was conducted on paddy type IR8 to study the the effect of sixgermination densities (25, 50,.....,150 kg seed per ha) in four blocks (IIV).Determine whether germination densities influences the paddy yield based onthe following results.

Yield of paddy variety IR8 at six germination densities

Ho: 1=2=3=4=5=6HA: 1 2 3 4 5 6Ho: b1=b2=b3=b4HA: b1 b2 b3 b4

Treatment

(kg seed/ha)

Paddy yield (kg/ha)

Block I Block II Block III Block IV

2550

75

100

125

150

5,1135,346

5,272

5,164

4,804

5,254

5,3985,952

5,713

4,831

4,848

4,542

5,3074,719

5,483

4,986

4,432

4,919

4,6784,264

4,749

4,410

4,748

4,098


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CF = (119030)2/24 = 590,339,204

SStotal = [(5113)2+(5398)2+.........+(4098)2]CF

= 4,801,068

SSblock = (30953)2 + (31284)2 + (29846)2 +(26947)2 - CF

6

= 1,944,361SStreatment = (20496)

2 + (20281)2 +.......+(18813)2 - CF

4

= 1,198,331

SSerror = SStotalSSblockSStreatment= 4,801,0681,944,3611,198,331= 1,658,376

Treatment (kg

seed/ha)

Paddy yield

(kg/ha)

Treatment

Block I Block II Block III Block IV total Mean

25

50

75

100

125

150

5,113

5,346

5,272

5,164

4,804

5,254

5,398

5,952

5,713

4,831

4,848

4,542

5,307

4,719

5,483

4,986

4,432

4,919

4,678

4,264

4,749

4,410

4,748

4,098

20,496

20,281

21,217

19,391

18,832

18,813

5,124

5,070

5,304

4,848

4,708

4,703

T Block 30,953 31,284 29,846 26,947

G Total 119,030

G Mean 4,960

M Block 5158.8 5214.0 4974.3 4491.2


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Analysis of variance on grain yield data

Source

Variation

DF SS MS Fcalculated Ftable5% 1%

Block

Treatment

Error

3

5

15

1,944,361

1,198,331

1,658,376

648,120

239,666

110,558

5.86**

2.17n.s.3.29 5.42

2.90 4.56

Total 23 4,801,068

Block Ftable= (3,15)

Treatment Ftable =(5,15)

Block= we accept Ha at both sig. Level 5% & 1%

Treatment= we accept Ho, no sig dif.......


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Exercise

Four researchers (P IP IV) were asked to measure the photosynthesis rate inan extensive experiment which involved five light treatments (T1- T5).

Results was reported as follows:

Using ANOVA and an appropriate method (at 1%), can the

differences among the treatments (if different) be accepted

without being sceptical on the researchersaccuracies?

Light

treatment

Photosynthesis rate (mol m-2 s-1)

P I P II P III P IV

T1 3.8 2.9 1.1 3.6

T2 6.7 6.8 3.2 7.0

T3 9.9 10.1 6.1 10.2

T4 12.5 11.8 7.4 13.0

T5 13.1 14.0 8.2 14.1


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P I P II P III P IV

T1 3.8 2.9 1.1 3.6

T2 6.7 6.8 3.2 7

T3 9.9 10.1 6.1 10.2

T4 12.5 11.8 7.4 13

T5 13.1 14 8.2 14.1

46 45.6 26 47.9

CF

SS total


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SSblock

SStreatment


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SV DF SS MS F F t (1%)

Block

treatment

error

Total

Conclusion:

Among light treatments: there are differences between mean (accept HA)

Among the researchers (Block): there are differences between mean (accept HA)


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LSD, Least Significant Difference

(Perbezaan Bererti Terkecil)

The conclusion in ANOVA is general:

e.g. If Ho is accepted, no further test is required as allmeans are the same.

e.g. If HA is accepted, the difference in means (which

pairs?) cannot be determined through F test in

ANOVA.


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Further step that is LSD test is carried out to determine which pairs aresignificantly different. If the treatments were conducted with the same no.

replications (=r), the formula to calculate LSD at the significance level of is as follows:

If no. replication are not the same (= r1 and r2 for two paired means), theformula above will be changed to:

LSD = t (2MSe/r)

LSD = t {MSe/(r1 + r2)}

t = ttabulated for 2 tail at significance level of anddegree of freedom of error (dferror), from ANOVAMSe = error mean square, from ANOVA


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Example 1 (CRD 1 factor)

One trial was conducted to determine the effectiveness of six types of pesticide, RS1-RS6 in fourreplications, 1-4, with an experimental design CRD. Does the pesticides increase the grainyield by preventing the pest deseases?

Hypothesis:Ho: All means are the same value

i.e. 1=2=3=4=5=6=controlHA: There are means which are not the same

Treatment Grain yield (kg ha-1)

1 2 3 4

RS1 2537 2069 2104 1797

RS2 3366 2591 2211 2544

RS3 2536 2459 2827 2385

RS4 2387 2453 1556 2116

RS5 1997 1679 1649 1859

RS6 1796 1704 1904 1320

Control 1401 1516 1270 1077


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Interpretation of Result

If the difference between means > LSDThe different is significant at a significance level of If the difference between means < LSDThe different is not significant at a significance level of Example 1: A trial on pesticides (CRD)

LSD0.05 = t 0.025(2 x 94773/4) = 453 (t 21, 0.025 = 2.08)

LSD0.01 = t 0.005 (2 x 94773/4) = 616 (t 21, 0.005 = 2.831)

Source of

variation

DF SS MS Fobserved Ftabulated

5% 1%

Treatment 6 5,587,175 931,196 9.82 2.57 3.81

Error 21 1,990,246 94,773

Total 27 7,577,421


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Conclusion:

Treatment Mean Different with control Significant

RS2 2678 1362 **

RS3 2552 1236 **

RS4 2128 812 **

RS1 2127 811 **

RS5 1796 480 *

RS6 1681 365 ns

Control 1316 -

** significant at 1%

* significant at 5%ns, the difference is not significant


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Example 2: An experiment on the effect of the light

treatment on photosynthesis rate

Determine the difference in mean pairs between the treatments at

significance level of 1%.

SV DF SS MS F Ftabulated 1%

Block 3 63 21 36.2 5.95

Treatment 4 242 60.5 104.3 5.41

Error 12 7 0.58

Total 19 312


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LSD0.01 = t 0.005 (2 x 0.58/4) = 1.65 (t 12, 0.005 = 3.055)

Concl:

....

LSD for block = 1.65......

on the table above, only the mean pair of T4 and T5 does not show asignificant difference at significant level of 1%. Other mean pairs showsignificant differences.

mean SignificantT1 vs. T2 3.08 **

T1 vs. T3 6.23 **

T2 vs. T3 3.15 **

.......

........

........

T4 vs. T5 1.17 n.s.


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USE LSD WHEN NECESSARY!

1. Use LSD when F test in ANOVA shows significantdifference between the mean.

2. DONT USE LSD to differentiate between means inwhich its treatments are more than FIVE.

3. If the experiment/trial is conducted with CONTROLtreatment, then the difference between means can be

done by comparing the CONTROL with other

treatments with no limit. However, if there is no

CONTROL treatment, find the difference for EACHpair as much as possible.


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Duncans Multiple Range Test (DMRT)(Ujian Berbilang Duncan)

Steps in DMRT:

1. Arrange all means in descending order.2. Calculate Standard Deviation for Mean Treatment, Sx

3. Calculate The Closest Significant Range (Julat BerertiTerdekat), Rp.

4. Arrange Rp in sequence next to its mean. (Rp must follow the meansequence i.e. descending order)

5. Calculate D = (MeanRp) . Draw a conclusion (see the followingexamples)

Sx = (MSe/r)

Rp = rp Sx

(rp = from New Multiple Range Test Table)


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Example 1: An experiment on pesticides

treatment grain yield (kg ha-1)

1 2 3 4

RS1 2537 2069 2104 1797

RS2 3366 2591 2211 2544

RS3 2536 2459 2827 2385

RS4 2387 2453 1556 2116

RS5 1997 1679 1649 1859

RS6 1796 1704 1904 1320

Control 1401 1516 1270 1077

Source of

variation

DF SS MS Fcalculated

Ftabulated

5% 1%

treatment 6 5,587,175 931,196 9.82 2.57 3.81

error 21 1,990,246 94,773total 27 7,577,421

DMRT method:

Sx = (MSe/r) = (94773/4) = 153.93Rp = rp Sxrp from table New Multiple Range Test for p= 2-7 (df =21) at significance

level of 5%as follows:


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p rp Rp = rp Sx2 2.94 453

3 3.09 4764 3.18 490

5 3.25 500

6 3.30 508

7 3.33 513

treatment Mean Rp D

Mean - Rp

RS2 2678 a 513 2165

RS3 2552 ab 508 2044

RS4 2128 bc 500 1628

RS1 2127 bc 490 1637

RS5 1796 c 476 1320

RS6 1681 cd 453 1228

Control 1316 d

For RS2 treatment, assume all means >2165 as same with this mean treatment. Thus,


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For RS2 treatment, assume all means >2165 as same with this mean treatment. Thus,mean treatments RS2 and RS3 are the same. These means are given same symbol(i.e. a).

For RS3 treatment, assume all means >2044 as same with this mean treatment. Thus,mean treatments RS3, RS4 and RS1 are the same. These means are given samesymbol (i.e. b).

For RS4 treatment, assume all means >1628 as same with this mean treatment. Thus,mean treatments RS4, RS1, RS5 and RS6 are the same. These means are givensame symbol (i.e. c).

For RS1 treatment, assume all means >1637 as same with this mean treatment. Thus,mean treatments RS1, RS5 and RS6 are the same.

Next, for RS5 treatment, assume all means >1320 as same with this mean treatment.Thus, mean treatments RS5 and RS6 are the same.

Lastly, for RS6 treatment, assume all means >1228 as same with this mean treatment.Thus, mean treatments RS6 and control are the same. These means are given samesymbol (i.e. d).

Treatment Mean

RS2 2678 a

RS3 2552 ab

RS4 2128 bc

RS1 2127 bc

RS5 1796 c

RS6 1681 cd

Kawalan 1316 d

Conclusion:


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An CRD experimental designed was conducted to examine the

effectiveness of plant hormonal treatments (H1, H2 dan H3) on

pineapple yield in four replications (R1-R4). The following results were

obtained:

Using ANOVA, test whether the hormonal treatment gives different yield

at a significance level of 1%:

Hormonal

treatment

Pineapple yield (seed ha-1 week-1)

R1 R2 R3 R4

H1 87 79 83 92

H2 80 73 75 71

H3 94 87 90 92

SV DF SS MS Fcalculated Ftable (0.01)

treatment 2 529 265 14.6 7.56

error 9 164 18.2total 11 693


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Conclusion: Fcalculated > Ftabulated. There are differences among the means, butwe dont know which treatments.

To solve this problem, used LSD or DMRT.

LSD methodLSD = t (2MSe/r)LSD 0.01 = t 0.005 x (2 x 18.2)/4 = 3.25 x 9.1 = 9.8Conclusion:

..........

meantreatment

Mean Significant

H1 & H2 10.5 **

H1 & H3 5.5 n.s.

H2 & H3 16 **


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DMRT method

Sx = (MSe/r) = (18.2)/4 = 2.13

rp at a significance level of 1%, df =9

Conclusion...........................

p rp* Rp = rp

Sx

2 4.60 9.8

3 4.86 10.4

Treatment Mean Rp D

Mean - Rp

H3 90.8 a 10.4 80.4

H2 85.3 a 9.8 75.5

H1 74.7 b


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Error Protection P = number of means for range being tested


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df level

2 3 4 5 6 7 8 9 10 12 14 16 18 20

1 0.05 18.0 18.0 18.0 18.0 18.0 18.0 18.0 18.0 18.0 18.0 18.0 18.0 18.0 18.0

0.01 90.0 90.0 90.0 90.0 90.0 90.0 90.0 90.0 90.0 90.0 90.0 90.0 90.0 90.0

2 0.05 6.09 6.09 6.09 6.09 6.09 6.09 6.09 6.09 6.09 6.09 6.09 6.09 6.09 6.09

0.01 14.0 14.0 14.0 14.0 14.0 14.0 14.0 14.0 14.0 14.0 14.0 14.0 14.0 14.0

3 0.05 4.50 4.50 4.50 4.50 4.50 4.50 4.50 4.50 4.50 4.50 4.50 4.50 4.50 4.50

0.01 8.26 8.5 8.6 8.7 8.8 8.9 8.9 9.0 9.0 9.0 9.1 9.2 9.3 9.3

4 0.05 3.93 4.01 4.02 4.02 4.02 4.02 4.02 4.02 4.02 4.02 4.02 4.02 4.02 4.02

0.01 6.51 6.8 6.9 7.0 7.1 7.1 7.2 7.2 7.3 7.3 7.4 7.4 7.5 7.5

5 0.05 3.64 3.74 3.79 3.83 3.83 3.83 3.83 3.83 3.83 3.83 3.83 3.83 3.83 3.83

0.01 5.70 5.96 6.11 6.18 6.26 6.33 6.40 6.44 6.5 6.6 6.6 6.7 6.7 6.8

6 0.05 3.46 3.58 3.64 3.68 3.68 3.68 3.68 3.68 3.68 3.68 3.68 3.68 3.68 3.68

0.01 5.24 5.51 5.65 5.73 5.81 5.88 5.96 6.0 6.0 6.1 6.2 6.2 6.3 6.3

7 0.05 3.35 3.47 3.54 3.58 3.60 3.61 3.61 3.61 3.61 3.61 3.61 3.61 3.61 3.61

0.01 4.95 5.22 5.37 5.45 5.53 5.61 5.69 5.73 5.8 5.8 5.9 5.9 6.0 6.0

8 0.05 3.26 3.39 3.47 3.52 3.55 3.56 3.56 3.56 3.56 3.56 3.56 3.56 3.56 3.56

0.01 4.74 5.00 5.14 5.23 5.23 5.40 5.47 5.51 5.5 5.6 5.7 5.7 5.8 5.8

9 0.05 3.20 3.34 3.41 3.47 3.50 3.52 3.52 3.52 3.52 3.52 3.52 3.52 3.52 3.52

0.01 4.60 4.86 4.99 5.08 5.17 5.25 5.32 5.36 5.4 5.5 5.5 5.6 5.7 5.7

10 0.05 3.15 3.30 3.37 3.43 3.46 3.47 3.47 3.47 3.47 3.47 3.47 3.47 3.47 3.48

0.01 4.48 4.73 4.88 4.96 5.06 5.13 5.20 5.24 5.28 5.36 5.42 5.48 5.54 5.55

11 0.05 3.11 3.27 3.35 3.39 3.43 3.44 3.45 3.46 3.46 3.46 3.46 3.46 3.47 3.48

0.01 4.39 4.63 4.77 4.86 4.94 5.01 5.06 5.12 5.15 5.24 5.28 5.34 5.38 5.39

12 0.05 3.08 3.23 3.33 3.36 3.40 3.42 3.44 3.44 3.46 3.46 3.46 3.46 3.47 3.48

0.01 4.32 4.55 4.68 4.76 4.81 4.92 4.96 5.02 5.07 5.13 5.17 5.22 5.24 5.26

13 0.05 3.06 3.21 3.30 3.35 3.38 3.41 3.42 3.44 3.45 3.45 3.46 3.46 3.47 3.47

0.01 4.26 4.48 4.62 4.69 4.74 4.84 4.88 4.94 4.98 5.04 5.08 5.13 5.14 5.15

14 0.05 3.03 3.18 3.27 3.33 3.37 3.39 3.41 3.42 3.44 3.45 3.46 3.46 3.47 3.47

0.01 4.21 4.42 4.55 4.63 4.70 4.78 4.83 4.87 4.91 4.96 5.00 5.04 5.06 5.07


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15 0.05 3.01 3.16 3.25 3.31 3.36 3.38 3.40 3.42 3.43 3.44 3.45 3.46 3.47 3.47

0.01 4.17 4.37 4.50 4.58 4.64 4.72 4.77 4.81 4.84 4.90 4.94 4.97 4.99 5.00

16 0.05 3.00 315 3.23 3.30 3.34 3.37 3.39 3.41 3.43 3.44 3.45 3.46 3.47 3.47

0.01 4.13 4.34 4.45 4.54 4.60 4.67 4.72 4.76 4.79 4.84 4.88 4.91 4.93 4.94

17 0.05 2.98 3.13 3.22 3.28 3.33 3.36 3.38 3.40 3.42 3.44 3.45 3.46 3.47 3.47

0.01 4.10 4.30 4.41 4.50 4.56 4.63 4.68 4.72 4.75 4.80 4.83 4.86 4.88 4.89

18 0.05 2.97 3.12 3.21 3.27 3.32 3.35 3.37 3.39 3.41 3.43 3.45 3.46 3.47 3.47

0.01 4.07 4.27 4.38 4.46 4.53 4.59 4.64 468 4.71 4.76 4.79 4.82 4.84 4.85

19 0.05 2.96 3.11 3.19 3.26 3.31 3.35 3.37 3.39 3.41 3.43 3.44 3.46 3.47 3.47

0.01 4.05 4.24 4.35 4.43 4.50 4.56 4.61 4.64 4.67 4.72 4.76 4.79 4.81 4.82

20 0.05 2.95 3.10 3.18 3.25 3.30 3.34 3.36 3.38 3.40 3.43 3.44 3.46 3.46 3.47

0.01 4.02 4.22 4.33 4.40 4.47 4.53 4.58 4.61 4.65 4.69 4.73 4.76 4.78 4.79

22 0.05 2.93 3.08 3.17 3.24 3.29 3.32 3.35 3.37 3.39 342 3.44 3.45 3.46 3.47

0.01 3.99 4.17 4.28 4.36 4.42 4.48 4.53 4.57 4.60 4.65 4.68 4.71 4.74 4.75

24 0.05 2.92 3.07 3.15 3.22 3.28 3.31 3.34 3.37 3.38 3.41 3.44 3.45 3.46 3.47

0.01 3.96 4.14 4.24 4.33 4.39 4.44 4.49 4.53 4.57 4.62 4.64 4.67 4.70 4.72

30 0.05 2.89 3.04 3.12 3.20 3.25 3.29 3.32 3.35 3.37 3.40 3.43 3.44 3.46 3.47

0.01 3.89 4.06 4.16 4.22 4.32 4.36 4.41 4.45 4.48 4.54 4.58 4.61 4.63 4.65

60 0.05 2.83 2.98 3.08 3.14 3.20 3.24 3.28 3.31 3.33 3.37 3.40 3.43 3.45 3.47

0.01 3.76 3.92 4.03 4.12 4.17 4.23 4.27 4.31 434 4.39 4.44 4.47 4.50 4.53


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lecture 8 anova

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