The Latin Square Design -...

Islamic University, Gaza - Palestine

The Latin Square Design

• Text reference, Section 4-2,

• These designs are used to simultaneously control (or eliminate) two sources of nuisance variability

• A significant assumption is that the three factors (treatments, nuisance factors) do not interact

• If this assumption is violated, the Latin square design will not produce valid results

• Latin squares are not used as much as the RCBD in industrial experimentation


The Rocket Propellant Problem – Latin Square Design

• This is a

• Page 140 shows some other Latin squares• Table 4-13 (page 140) contains properties of Latin squares• Statistical analysis?

5 5 Latin square design


Statistical Analysis of the Latin Square Design

• The statistical (effects) model is 1,2,...,

1, 2,...,

1, 2,...,ijk i j k ijk

i p

y j p

k p

• The statistical analysis (ANOVA) is much like the analysis for the RCBD.

• See the ANOVA table, page 137 (Table 4-9)• The analysis for the rocket propellant example is presented on

text pages 138 & 139


Balance

• An experiment is balanced if all treatments have the same number of experimental units.

• An experiment is dye balanced if all treatments have an equal number of experimental units dyed with each dye.

4

• An incomplete block design is a balanced incomplete block design if all treatments have the same number of experimental units and each pair of treatments appears together in the same block an equal number of times.


• If the experimental units are not uniform, blocking is necessary to separate experimental error, therefore, block is considered as a random factor

Randomized Complete Block Design (RCBD)

Block 1

Block 2

Block 3

Canal


• Two factors (Trts & Block) - Two-way ANOVA • Most widely used • Block can be:

– Spatial: pond, row, districts/village, households, etc.– Time: week, month or year– Age: different age group of animals

Randomized Complete Block Design (RCBD)

– Age: different age group of animals– Height: tree trunk, shelf in the drier etc.– Soil heterogeneity, slope, insect migration etc.

Two null hypotheses (H0)• no effects of block• no effects of treatment


Freq.A

Comparisons of five means

Freq.

Block 2Block 1 Block 3

A AA


1. Randomized Complete Block Design (RCBD)all the experimental units within the block are considered

uniform or identicaltreatment allocation in each block is completely randomBlocks are considered as replications

Experimental designs:

Block 1Block 2Block 3

Canal


Two hypothesis are tested by comparing the variation, therefore, called as Two-way Analysis of Variance (ANOVA)

• between treatments with the variation among treatments and blocks

• Effect of block is separated

• If variation between treatments (“T”reatment effect) is higher than • If variation between treatments (“T”reatment effect) is higher than the variation within treatment (i.e. “R”andom error), there is a significant difference

Model: Yi = + Ti + Bi + Ri


Separation of variation

Yi = Ri

+ B

If Ti > Ri treatment effect is significant after separationOf block effects

175

200

225

No

. o

f e

gg

s/t

an

k

Random errors

+ Bi

+ Ti

+

0

25

50

75

100

125

150

Control Cooling

Treatments

No

. o

f e

gg

s/t

an

k

Treatment effect

Block effect


Also called as “Two factor experiment/analysis”

Examples: 1. Fertilization trials in different types of soil

- Organic, in-organic and combination- 0, 40 and 60 kg N/ha/week etc.

2. Crop/vegetable/fruits varieties in different districts2. Crop/vegetable/fruits varieties in different districts3. Animal breeds in different villages/families4. Drug efficacy on different age groups5. Tilapia seed production by week or month6. Weekly DO and Temp differences7. Fish growth trial in cages in ponds


Randomization and layout

1. Determine the total number of experimental units (cages) = t x b e.g. to test 6 feeds in 4ponds, you will need 24 cages (6/pond)

2. Assign cage number in each pond (1 to 6)2. Assign cage number in each pond (1 to 6)

3. Assign all the treatments in each block (pond) randomly by using lottery or random table


Randomization: Block T1 T2 T3 T4 T5 T6

1

2

3

4

Pond 1 Pond 2 Pond 2 Pond 4

T3 T4

T2 T6

T5 T4

T6 T3

T5 T1

T2 T4

T2 T5

T4 T3

T6 T1

T2 T1

T6 T5

T3 T4


Data analysis: 1. Group the data by treatments and calculate the treatment totals (T), block totals (B) and grand total (G), the grand mean and the coefficient of variation (c.v.) etc.

2. Using number of treatments (t) and the number of blocks (b) determine the degree of freedom (d.f.) for each source of (b) determine the degree of freedom (d.f.) for each source of variation

3. Construct an outline/table (next slide) of the analysis of variance


ANOVA table of a RCBD experiment

Source of variation

Sum of squares

d.f. Mean square

F p

Trts SST t-1 SST= MST

t-1

MST

MSE

p<0.05 *

P<0.01 **t-1 MSEP>0.05 ns

Block SSB b-1 SSB= MSB

b-1

MSB

MSE

p<0.05 *

P<0.01 **

P>0.05 nsResidual error

SSE Total –others

SSE = MSE

df (E)

Total TSS txb-1

t = number of treatments; b = number of blocks/replicates per treatment


4. Using Xi to represent the measurement of the ith plot, Ti as the total of the ith treatment, and n as the total number of experimental plots [ i.e. n = (r) (t) ], calculate the correction factor (CF) and the various sums of square (SS)

5. Calculate the mean square (MS) for each source of variation by dividing SS by their corresponding d.f.

6. Calculate the F- values (R.A. Fisher) for testing significance of the treatment difference (F = MST/MSE and MSB/MSE)

7. Enter all the values computed in the ANOVA table


Example: Four different feeds were tested on 20pigs. Following were the mean final weights (kg) of 19 pigs (1 pig died).

Here, H0 = 1 = 2 = 3 = 4 and there is no effect of block

Family Feed 1 Feed 2 Feed 3 Feed 4 TotalFamily Feed 1 Feed 2 Feed 3 Feed 4 Total1 60.8 68.7 102.6 87.9 320

2 57.0 67.7 102.1 84.2 311

3 65.0 74.0 100.2 83.1 322

4 58.6 66.3 96.5 85.7 307

5 61.7 69.8 - 90.3 222

Total 303.1 346.5 401.4 431.2 1482.2

n 5 5 4 5 19

Mean 60.6 69.3 100.4 86.2


Step 1: Calculate Missing value

Missing value estimation

= (t * treatment total + b*block total) – Grand total

(t-1)(b-1)

= [(4*401.4+5*222) – 1482.2]/(3*4) = 102.7= [(4*401.4+5*222) – 1482.2]/(3*4) = 102.7

Notes:

- Missing values should not be more than 10% and it should be used only lost data due to unavoidable circumstances

- In case of more than 1 missing data, guess others and estimate first one. Then use it to estimate others/guessed ones – iterate it until all the values become stable


Example: Four different feeds were tested on 20pigs. Following were the mean final weights (kg) of 19 pigs (1 pig died).

Here, H0 = 1 = 2 = 3 = 4 and there is no effect of block

Family Feed 1 Feed 2 Feed 3 Feed 4 TotalFamily Feed 1 Feed 2 Feed 3 Feed 4 Total1 60.8 68.7 102.6 87.9 320

2 57.0 67.7 102.1 84.2 311

3 65.0 74.0 100.2 83.1 322

4 58.6 66.3 96.5 85.7 307

5 61.7 69.8 102.7 90.3 325

Total 303.1 346.5 504.1 431.2 1584.9

n 5 5 5 5 20

Mean 60.6 69.3 100.8 86.2


Step 1: Calculate sum of squares

Correction factor/Intercept ( C )

= (Grand total)2 /n = (1584.9)2 /20 = 125,595

Total SS = (60.8) 2 + (57.0)2 +--+ (90.3)2 - C = 4,934

Treatment SS = (Treatment total)2/t – CTreatment SS = (Treatment total) /t – C

= (303.1)2 /5+ (346.5)2 /5 + (504.1)2 /5 + (431.2)2 /5 – 125,595 = 4,801

Block SS = (Block total)2/n – C

= (320)2/4+(311)2/4+3222/4+(307)2/4+(325)2/4

-125,595 = 57

Error SS=Total SS–SST-SSB = 4,934–4,801– 57=76


Sources of variation

Sum of squares (SS)

df Mean SS = SS/df

F = MST/MSE

P Significance

Total 4934 19

Treatment 4801 3 1600 253 <0.01 **

Step 2: Prepare an ANOVA table

Block 57 4 14 2.25 >0.05 ns

Error or residual

76 12 6

Note: Numerator df = 3, Denominator d.f. = 12 for treatment F, and 4and 12 respectively for the block F.Reject H0 for Treatment but accept H0 for block orFamily had no effects on pig growth but type of feed has effects


Tests of Between-Subjects Effects

Dependent Variable: WT

125595.401 1 125595.401 8838.989 .000

56.837 4 14.209a

4801.022 3 1600.341 252.915 .000

SourceHypothesis

Error

Intercept

HypothesisFEED

Type II Sumof Squares df Mean Square F Sig.

Results from SPSS Analysis

4801.022 3 1600.341 252.915 .000

75.931 12 6.328b

56.837 4 14.209 2.246 .125

75.931 12 6.328b

Hypothesis

Error

FEED

Hypothesis

Error

BLOCK

MS(BLOCK)a.

MS(Error)b.


Two-way (Multi-factor) ANOVA to analyze

Check the block effect first,

1. If there is an effect of block - treatment effects need to be compared within each block separately

2. If there is no significant block effect data can be 2. If there is no significant block effect data can be analyzed using one-way ANOVA and go for t-test and Multiple range tests to compare treatments (Tukey’s test is widely accepted and most common)

# Note: If ANOVA shows no significant difference then multiple range tests are not necessary


Pairwise Comparisons

Dependent Variable: WT

-8.680* 1.591 .000 -12.146 -5.214

-40.200* 1.591 .000 -43.666 -36.734

-25.620* 1.591 .000 -29.086 -22.154

8.680* 1.591 .000 5.214 12.146

-31.520* 1.591 .000 -34.986 -28.054

-16.940* 1.591 .000 -20.406 -13.474

(J) FEED 2.00

3.00

4.00

1.00

3.00

4.00

(I) FEED 1.00

2.00

MeanDifference

(I-J) Std. Error Sig.a

Lower Bound Upper Bound

95% Confidence Interval forDifference

a

-16.940* 1.591 .000 -20.406 -13.474

40.200* 1.591 .000 36.734 43.666

31.520* 1.591 .000 28.054 34.986

14.580* 1.591 .000 11.114 18.046

25.620* 1.591 .000 22.154 29.086

16.940* 1.591 .000 13.474 20.406

-14.580* 1.591 .000 -18.046 -11.114

4.00

1.00

2.00

4.00

1.00

2.00

3.00

3.00

4.00

Based on estimated marginal means

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

Adjustment for multiple comparisons: Least Significant Difference (equivalent to noadjustments).

a.


WT

5 60.6200

5 69.3000

5 86.2400

5 100.8200

1.000 1.000 1.000 1.000

5 60.6200

5 69.3000

FEED 1.00

2.00

4.00

3.00

Sig.

1.00

2.00

Tukey HSDa,b

Duncana,b

N 1 2 3 4

Subset

5 86.2400

5 100.8200

1.000 1.000 1.000 1.000

5 60.6200

5 69.3000

5 86.2400

5 100.8200

1.000 1.000 1.000 1.000

4.00

3.00

Sig.

1.00

2.00

4.00

3.00

Sig.

Scheffea,b

Means for groups in homogeneous subsets are displayed.Based on Type II Sum of SquaresThe error term is Mean Square(Error) = 6.328.

Uses Harmonic Mean Sample Size = 5.000.a.

Alpha = .05.b.

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