Graeco Latin Square Design lecture 19

Graeco Latin Square Design

lecture - 19 

Graeco Latin Square Design

The Graeco-Latin square is an extension of a Latin square and can simultaneously manage to control three sources of nuisance variability other than treatment. The experimental units grouped in three different ways, by rows, by columns, by Greek letters. The Graeco Latin square design is also called triple grouping design.

Construction of Graeco Latin Square

The Graeco Latin square is created by superimposing the two Latin squares (Latin letters & Greek letters) of the identical size and each Latin letter each Greek letter absolutely correspond once.  The construction of the  Graeco Latin square design involves using the  Latin square,  where treatments are denoted by Latin letters and superimposing on it a  second Latin square in which variable levels are denoted by Greek letters. The Greek letters are usually assigned in a way that each Greek letter occurs once and only once in each row, column and each Latin letter.

Let us create two Latin squares, one using Latin letters and the other using Greek letters.

A 3 X 3 Graeco Latin square is created by superimposing the two Latin squares.

Imagine that you build two 4 X 4 Latin squares one with Latin letters and the other with the Greek characters.

Superimposing the above two Latin squares

Statistical Model and Analysis

The Graeco Latin square design is represented by the following linear statistical model.

Yijkl = μ + ρi +βj + τk + σl + ϵijkl        i, j, k , l =1, 2, ..., P

ANOVA Table:

SV	df	SS	MS	F
Rows	P - 1	SSR	MSR	F1
Columns	P - 1	SSC	MSC	F2
Greek letters	P - 1	SSG	MSG	F3
Treatments	P - 1	SST	MST	F4
Error	(P – 1) (P – 3)	SSE	MSE
Total	P x P - 1	SSTotal

Short cut to compute various sums of squares:

Short cut to compute various sums of squares:

Example:

Tests are done on four cars with four drivers over the course of four days in order to compare four different petrol additives. Each day, a maximum of four runs may be completed. The total amount of vehicle emissions is the response.

 Treatment factor: petrol additive, denoted by A, B, C and D.

Block factor 1: driver, denoted by 1, 2, 3, 4.

Block factor 2: day, denoted by 1, 2, 3, 4.

Block factor 3: car, denoted by α, β, γ, δ.

Write the appropriate statistical model for the experiment and test the significance of petrol additive consumption at 5 % significance level.

Solution:  let Yijkl is the consumption of petrol additive, μ is the mean consumption of petrol additive (historical). The effect of four drivers (Ri), four days (Ci) , four types of cars (Greek letters) and four petrol additives (Latin letters) on Petrol consumption is represented by the following linear model:

Yijkl = μ + Ri +Cj + τk + σl + ϵijkl        i, j, k , l =1, 2, 3, 4.

 Setup our hypothesis as:

i.H0: The four petrol additives are consumed in the same amounts.

Vs.

   H1: The four petrol additives are not consumed in the same way.

ii. The significance level; α = 0.05

iii. The test statistic:

F = MST / MSE ~ F(3, 3)

iv. Reject H0, if F >= 9.28

v. Computation:

	Treatment (Gasoline)				Total		Car
	A	B	C	D
	32	24	28	34			32	28	34	24
	36	25	35	30			30	25	36	35
	23	29	31	20			29	20	31	23
	33	31	25	27			25	33	31	27
	124	109	119	111			116	106	132	109
	15376	11881	14161	12321	53739		13456	11236	17424	11881	53997

ANOVA Table:

SV	df	SS	MS	F
Rows	3	90.6875	30.229	2.50
Columns	3	68.1875	22.73	1.88
Greek letters	3	101.1875	33.73	2.80
Treatments	3	36.6875	12.33	0.27
Error	3	36.1875	12.06
Total	15	332.9375

F = MST / MSE = 0.27

vi. Remarks:

he calculated F value falling in the acceptance region, we have not sufficient evidence to reject H0 . Thus, it is concluding that the 4 petrol additive consumption is identical.

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