BIBD Statistical Model & Analysis Lecture - 41

 Statistical Model 

& 

Analysis of BIBD

Lecture - 41

Statistical Model

Let Yij is response of a BIBD using RCBD, and then it will be represented by the following linear model.

$$

Yij = μ + βi + τj + ϵij      i =1, 2, ..., b      j =1, 2, ..., t

Where:

μ is the general mean effect

 βi is the fixed additive i^th block effect

 τj is the fixed additive j^th treatment effect and

ϵij is the iid random error with ϵij ~ N(0, σ2).

Statistical Analysis

 Let Yij represent the yield of t treatments is distributed in b blocks in a balance incomplete block design further more each treatment is replicated r times and k number of treatments in block and each is replicated λ times.

ANOVA Table

Example: 

A small chemical lab wants to check the performance of four catalysts (A, B, C, and D) on raw materials but due to some constraints they only available 3 raw material for four catalysts. The investigator using the technique of BIBD and assign four catalysts to three raw material and data is obtained given below:

Catalyst	Raw Material
	1	2	3	4
A	73	74		71
B		75	67	72
C	73	75	68
D	75		72	75

Test the hypothesis about the significance of four catalysts by catalysts adjustment in blocks.

Solution:

i.   H0: τ1 = τ2 = τ3 = τ4 Vs. H1: τ1 ≠  τ2 ≠ τ3 ≠ τ4

ii. The significance level; α = 0.05

iii. The Test statistic:

iv. Reject H0, if F => F0.05 (3, 5) = 5.41

v. Computation:

Treatment adjustment

ANOVA Table

SV	df	SS	MS	F
Blocks	3	55.00	18.33
Treatment adj	3	22.71	7.57	11.50
Error	5	3.29	0.658
Total	12	81.00

vi.  Remarks

The four catalysts are significant.

Example:

An engineering research approach to increase a chemical process's yield. The factors of interest for treatments in four distinct chemical formulations (A, B, C, and D). Blocks are four-level groupings of raw materials that are the factors. The engineer intends to investigate how four distinct chemical compositions affect the yield process. because there are only three chemical formulations that can be used due to the lack of raw ingredients. This experiment's design, which is displayed below, is a BIBD.

Chemical Formulation	Batches of Raw Material
	1	2	3	4
A	95	101		90
B		111	110	107
C	119	117	113
D		95	93	102

Solution:

This is a BIBD (4, 4, 3, 3, 2)

i.                    State null and alternative hypothesis:

H0: μA = μB = μC = μD   Vs.  H1: μA ≠ μB ≠ μC ≠ μD

ii. The significance level; α = 0.05

iii. The test statistics:

   iv. Reject H0, if F => F0.05 (3, 5) = 5.41

   v. Computation:

Treatment adjustment

ANOVA Table:

SV	df	SS	MS	F
SST adj	3	808.48	269.49	0.57
Block	3	159	53
Error	5	2331.52	466
Total	11	3299

vi.                    The treatments are insignificant. 

Experiment

A balanced incomplete-block design (BIBD) that is symmetric was used to study the yield on four varieties of cassava with four different rates of NPK. These rates were administered in addition to the natural manure. The data collected from the experiment are tabulated in the layout of table given below:

Qualitative layout of the expt.					Tr#	A	B	C	D
Block 1	B	C	D			2		20	7
Block 2	A	C	D				32	14	3
Block 3	A	B	C			4	13	31
Block 4	A	B	D			0	23		11

Solution: clearly it is BIBD. with t = 4 , r = 3, k = 3

AB	2 times
AC	2 times
AD	2 times
BC	2 times
BD	2 times
CD	2 times

ANOVA Table

SV	df	SS	MS	F	Ftab
SS Block	3	100.66	33.55
	3	880.3275	293.4425	4.05	3.41
SSE	5	363.2725	72.73
SSTotal	11	1344.66

The treatments are significant.

Let the block adjusted

ANOVA Table

When block adjusted

SV	df	SS	MS	F	F tab
SST	3	975.34	325.11	4.47	3.41
SSBadj	3	6.17	2.05
SSE	3	363.6725	72.73
SSTotal	11	1344.67

The treatments are significant even block is adjusted.

• Read More: Construction Methods of BIBD