ANCOVA Model with RCBD & Statistical Analysis lecture - 21

 

 ANCOVA Model RCBD 

&

Statistical Analysis

lecture 21

ANCOVA MODEL with RCBD

Let Yij denote the yield in i^th block of j^th treatment and is the corresponding covariate on dependent variable.

The ANCOVA model in the case of RCBD is given by:

Analysis of two Ways ANCOVA

Let (Yij, Xij) be the pair of observations of j^th treatment in the i^th block of dependent variable and covariate, and each pair is replicated r times. The data may be arranged as shown in the following table:

The table of sum of squares and sum of products:

Un adjusted ANOVA

Case 1: If H0: β = 0 is accepted. Then covariance adjustment has no real effect. The interpretation is developing on the unadjusted ANOVA.

Case 2: If H0: β = 0 is rejected, then continue the covariance analysis.

Adjust the forinear relationship between X and Y. Then completes the ANOVA for adjusted treatment effect on Y’s.

Adjusted ANOVA

Example

 The data on 3 treatments is given below:

Block	Treatment
	A		B		C
	X	Y	X	Y	X	Y
1	5	17	6	23	4	29
2	15	16	8	16	10	25
3	12	12	15	18	15	24

 Perform 2 ways ANCOVA and test the hypothesis that 3 treatments are significant at 5%.

Solution: We setup our hypothesis as:

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

ii. The significance Level;   α = 0.05

iii. The Test statistic: ANCOVA RCBD

iv. Calculation:

Computation for Y:

Computation for X:

Computation for XY:

Table of the sum of squares and product

The calculated F value falls in the acceptance region; thus, it is concluded that Y (yield) and X (shadow) are independent. We place our conclusion on an unadjusted ANOVA.

Reject  if H0; F => 6.94

The calculated F value (F = 31.00) is more than the F table value (F table = 6.94), thus the F calculated value falls in the rejection region. It is concluding that the effect of 3 treatments is significant.

Example

The entomologist wanted to test the effect of seven new chemicals on the control of maggots in onions. He used RCBD with blocks of 7 plots each. He planted each plot uniformly within blocks. He believed that chemicals would affect yield by controlling the maggots; however, he did not know whether they would affect germination. On one month after treatment, he counted the number of onions in each plot. At harvest, he graded the onions and counted the number of saleable onions per plot.

X: number of emerging onions (stand) and Y: number of saleable onions.

Block	Treatment
	1		2		3		4		5		6		7
	X	Y	X	Y	X	Y	X	Y	X	Y	X	Y	X	Y
I	103	88	86	70	135	112	128	103	113	94	154	118	65	59
II	77	68	74	68	157	124	141	111	120	105	155	135	126	102
III	90	81	63	61	132	95	118	98	124	105	143	110	93	85

Solution: We setup our hypothesis as:

i.     H0 : μ1 = μ2 = μ3 = μ4 = μ5 = μ6 = μ7   Vs.   H1 : μ1  ≠ μ2 ≠ μ3 ≠ μ4  ≠ μ5  ≠ μ6  ≠ μ7 

ii. The significance Level;   α = 0.05

iii. The Test statistic: ANCOVA RCBD

iv. Calculation:

Block	Treatment
	1		2		3		4		5		6		7
	X	Y	X	Y	X	Y	X	Y	X	Y	X	Y	X	Y
I	103	88	86	70	135	112	128	103	113	94	154	118	65	59
II	77	68	74	68	157	124	141	111	120	105	155	135	126	102
III	90	81	63	61	132	95	118	98	124	105	143	110	93	85

Now consider blocks.

Computation for Y:

Computation for X:

Computation for XY:

Table for the sum of squares and products:

Un adjusted ANOVA

Now test the slope of the covariate:

The calculated F value falls in the rejection region. We do not have sufficient evidence to accept H0. The covariate (X) has a significant effect on the response variable (Y).

 This conclusion directing that effect of covariate needs to be removed as follows:

Adjusted ANOVA table:

SV	df	S.S	M.S	F
Adjusted Treatment	6	76.48	12.746	0.46
Error	11	299.26	27.20
Total +Error	17	375.74

v.  Reject  if F => 3.09

vi. Remarks: 

The calculated F value falls in the acceptance region. So. We have not sufficient evidence to reject H0.Thus, it is concluded that the effect of 7 chemicals has no significant effect at the 5% significance level.

• Read More: Introduction to Factorial Experiment