he following data is the data pertaining to a feeding trial on sheep.

Question

he following data is the data pertaining to a feeding trial on sheep. Treatments

A: Grazing only

B: Grazing + Maize Supplements

C: Grazing + Maize + Protein Supplement P₁

D: Grazing + Maize + Protein Supplement P₂

E: Grazing + Maize + Protein Supplement P₃

Layout and Wool Yield (100 gm) is given as:

32(D)	33(E)	30(C)	28(B)	24(A)
51(C)	45(D)	41(A)	45(E)	29(B)
41 (E)	29 (A)	24 (B)	36 (D)	35 (C
38 (B	39(C)	42(E	23(A)	37(D)
38(A	24(B)	21(D)	29(C)	26(E)

Analyse the design with appropriate method and calculate the Critical Difference for the treatment mean yield

29 Apr 2025

Answer :

Word Count : 920

We are given data from a feeding trial on sheep with 5 treatments (A to E) and 5 blocks/replications, where each block has one of each treatment. This is a Randomized Block Design (RBD). --- ### Step 1: Arrange the data by Treatments and Replications From the table: | Replication | A | B | C | D | E | | ----------- | -- | -- | -- | -- | -- | | 1 | 24 | 28 | 30 | 32 | 33 | | 2 | 41 | 29 | 51 | 45 | 45 | | 3 | 29 | 24 | 35 | 36 | 41 | | 4 | 23 | 38 | 39 | 37 | 42 | | 5 | 38 | 24 | 29 | 21 | 26 | --- ### Step 2: Compute Totals and Means #### Treatment Totals and Means | Treatment | Total | Mean | | --------- | ------------------------ | ---------------- | | A | 24+41+29+23+38 = 155 | 155/5 = 31.0 | | B | 28+29+24+38+24 = 143 | _______ ________ __________ _________ ________ __________ _______ ___ __________.
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Treatments	Blocks
Treatments	I	II	III	IV
1	105	114	108	109
2	112	113	Y	112
3	106	114	105	109

Levels of FYM	Replications
Levels of FYM	I	II	III	IV
1	6.90	4.60	4.40	4.81
2	6.48	5.57	4.28	4.45
3	6.52	7.60	5.30	4.45
4	6.90	6.65	6.75	7.75
5	6.00	6.18	6.50	5.50
6	7.90	7.57	6.80	6.62

Gasoline A:	19	21	20	18	21	21
Gasoline B:	23	20	22	20	24	23
Gasoline C:	20	17	21	19	20	17

Village	1	2	3	4	5	6	7	8	9	10
No. of Households	35	28	20	25	30	19	10	12	18	15