The 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
We are given a feeding trial on sheep with 5 treatments (A, B, C, D, E) arranged in a Randomized Complete Block Design (RCBD) over 5 blocks (rows). We are to calculate the Critical Difference (CD) for treatment means. Step 1: Arrange the data by treatments | Treatment | Observations | Sum | Mean | | --------- | ------------------ | --- | ---- | | A | 24, 41, 29, 23, 38 | 155 | 31.0 | | B | 28, 29, 24, 38, 24 | 143 | 28.6 | | C | 30, 51, 35, 39, 29 | 184 | 36.8 | | D | 32, 45, 36, 37, 21 | 171 | 34.2 | | E | 33, 45, 41, 42, 26 | 187 | 35.8 | Step 2: Compute Block (row) totals and overall mean * Block 1: 32+33+30+28+24 = 147 * _____ ______ _____ ________ ____ ____ ___ _____ _________.
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For the given data the yield of the treatment 2 in 3rd block is missing. Estimate the missing value and analyse the data.
| Treatments | Blocks | |||
| I | II | III | IV | |
| 1 | 105 | 114 | 108 | 109 |
| 2 | 112 | 113 | Y | 112 |
| 3 | 106 | 114 | 105 | 109 |
A sample of 100 employees is to be drawn from a population of collages A and B. The population means and population mean squares of their monthly wages are given below:
| Village | Ni | ||
| Collage A | 400 | 60 | 20 |
| Collage B | 200 | 120 | 80 |
In a population of size N = 5, the values of the population characteristics are 1, 3, 5, 7, 9, a sample of size 2 is drawn. Verify that is an unbiased estimate of
and
is equal to
In a class of Statistics, total number of students is 30. Select the linear and circular systematic random samples of 12 students. The age of 30 students is given below:
Age: 22 25 22 21 22 25 24 23 22 21 20 21 22 23 25 23 24 22 24 24 21 20
23 21 22 20 20 21 22 25
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) |
For the given data the yield of the treatment 2 in 3rd block is missing. Estimate the missing value and analyse the data.
| Treatments | Blocks | |||
| I | II | III | IV | |
| 1 | 105 | 114 | 108 | 109 |
| 2 | 112 | 113 | Y | 112 |
| 3 | 106 | 114 | 105 | 109 |
A manurial trial with six levels of Farmyard Manure (FYM) was carried out in a randomised block design with 4 replications at the experimental station Junagarh with a new study the rate of decomposition of organic matters in soil and its synthetic capacity in soil on cotton crop. The yield per plot in kg for different levels of FYM and replications is given below:
Cotton Yield Per Plot (in Kg)
| Levels of FYM | Replications | |||
| 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 |