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 | 5.30 |
| 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 |
Carry out the Analysis of Variance test and draw the conclusions.
To carry out the Analysis of Variance (ANOVA) test for the given data, we need to follow these steps:
Step 1: Calculate the total sum of squares (TSS)
First, we need to find the grand total (GT) of all observations.
GT = 6.90 + 4.60 + 4.40 + 4.81 + 6.48 + 5.57 + 4.28 + 4.45 + 6.52 + 7.60 + 5.30 + 5.30 + 6.90 + 6.65 + 6.75 + 7.75 + 6.00 + 6.18 + 6.50 + 5.50 + 7.90 + 7.57 + 6.80 + 6.62
GT = 150.93
TSS = Σ(Observation - Grand Mean)^2
Grand Mean = GT / (Number of Treatments × Number of Replications)
= 150.93 / (6 × 4)
= 6.29
TSS = (6.90 - 6.29)^2 + (4.60 - 6.29)^2 + ... + (6.62 - 6.29)^2
TSS = 10.8041
Step 2: Calculate the sum of squares due to treatments (SST)
Treatment Totals:
T1 = 4.81 + 4.45 + 4.40 + 4.60 _________ _____ ______ ________ ___ _______ ________ ________ ________ ____ ____.
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A population consists of 10 villages with a total of 212 households. The second column of the accompanying table shows the number of households corresponding to each village. Select a PPS with replacement sample of 6 villages by using the Cumulative Total method:
| Village | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| No. of Households | 35 | 28 | 20 | 25 | 30 | 19 | 10 | 12 | 18 | 15 |
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 |
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:
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 |
State whether the following statements are true or false and also give the reason in support of your answer:
(a) lies between
and.
prop st Random st
(b) The total number of all possible samples of size 3 without replacement from a population of size 7 is 21.
(c) While analysing the data of a 5 × 5 Latin Square design the d.f. for ESS is equal to 16.
(d) In a Two-way Analysis of Variance test with 5 observations per cell having 4 blocks and 4 treatments the degree of freedom for the total variation is 64.
(e) The probability of selection of a sample of n from the population by SRSWOR is 1/ N.
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
andV(
) is equal to
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 |