2. Estimation of variability parameters - Final.pptx
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1) The document presents lint index data for 25 cotton genotypes evaluated in a randomized block design with 3 replications. 2) It calculates components of variance, including phenotypic variance (Vph), genotypic variance (Vg), and environmental variance (Ve), for the lint index data. 3) The phenotypic coefficient of variation (PCV) and genotypic coefficient of variation (GCV) were found to be moderate, indicating a moderate level of variability and low environmental influence.
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FACTOR: 30%
See below.
NAME OF STAFF SETTING ASSIGNMENT: Dr B L Shea
0
MANCHESTER METROPOLITAN UNIVERSITY
FACULTY OF SCIENCE AND ENGINEERING
SCHOOL OF COMPUTING, MATHEMATICS & DIGITAL TECHNOLOGY
ACADEMIC YEAR 2013-2014:
REFERRED COURSEWORK
BSC(HONS) FINANCIAL MATHEMATICS
BSC(HONS) MATHEMATICS
YEAR/STAGE THREE
UNIT 6G6Z3005 : APPLIED REGRESSION AND MULTIVARIATE ANALYSIS
Answer ALL questions.
The pass mark is 40% which corresponds to a minimum of 72
marks out of a possible 180 marks.
The deadline is 8th August 2014.
SECTION A
1. (a) Three measurementsx1, x2 andx3 have the following sample covariance matrix.
∑̂ =
9 2 0
2 4 1
0 1 4
(i) Verify that the corresponding sample correlation matrix C, is given by
C =
1 13 0
1
3 1
1
4
0 14 1
[2]
(ii) Given that one of the eigenvalues of C is equal to one, calculate the other two
eigenvalues and determine the proportion of the variation in the data explained
by the first principal component.
[6]
(iii) Using the sample correlation matrix C, calculate the first principal component.
[6]
(b) A Principal Components Analysis of the prices of food items in 23 cities was carried
out with a view to forming a measure of the Consumer Price Index(CPI). A Minitab
analysis of this data is attached.
(i) Explain why Principal Components Analysis was performedon the correlation
matrix instead of the covariance matrix.
[2]
(ii) If the first Principal Component is taken as a measure of the CPI calculate, to
one decimal place, the value of the index for Atlanta.
[2]
(iii) Which is the most expensive city and which is the least expensive city?
[2]
(Question 1 continued overleaf)
1
(Question 1 continued)
Minitab output for Question 1
Descriptive Statistics: bread, burger, milk, oranges, tomatoes
Variable N Mean Median TrMean StDev SE Mean
bread 23 25.291 25.300 25.267 2.507 0.523
burger 23 91.86 91.00 91.63 7.55 1.58
milk 23 62.30 62.50 61.96 6.95 1.45
oranges 23 102.99 105.90 102.90 14.24 2.97
tomatoes 23 48.77 46.80 48.74 7.60 1.59
Principal Component Analysis: bread, burger, milk, oranges, tomatoes
Eigenanalysis of the Correlation Matrix
Eigenvalue 2.4225 1.1047 0.7385 0.4936 0.2408
Proportion 0.484 0.221 0.148 0.099 0.048
Cumulative 0.484 0.705 0.853 0.952 1.000
Variable PC1 PC2 PC3 PC4 PC5
bread 0.496 -0.309 0.386 -0.509 -0.500
burger 0.576 -0.044 0.262 0.028 0.773
milk 0.340 -0.431 -0.835 -0.049 0.008
oranges 0.225 0.797 -0.292 -0.479 -0.006
tomatoes 0.506 0.287 0.012 0.713 -0.391
(Question 1 continued overleaf)
2
(Question 1 continued)
Data Display
Row c ...
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- 1. Estimation of variability parameters (Components of variance) Estimation of PCV and GCV for replicated data Problem : 1 Lint index of an experiment with 25 cotton genotypes is given below. These genotypes have been evaluated in RBD with 3 replications. Workout components of variance for lint index.
- 2. Genotypes Replication I Replication II Replication III Total 1 36.5 36.0 36.4 108.9 2 51.0 54.0 55.5 160.5 3 42.1 44.8 42.9 129.8 4 46.5 51.0 49.5 147.0 5 49.5 48.0 47.3 144.8 6 51.0 49.5 51.5 152.0 7 50.0 49.0 49.8 148.8 8 54.5 52.0 55.5 162.0 9 50.5 48.5 52.0 151.0 10 53.8 54.0 53.5 161.3 11 57.0 56.0 55.0 168.0 12 52.5 51.0 50.0 153.5
- 3. 13 47.5 48.0 48.5 144.0 14 57.0 56.0 57.5 170.5 15 39.0 37.0 40.0 116.0 16 47.5 49.0 48.0 144.5 17 60.0 61.0 62.0 183.0 18 61.0 62.5 61.5 185.0 19 46.5 48.5 45.0 140.0 20 47.0 47.5 48.5 143.0 21 54.0 55.0 55.5 164.5 22 48.0 48.5 49.5 146.0 23 47.0 45.5 47.0 139.5 24 58.5 58.0 57.5 174.0 25 55.5 57.5 56.5 169.5 Total 1263.4 1267.8 1275.9 3807.1
- 4. Estimation of variability parameters (Components of variance) Estimation of PCV and GCV for replicated data Correction factor = = = 193253.46 Raw sum of squares = 36.52 + …………. 56.52 = 196058.55 (Grand Total)2 N (3807.1)2 75 ( ∑ x 2 )
- 5. Total sum of squares = Raw S.S – C.F = 196058.55 – 193253.46 = 2805.09 Treatment sum of squares = 108.92 + ……. + 169.52 – C.F = 2708.59 Sum of squares of Treatment Total No. of rep = – C.F 587886.17 3 = – 193253.46
- 6. Replication sum of squares = Error SS = Total SS – Treat SS – Rep SS = 93.28 Sum of squares of Rep.Total No. of Treatments = 1263.42 + …… 1275.92 25 = – 193253.46 3.22 – C.F
- 7. Anova Sources of Variation df SS MSS SS df F.ratio Replication (r-1) 3-1 = 2 3.22 1.61 1.61 1.94 Treatment (t-1) 25-1 = 24 2708.59 112.86 112.86 1.94 Error (t-1) (r-1) 24 x 2 = 48 93.28 1.94 Total 74 2805.09 = 0.83 = 58.18
- 8. F Table (48, 24) 5% 1.74 1% 2.20 If the calculated value is higher than the table value, it is significant. Treatments are differing significantly.
- 9. Environmental variance (Ve) = EMS (Error Mean square) = 1.94 Genotypic variance (Vg) = = = 36.97 (Ms due to treatment – EMS) r 112.86 – 1.94 3
- 10. Phenotypic variance (Vph) = Vg + Ve = 36.97 + 1.94 = 38.91 GCV = = 50.76 = = = 11.98% √Vg x x 100 3807.1 x = 75 √36.97 x 100 50.76 6.08 x 100 50.76
- 11. PCV = = = 12.29% √Vph x √38.91 x 100 50.76 x 100
- 12. ECV = = = 2.76% < 10 - low 10 - 20 - Moderate above 20 - High The PCV and GCV of lint index are moderate. It means that the variability is moderate. The influence of environment is low. √Ve x √1.94 x 100 50.76 x 100
- 13. Heritability = = = 95.01% GA = K x √Vph x = 2.06 x √38.91 x = 2.06 x 6.24 x 0.95 = 12.21 Vg Vph x 100 36.97 38.91 x 100 Vg Vph 36.97 38.91
- 14. GA as percent of Mean = = = 24.05 Inference 1. If the GCV is higher than PCV, it indicates that there is little influence of environment on the expression of character. Selection for such character will be rewarding. GA x x 100 12.21 50.76 x 100
- 15. 2. If PCV is higher than GCV, it means that the apparent variation is not only due to genotypes but also due to influence of environment. Selection for such traits may be misleading. 3. If ECV is higher than PCV & GCV, it means the environment is playing a significant role in the expression of such character. Selection for the improvement of such character will be ineffective.
- 16. R1 R2 R3 Total T1 65 95 84 244 T2 46 54 58 158 T3 62 68 67 197 T4 67 61 63 191 T5 75 77 71 223 T6 47 51 53 151 T7 61 66 69 196 Total 423 472 465 1360 Problem : 2 Calculate the components of variance.