Research Methodology - Dr. Krishan K. Pandey

Analysis of Variance and Co-variance 279
198
= =
274. 40
0. 7216
Deviation of initial group means from Final means of groups in X (unadjusted)
general mean (= 14) in case of Y
Group I –7.40 9.80
Group II 0.40 22.80
Group III 7.00 35.00
Adjusted means of groups in X = (Final mean) – b (deviation of initial mean from general mean
in case of Y)
Hence,
Adjusted mean for Group I = (9.80) – 0.7216 (–7.4) = 15.14
Adjusted mean for Group II = (22.80) – 0.7216 (0.40) = 22.51
Adjusted mean for Group III = (35.00) – 0.7216 (7.00) = 29.95
Questions
1. (a) Explain the meaning of analysis of variance. Describe briefly the technique of analysis of variance for
one-way and two-way classifications.
(b)State the basic assumptions of the analysis of variance.
2. What do you mean by the additive property of the technique of the analysis of variance? Explain how
this technique is superior in comparison to sampling.
3. Write short notes on the following:
(i) Latin-square design.
(ii) Coding in context of analysis of variance.
(iii) F-ratio and its interpretation.
(iv) Significance of the analysis of variance.
4. Below are given the yields per acre of wheat for six plots entering a crop competition, there of the plots
being sown with wheat of variety A and three with B.
Variety Yields in fields per acre
1 2 3
A 30 32 22
B 20 18 16
Set up a table of analysis of variance and calculate F. State whether the difference between the yields of
two varieties is significant taking 7.71 as the table value of F at 5% level for v 1 = 1 and v 2 = 4.
(M.Com. II Semester EAFM Exam., Rajasthan University, 1976)
5. A certain manure was used on four plots of land A, B, C and D. Four beds were prepared in each plot and
the manure used. The output of the crop in the beds of plots A, B, C and D is given below: