Research Methodology - Dr. Krishan K. Pandey
Research Methodology - Dr. Krishan K. Pandey Research Methodology - Dr. Krishan K. Pandey
Analysis of Variance and Co-variance 261 (iii) Find out the square of all the item values one by one and then take its total. Subtract the correction factor from this total and the result is the sum of squares for total variance. Symbolically, we can write: 2 2 bg T Total SS =∑Xij − i = 1, 2, 3, … n j = 1, 2, 3, … (iv) Obtain the square of each sample total (T ) j 2 and divide such square value of each sample by the number of items in the concerning sample and take the total of the result thus obtained. Subtract the correction factor from this total and the result is the sum of squares for variance between the samples. Symbolically, we can write: 2 di bg Tj SS between = ∑ − n j T n 2 j = 1, 2, 3, … where subscript j represents different samples or categories. (v) The sum of squares within the samples can be found out by subtracting the result of (iv) step from the result of (iii) step stated above and can be written as under: R S| T| 2 ij SS within = ∑X− T 2 j =∑Xij −∑ n 2 T U R 2 bg Tj T V| −S| di bg ∑ − n n j n W| T| di j 2 After doing all this, the table of ANOVA can be set up in the same way as explained earlier. CODING METHOD Coding method is furtherance of the short-cut method. This is based on an important property of F-ratio that its value does not change if all the n item values are either multiplied or divided by a common figure or if a common figure is either added or subtracted from each of the given n item values. Through this method big figures are reduced in magnitude by division or subtraction and computation work is simplified without any disturbance on the F-ratio. This method should be used specially when given figures are big or otherwise inconvenient. Once the given figures are converted with the help of some common figure, then all the steps of the short-cut method stated above can be adopted for obtaining and interpreting F-ratio. Illustration 1 Set up an analysis of variance table for the following per acre production data for three varieties of wheat, each grown on 4 plots and state if the variety differences are significant. 2 U V| W|
262 Research Methodology Per acre production data Plot of land Variety of wheat A B C 1 6 5 5 2 7 5 4 3 3 3 3 4 8 7 4 Solution: We can solve the problem by the direct method or by short-cut method, but in each case we shall get the same result. We try below both the methods. Solution through direct method: First we calculate the mean of each of these samples: X 1 X 2 X 3 Mean of the sample means or X 6 + 7 + 3 + 8 = = 6 4 5 + 5 + 3 + 7 = = 5 4 5 + 4 + 3 + 4 = = 4 4 X + X + X = k 1 2 3 6 + 5 + 4 = = 5 3 Now we work out SS between and SS within samples: F H I K 2 SS between = n1 X1 − X + n2 X 2 − X + n3 X 3 − X F H I K 2 F H I K = 4(6 – 5) 2 + 4(5 – 5) 2 + 4(4 – 5) 2 = 4 + 0 + 4 = 8 2 2 2 d 1i 1i d 2i 2i d 3i 3i , i = 1, 2, 3, 4 SS within = ∑ X − X + ∑ X − X + ∑ X − X = {(6 – 6) 2 + (7 – 6) 2 + (3 – 6) 2 + (8 – 6) 2 } + {(5 – 5) 2 + (5 – 5) 2 + (3 – 5) 2 + (7 – 5) 2 } + {(5 – 4) 2 + (4 – 4) 2 + (3 – 4) 2 + (4 – 4) 2 } = {0 + 1 + 9 + 4} + {0 + 0 + 4 + 4} + {1 + 0 + 1 + 0} = 14 + 8 + 2 = 24 2
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262 <strong>Research</strong> <strong>Methodology</strong><br />
Per acre production data<br />
Plot of land Variety of wheat<br />
A B C<br />
1 6 5 5<br />
2 7 5 4<br />
3 3 3 3<br />
4 8 7 4<br />
Solution: We can solve the problem by the direct method or by short-cut method, but in each case<br />
we shall get the same result. We try below both the methods.<br />
Solution through direct method: First we calculate the mean of each of these samples:<br />
X 1<br />
X 2<br />
X 3<br />
Mean of the sample means or X<br />
6 + 7 + 3 + 8<br />
=<br />
= 6<br />
4<br />
5 + 5 + 3 + 7<br />
=<br />
= 5<br />
4<br />
5 + 4 + 3 + 4<br />
=<br />
= 4<br />
4<br />
X + X + X<br />
=<br />
k<br />
1 2 3<br />
6 + 5 + 4<br />
= = 5<br />
3<br />
Now we work out SS between and SS within samples:<br />
F<br />
H<br />
I<br />
K<br />
2<br />
SS between = n1 X1 − X + n2 X 2 − X + n3 X 3 − X<br />
F<br />
H<br />
I<br />
K<br />
2<br />
F<br />
H<br />
I<br />
K<br />
= 4(6 – 5) 2 + 4(5 – 5) 2 + 4(4 – 5) 2<br />
= 4 + 0 + 4<br />
= 8<br />
2<br />
2<br />
2<br />
d 1i 1i<br />
d 2i 2i<br />
d 3i 3i<br />
, i = 1, 2, 3, 4<br />
SS within = ∑ X − X + ∑ X − X + ∑ X − X<br />
= {(6 – 6) 2 + (7 – 6) 2 + (3 – 6) 2 + (8 – 6) 2 }<br />
+ {(5 – 5) 2 + (5 – 5) 2 + (3 – 5) 2 + (7 – 5) 2 }<br />
+ {(5 – 4) 2 + (4 – 4) 2 + (3 – 4) 2 + (4 – 4) 2 }<br />
= {0 + 1 + 9 + 4} + {0 + 0 + 4 + 4} + {1 + 0 + 1 + 0}<br />
= 14 + 8 + 2<br />
= 24<br />
2