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Alternatively, it can also be calculated in terms of SS for total variance:
We can now set up the ANOVA table for this problem:
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(a)
(i)
(ii)
(iii)
(iv)
The table shows that the calculated value of F is 1.5 which is less than the table value of 4.26
at 5% level with degree of freedom, v 1 =2 and v 2 =9. Hence, it could arise due to chance.
Thus, null hypothesis of no difference is sample means. Therefore, it is concluded that the
difference in wheat output due to varieties is insignificant and is just a matter of chance.
Two-way ANOVA
When the data are classified because of two factors, two-way ANOVA is used. For example,
the output of agriculture may be classified based on different varieties of seeds and on the
basis of different varieties of fertilizers used. Thus, in a business firm, its sales data may be
classified based on different salesmen andbased on sales in different regions. During a
particular period, various units of a product is manufactured can be classified because of
different types of machines used and because of different grades of labor. Such a two-way
design may have repeated measurements of each factor or may have repeated values. In case
of repeated measurements, the ANOVA technique is little different. In such measurements
the interaction variation should also be calculated. In the context of both these designs the
two-way ANOVA technique shall be discussed here with the help of relevant example.
In the context of two-way design when repeated values are not there: Since the values
are not repeated, the sum of squares within the samples cannot be calculated as can be
done in case of one-way ANOVA. Thus, the residual or error variation by subtraction
is calculated after calculating the sum of squares for total variance and for variance
between varieties of one treatment as also for variance between varieties of other
treatment.
The various steps involved are:
The coding device should be used, if the same can simplify the task.
The total of values of individual items is to be taken in all samples; let us call it T.
The correction factor is calculated as under: Correction factor =
The square of all the item values are calculated one by one and then totaling is done.
From this total, the correction factor is to be subtracted to get the sum of squares of
deviations for total variance.
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(v)
(vi)
(vii)
(viii)
(ix)
Sum of squares of deviations for total variance or total SS =
The total of different columns is taken to obtain the square of each column total. The
total of such squared values of each column is divided by the number of items in that
column and the total of the result thus obtained is taken. Thereafter, the correction
factor is to be subtracted from this total. Thus, the sum of squares of deviations for
variance between columns or SSB is obtained.
The total of different rows is taken to obtain thesquare of each row total. Such squared
values of each row are divided by the number of items in the corresponding row and
the total of the result thus obtained is taken to subtract the correction factor from the
total. Thus, the sum of squares of deviations for variance between the rows or SS
between the rows is obtained.
The sum of squares of deviations for residual or error variance can be calculated by
subtracting the result of sum obtained in step (v) and (vi) from the result of step (iv)
mentioned above. Thus, Total SS – (SS B + SS between rows) = SS for residual or
error variance.
The degrees of freedom (d.f) can be calculated as mentioned below:
The d.f for total variance = (c×r – 1)
The d.f for variance between the columns = (c – 1)
The d.f for variance between the rows = (r – 1)
The d.f for residual variance = (c – 1)(r – 1)
Where, c is the number of columns, r is the number of rows
ANOVA table can be prepared in the common fashion as shown below:
Table 6.3 Analysis of variance Two-way ANOVA
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Example 11: Set up ANOVA table for the following information about reduction of blood
pressure in mm of Hg by three drugs to evaluate the effectiveness for three different groups
of people:
Do the drugs act differently?
Are the different groups of people affected differently?
Is the interaction term significant?
Answer the above questions taking a significant level of 5%.
Solution: Let us first make all the required calculations as shown below:
We can set up ANOVA table as shown
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Step (v) SS within samples = (14 – 14.5) 2 + (15 – 14.5) 2 + (10 – 9.5) 2 + (9 – 9.5) 2 + (11 –
11) 2 + (11 – 11) 2 + (12 – 11.5) 2 + (11 – 11.5) 2 + (7 – 7.5) 2 + (8 – 7.5) 2 + (10 – 10.5) 2 + (11
– 10.5) 2 + (10 – 10.5) 2 + (11 – 10.5) 2 + (11 – 11) 2 + (11 – 11) 2 + (8 – 7.5) 2 + (7 – 7.5)
2
= 3.50
Step (vi) SS for interaction variation = 76.28 – [28.77 + 14.78 + 3.50]
= 29.23
ANOVA table
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1.
2.
3.
4.
5.
* These figures are left-over figures and have been obtained by subtracting the total of all
other value in the said columnfrom thetotal column.
Thus, interaction SS = (76.28) – (28.77 + 14.78 + 3.50) = 29.23 and
Interaction degrees of freedom = (17) – (2 + 2 + 9) = 4.
The above table shows that all the three F-ratios are significant of 5% level which means that
the drugs act differently. Different groups of people are affected differently, and the
interaction term is significant. In fact, if the interaction term happens to be significant, it is
pointless to talk about the differences between various treatments; that is, differences
between drugs or differences between groups of people in the given case.
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