Page 175 - Vogel's TEXTBOOK OF QUANTITATIVE CHEMICAL ANALYSIS

COMPARISON OF MORE THAN TWO MEANS (ANALYSIS OF VARIANCE) 4.18
An Analysis of Variance calculation is best illustrated by using specific values
in situation (b) just referred to.
Example II. Three analysts were each asked to perform four replicate titrations
using the same solutions. The results are given below.

Titre (mL)
Analyst A Analyst B Analyst C

To simplify the calculation it is sound practice to subtract a common number,
e.g. 22.50, from each value. The sum of each column is then determined. Note:
this will have no effect on the final values.

Analyst A Analyst B Analyst C

0.03 - 0.02 0.07
o. 10 -0.10 0.12
0.04 - 0.02 0.1 1
0.12 - 0.07 0.15
- -
Sum = 0.29 -0.21 0.45

The following steps have to be made in the calculation:
(a) The grand total
T = 0.29 - 0.21 + 0.45
= 0.53
(b) The correction factor (C.F.)

where N is the total number of results.
(c) The total sum of squares. This is obtained by squaring each result, summing
the totals of each column and then subtracting the correction factor (C.F.).

Analyst A Analyst B Analyst C
0.0009 0.0004 0.0049
0.0100 0.0 1 O0 0.0 144
0.001 6 0.0004 0.0121
0.0144 0.0049 0.0225
Sum = 0.0269 0.01 57 0.0539