Information Technology Reference
In-Depth Information
To obtain our sum of squares with the computational formula, we
must calculate two quantities. The first,
Y
2
or the sum of the squared
scores, is calculated by squaring each
Y
i
score and then summing these
squared scores. These values are found on the right side of Table 2.4. Their
sum,
Y
2
, is equal to 639.
The second quantity we need to calculate is (
Y
)
2
n
,orthesquared
total sum of the
Y
scores divided by the number of scores. In Table 2.4,
we see that the sum of the
Y
scores is equal to 55 or
/
Y
=
55. When we
Y
)
2
×
=
square this total (55
55), we arrive at (
3,025. We now have the
constituents for computing the sum of squares.
Y
)
2
n
(
Y
2
=
−
SS
3025
5
SS
=
639
−
SS
=
639
−
605
=
34
.
(2.4)
As we noted previously, by dividing the sum of squares by the degrees
of freedom,
SS
/
df
, we obtain the variance. Hence, 34
/
4
=
8
.
5. This is the
same value that we obtained with the defining formula.
2.8 STANDARD DEVIATION AS A MEASURE OF VARIABILITY
You will recall that in order to remove the problem of our deviations from
the mean always summing to zero, we squared the deviations to eliminate
the negative values. While this numerical operation sidesteps the zero
sum problem, it makes it difficult to “intuitively” interpret the value of the
variance. For example, what exactly does a variance of 8.5 really mean?
This value reflects the squaring operation and, technically, informs us that
the five scores deviate from the mean by 8.5 “squared units.”
This description is not particularly helpful, and so most researchers
compute an additional measure of variability known as the
standard devi-
ation
,symbolizedaseither
s
or
SD
. The formula for the standard deviation
is simply the square root of the variance.
=
√
variance
SD
.
(2.5)
Thus in the present example,
√
8
92. By computing the
square root of the variance, we are literally “unsquaring” the variance,
which allows us to interpret the variability of the scores in the original units
of measurement. For example, if the original (raw) scores in Tables 2.3
and 2.4 were five
ho
urly wages, we can now say that the average wage was
$11.00 per hour (
Y
.
5
=
SD
=
2
.
11) and that on the average these wages deviated or
varied above or below the mean by $2.92. This is more informative than
to say the wages deviated by 8.5 squared dollars and cents.
=
