Page 118 - DECO504_STATISTICAL_METHODS_IN_ECONOMICS_ENGLISH

Statistical Methods in Economics

Notes Limitations

(i) As compared to other measures it is difficult to compute. However, it does not reduce the
importance of this measure because of high degree of accuracy of results it gives.
(ii) It gives more weight to extreme items and less to those which are near the mean. It is because of
the fact that the squares of the deviations which are big in size would be proportionately greater
than the squares of those deviations which are comparatively small. The deviations 2 and 8 are
in the ratio of 1: 4 but their sqaures, i.e., 4 and 64, would be in the ratio of 1: 16.
Correcting Incorrect Values of Standard Deviation

Mistakes in calculations are always possible. Sometimes it so happens that while calculating mean
and standard deviation we unconsciously copy out wrong items. For example, an item 21 may be
copied as 12. Similarly one item 127 may be taken as only 27. In such cases if the entire calculations
are done again, it would become too tedious a task. By adopting a very simple procedure we can
correct the incorrect values of mean and standard deviation. For obtaining correct mean we find out
correct ΣX by deducting from the original ΣX the wrong items and adding to it the correct items.
2
Similarly for calculating correct standard deviation we obtain the value of correct ΣX . The following
illustration shall clarify the calculations.
Example 17: A student obtained the mean and standard deviation of 100 observations as 40 and
5.1 respectively. It was later found that one observation was wrongly copied as 50,
the correct figure being 40. Find the correct mean and standard deviation.
Solution:
Correct Mean:
We are given X = 40, σ = 5.1, N = 100

ΣX
X = N

ΣX
40 = or ΣX = 4000
100
But correct ΣX = ΣX – Wrong items + Correct items = 4000 – 50 + 40 = 3990.

Correct X 3990
Σ
Correct X = = = 39.9.
N 100
Correct Standard Deviation

()
ΣX 2 2
σ = –X
N
ΣX 2
5.1 = ( –40 ) 2
100
Squaring, we get

ΣX 2
26.01 = –1600
100
2
2
2601 = ΣX – 1,60,000 or ΣX = 2601 + 1, 60, 000 = 1, 62, 601.
2
2
Correct ΣX = Incorrect ΣX – wrong item square + Correct item square

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