252oneaex5 2/8/08
Review of Computation of the Sample Variance and Standard Deviation
There are two formulas for the sample variance. The definitional formula s 2 
 x  x 
2
is
n 1
based on the statement that a population mean is the average squared deviation of observations from the
mean and the fact that the expected value of the sample mean is equal to the population mean E s 2   2 .
(See 251dscr_B.) The computational formula s 2 
equalities in statistics, that
x
 x      x  N
2
2
 
2
 nx
2
n 1
2
is based on one of the most potent
. (See 251dscr_C.)
The computational formula is one of the most important formulas you will learn. Note that
x
2
is not the same as
 x  . For example, if x is 2,3,5 ,  x
2
 2 2  3 2  5 2  4  9  25  38 ,
2
not 2  3  52  10 2  100 .
Example: Use x  2,3,5
Computational Method
x2
x
2
4
3
9
5
25
10
38
From this we find
 x  10,  x
2
 38, x 
Definitional Method
x
x  x 
2
-1.33333
3
-0.33333
5
1.66667
10
0.00001
 x  10  3.33333
n
3
 x  x 2
1.77778
0.11111
2.77778
4.66667
 x  x 
2
and
 4.66667 Note that
 x  x  should be zero, but will often be slightly different because of rounding. Now, if we use the
 x  nx  38  33.33333   4.6667  2.3333 . (Some texts
computational method, we can use s 
2
2
2
2
prefer s 2 

 x 
1
x2 
n
n 1
2
n 1
3 1
2
1
38  10 2
4.66666667
3


 2.33333 which gives us a little more
3 1
2
 x  x 

2
accuracy for a little more work.) If we use the definitional method s
2

4.66667
2
n 1
 2.33333 , but note that we had to do three subtractions instead of 1.
In any case the sample standard deviation is the square root of the sample variance
s  4.66667  1.527525 and is actually less affected by rounding error than the variance. The standard
error of the sample mean is the standard deviation divided by the square root of the sample size.
sx 
s
n

1.527525
3

s2

n
2.33333
 0.77778  0.8819 .
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