Download Example 2

Mean square deviation
Root mean square deviation
Variance
Standard deviation
Choose one of the following:
•
•
•
•
•
•
Example 1 : Method A
Raw data using defintion
Example 1 : Method B
Raw data using alternative
Example 2 : Method A
Frequency distribution using defintion
Example 2 : Method B
Frequency distribution using alternative
Summary of formulae
Notes on this presentation
Example 1 : Method A
Extracts were taken from nine leaf cells and the pH of each was measured. The results
were as follows:
6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9
 Tabulate the values (x) and find the mean:
1
2
3
4
5
6
7
8
9
Total
x
6.5
5.9
5.4
6.0
6.1
5.9
5.8
5.6
5.9
53.1
The sample mean x
x
n
53.1
=
9
=
= 5.9
x
Example 1 : Method A
Extracts were taken from nine leaf cells and the pH of each was measured. The results
were as follows:
6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9
 Find the deviation from the mean, for each x value, x – x :
x = 5.9
1
2
3
4
5
6
7
8
9
Total
x
6.5
5.9
5.4
6.0
6.1
5.9
5.8
5.6
5.9
53.1
x–x
0.6
0.0
-0.5
0.1
0.2
0.0
-0.1
-0.3
0.0
Example 1 : Method A
Extracts were taken from nine leaf cells and the pH of each was measured. The results
were as follows:
6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9
 Square the deviations from the mean and find the ‘sum of squares’ Sxx:
x = 5.9
1
2
3
4
5
6
7
8
9
Total
x
6.5
5.9
5.4
6.0
6.1
5.9
5.8
5.6
5.9
53.1
x–x
0.6
0.0
-0.5
0.1
0.2
0.0
-0.1
-0.3
0.0
(x – x )2
0.36
0.00
0.25
0.01
0.04
0.00
0.01
0.09
0.00
0.76
msd & rmsd
Variance &
standard
deviation
Sxx =
 ( x  x )2
Example 1 : Method A
Mean square deviation
Extracts were taken from nine leaf cells and the pH of each was measured. The results
were as follows:
6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9
 Divide the ‘sum of squares’ Sxx by n :
x = 5.9
1
2
3
4
5
6
7
8
9
Total
x
6.5
5.9
5.4
6.0
6.1
5.9
5.8
5.6
5.9
53.1
x–x
0.6
0.0
-0.5
0.1
0.2
0.0
-0.1
-0.3
0.0
(x – x )2
0.36
0.00
0.25
0.01
0.04
0.00
0.01
0.09
0.00
0.76
Mean square deviation
=
S xx
n
 ( x  x )2
=
n
= 0.76
9
= 0.0844 (to 3 s.f.)
 ( x  x )2
Example 1 : Method A
Root mean square deviation
Extracts were taken from nine leaf cells and the pH of each was measured. The results
were as follows:
6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9
 Divide the ‘sum of squares’ Sxx by n and take the square root :
x = 5.9
1
2
3
4
5
6
7
8
9
Total
x
6.5
5.9
5.4
6.0
6.1
5.9
5.8
5.6
5.9
53.1
x–x
0.6
0.0
-0.5
0.1
0.2
0.0
-0.1
-0.3
0.0
(x – x )2
0.36
0.00
0.25
0.01
0.04
0.00
0.01
0.09
0.00
0.76
Root mean square
deviation (rmsd)
=
=
S xx
=
n
 ( x  x )2
n
0.76
9
= 0.291 (to 3 s.f.)
 ( x  x )2
RETURN
Example 1 : Method A
Variance
Extracts were taken from nine leaf cells and the pH of each was measured. The results
were as follows:
6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9
 Divide the ‘sum of squares’ Sxx by n – 1 :
x = 5.9
1
2
3
4
5
6
7
8
9
Total
x
6.5
5.9
5.4
6.0
6.1
5.9
5.8
5.6
5.9
53.1
x–x
0.6
0.0
-0.5
0.1
0.2
0.0
-0.1
-0.3
0.0
(x – x )2
0.36
0.00
0.25
0.01
0.04
0.00
0.01
0.09
0.00
0.76
Variance
S xx
 ( x  x )2
=
=
n 1
n 1
=
0.76
8
= 0.095
 ( x  x )2
Example 1 : Method A
Standard deviation
Extracts were taken from nine leaf cells and the pH of each was measured. The results
were as follows:
6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9
 Divide the ‘sum of squares’ Sxx by n – 1 and take the square root :
x = 5.9
1
2
3
4
5
6
7
8
9
Total
x
6.5
5.9
5.4
6.0
6.1
5.9
5.8
5.6
5.9
53.1
x–x
0.6
0.0
-0.5
0.1
0.2
0.0
-0.1
-0.3
0.0
(x – x )2
0.36
0.00
0.25
0.01
0.04
0.00
0.01
0.09
0.00
0.76
Standard deviation s
=
=
S xx
=
n 1
 ( x  x )2
n 1
0.76
8
= 0.308 (to 3 s.f.)
 ( x  x )2
RETURN
Example 1 : Method B
Extracts were taken from nine leaf cells and the pH of each was measured. The results
were as follows:
6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9
 Tabulate the values (x) and find the mean:
1
2
3
4
5
6
7
8
9
Total
x
6.5
5.9
5.4
6.0
6.1
5.9
5.8
5.6
5.9
53.1
The sample mean x
x
n
53.1
=
9
=
= 5.9
x
Example 1 : Method B
Extracts were taken from nine leaf cells and the pH of each was measured. The results
were as follows:
6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9
 Tabulate the squares of the values (x2) and find the total:
1
2
3
4
5
6
7
8
9
Total
x
6.5
5.9
5.4
6.0
6.1
5.9
5.8
5.6
5.9
53.1
x2
42.25
34.81
29.16
36.00
37.21
34.81
33.64
31.36
34.81
314.05
x2
Example 1 : Method B
Extracts were taken from nine leaf cells and the pH of each was measured. The results
were as follows:
6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9
 Find the ‘sum of squares’ Sxx by subtracting nx 2 from x2:
x = 5.9
1
2
3
4
5
6
7
8
9
Total
x
6.5
5.9
5.4
6.0
6.1
5.9
5.8
5.6
5.9
53.1
x2
42.25
34.81
29.16
36.00
37.21
34.81
33.64
31.36
34.81
314.05
Sxx =  x 2  nx 2
=
314.05 – 9  5.92
=
0.76
msd &
rmsd
x2
Variance &
standard
deviation
Example 1 : Method B
Mean square deviation
Extracts were taken from nine leaf cells and the pH of each was measured. The results
were as follows:
6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9
 Divide the ‘sum of squares’ Sxx by n :
x = 5.9
1
2
3
4
5
6
7
8
9
Total
x
6.5
5.9
5.4
6.0
6.1
5.9
5.8
5.6
5.9
53.1
x2
42.25
34.81
29.16
36.00
37.21
34.81
33.64
31.36
34.81
314.05
Mean square deviation
=
S xx
n
=
 x 2  nx 2
n
= 0.76
9
= 0.0844 (to 3 s.f.)
x2
Example 1 : Method B
Root mean square deviation
Extracts were taken from nine leaf cells and the pH of each was measured. The results
were as follows:
6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9
 Divide the ‘sum of squares’ Sxx by n and take the square root :
x = 5.9
1
2
3
4
5
6
7
8
9
Total
x
6.5
5.9
5.4
6.0
6.1
5.9
5.8
5.6
5.9
53.1
x2
42.25
34.81
29.16
36.00
37.21
34.81
33.64
31.36
34.81
314.05
Root mean square
deviation (rmsd)
=
=
S xx
=
n
 x 2  nx 2
n
0.76
9
= 0.291 (to 3 s.f.)
x2
RETURN
Example 1 : Method B
Variance
Extracts were taken from nine leaf cells and the pH of each was measured. The results
were as follows:
6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9
 Divide the ‘sum of squares’ Sxx by n – 1 :
x = 5.9
1
2
3
4
5
6
7
8
9
Total
x
6.5
5.9
5.4
6.0
6.1
5.9
5.8
5.6
5.9
53.1
x2
42.25
34.81
29.16
36.00
37.21
34.81
33.64
31.36
34.81
314.05
Variance
=
S xx
n 1
= 0.76
8
= 0.095
x2
=
 x 2  nx 2
n 1
Example 1 : Method B
Standard deviation
Extracts were taken from nine leaf cells and the pH of each was measured. The results
were as follows:
6.5, 5.9, 5.4, 6.0, 6.1, 5.9, 5.8, 5.6, 5.9
 Divide the ‘sum of squares’ Sxx by n – 1 and take the square root :
x = 5.9
1
2
3
4
5
6
7
8
9
Total
x
6.5
5.9
5.4
6.0
6.1
5.9
5.8
5.6
5.9
53.1
x2
42.25
34.81
29.16
36.00
37.21
34.81
33.64
31.36
34.81
314.05
Standard deviation s
=
=
S xx
n 1
=
 x 2  nx 2
n 1
0.76
8
= 0.308 (to 3 s.f.)
x2
RETURN
Example 2 : Method A
The number of children per family, x, for a random selection of 100 families, is given
by the following table:
No. of children, x
0
1
2
3
4
5
6
>6
Frequency, f
9
24
35
19
8
3
2
0
 First tabulate the xf values and find their mean:
n = f
x
f
xf
1
0
9
0
2
1
24
24
3
2
35
70
4
3
19
57
5
4
8
32
xf
n
210
=
100
6
5
3
15
= 2.1
7
6
2
12
Total
21
100
210
The sample mean x
=
xf
Example 2 : Method A
The number of children per family, x, for a random selection of 100 families, is given
by the following table:
No. of children, x
0
1
2
3
4
5
6
>6
Frequency, f
9
24
35
19
8
3
2
0
 Find the deviation from the mean, for each x value, x  x :
x = 2.1
x
f
xf
xx
1
0
9
0
-2.1
2
1
24
24
-1.1
3
2
35
70
-0.1
4
3
19
57
0.9
5
4
8
32
1.9
6
5
3
15
2.9
7
6
2
12
3.9
Total
21
100
210
Example 2 : Method A
The number of children per family, x, for a random selection of 100 families, is given
by the following table:
No. of children, x
0
1
2
3
4
5
6
>6
Frequency, f
9
24
35
19
8
3
2
0
 Square the deviations from the mean ( x  x )2 :
x = 2.1
x
f
xf
x  x ( x  x )2
1
0
9
0
-2.1
4.41
2
1
24
24
-1.1
1.21
3
2
35
70
-0.1
0.01
4
3
19
57
0.9
0.81
5
4
8
32
1.9
3.61
6
5
3
15
2.9
8.41
7
6
2
12
3.9
15.21
Total
21
100
210
Example 2 : Method A
The number of children per family, x, for a random selection of 100 families, is given
by the following table:
No. of children, x
0
1
2
3
4
5
6
>6
Frequency, f
9
24
35
19
8
3
2
0
 Find the ‘sum of squares’ Sxx, the sum of ( x  x )2 f :
1
x = 2.1
x
f
xf
xx
0
9
0
-2.1
( x  x )2 ( x  x )2 f
4.41
39.69
2
1
24
24
-1.1
1.21
29.04
3
2
35
70
-0.1
0.01
0.35
4
3
19
57
0.9
0.81
15.39
5
4
8
32
1.9
3.61
28.88
6
5
3
15
2.9
8.41
25.23
7
6
2
12
3.9
15.21 30.42
Total
21
100
210
169
msd & rmsd
Variance &
standard
deviation
 ( x  x )2 f
Example 2 : Method A
Mean square deviation
The number of children per family, x, for a random selection of 100 families, is given
by the following table:
No. of children, x
0
1
2
3
4
5
6
>6
Frequency, f
9
24
35
19
8
3
2
0
 Divide the ‘sum of squares’ Sxx by n :
( x  x )2 ( x  x )2 f
x
f
xf
xx
1
0
9
0
-2.1
4.41
39.69
2
1
24
24
-1.1
1.21
29.04
3
2
35
70
-0.1
0.01
0.35
4
3
19
57
0.9
0.81
15.39
5
4
8
32
1.9
3.61
28.88
6
5
3
15
2.9
8.41
25.23
7
6
2
12
3.9
15.21 30.42
Total
21
100
210
169
Mean square deviation
=
S xx
n
=
169
100
= 1.69
 ( x  x )2 f
=
n
Example 2 : Method A
Root mean square deviation
The number of children per family, x, for a random selection of 100 families, is given
by the following table:
No. of children, x
0
1
2
3
4
5
6
>6
Frequency, f
9
24
35
19
8
3
2
0
 Divide the ‘sum of squares’ Sxx by n and take the square root :
( x  x )2 ( x  x )2 f
x
f
xf
xx
1
0
9
0
-2.1
4.41
39.69
2
1
24
24
-1.1
1.21
29.04
3
2
35
70
-0.1
0.01
0.35
4
3
19
57
0.9
0.81
15.39
5
4
8
32
1.9
3.61
28.88
6
5
3
15
2.9
8.41
25.23
7
6
2
12
3.9
15.21 30.42
Total
21
100
210
Root mean square
deviation
=
=
S xx
=
n
 ( x  x )2 f
n
169
100
= 1.3
169
RETURN
Example 2 : Method A
Variance
The number of children per family, x, for a random selection of 100 families, is given
by the following table:
No. of children, x
0
1
2
3
4
5
6
>6
Frequency, f
9
24
35
19
8
3
2
0
 Divide the ‘sum of squares’ Sxx by n – 1 :
( x  x )2 ( x  x )2 f
x
f
xf
xx
1
0
9
0
-2.1
4.41
39.69
2
1
24
24
-1.1
1.21
29.04
3
2
35
70
-0.1
0.01
0.35
4
3
19
57
0.9
0.81
15.39
5
4
8
32
1.9
3.61
28.88
6
5
3
15
2.9
8.41
25.23
7
6
2
12
3.9
15.21 30.42
Total
21
100
210
169
Variance
=
S xx
 ( x  x )2 f
=
n 1
n 1
=
169
99
= 1.71 (to 3 s.f.)
Example 2 : Method A
Standard deviation
The number of children per family, x, for a random selection of 100 families, is given
by the following table:
No. of children, x
0
1
2
3
4
5
6
>6
Frequency, f
9
24
35
19
8
3
2
0
 Divide the ‘sum of squares’ Sxx by n – 1 and take the square root :
( x  x )2 ( x  x )2 f
x
f
xf
xx
1
0
9
0
-2.1
4.41
39.69
2
1
24
24
-1.1
1.21
29.04
3
2
35
70
-0.1
0.01
0.35
4
3
19
57
0.9
0.81
15.39
5
4
8
32
1.9
3.61
28.88
6
5
3
15
2.9
8.41
25.23
7
6
2
12
3.9
15.21 30.42
Total
21
100
210
Standard deviation s
=
S xx
=
n 1
=
169
99
 ( x  x )2 f
n 1
= 1.31 (to 3 s.f.)
169
RETURN
Example 2 : Method B
The number of children per family, x, for a random selection of 100 families, is given
by the following table:
No. of children, x
0
1
2
3
4
5
6
>6
Frequency, f
9
24
35
19
8
3
2
0
 Tabulate the xf values and find their mean:
n = f
x
f
xf
1
0
9
0
2
1
24
24
3
2
35
70
4
3
19
57
5
4
8
32
xf
n
210
=
100
6
5
3
15
= 2.1
7
6
2
12
Total
21
100
210
The sample mean x
=
xf
Example 2 : Method B
The number of children per family, x, for a random selection of 100 families, is given
by the following table:
No. of children, x
0
1
2
3
4
5
6
>6
Frequency, f
9
24
35
19
8
3
2
0
 Tabulate the squares of the values (x2f) and find the total:
n = f
x
f
xf
x2f
1
0
9
0
0
2
1
24
24
24
3
2
35
70
140
4
3
19
57
171
5
4
8
32
128
6
5
3
15
75
7
6
2
12
72
Total
21
100
210
610
x2f
Example 2 : Method B
The number of children per family, x, for a random selection of 100 families, is given
by the following table:
No. of children, x
0
1
2
3
4
5
6
>6
Frequency, f
9
24
35
19
8
3
2
0
 Find the ‘sum of squares’ by subtracting nx from x2f :
2
x = 2.1
n = f
x
f
xf
x2f
1
0
9
0
0
2
1
24
24
24
3
2
35
70
140
4
3
19
57
171
5
4
8
32
128
6
5
3
15
75
7
6
2
12
72
Total
21
100
210
610
Sxx =
 x 2 f  nx 2
= 610 – 100 2.12
= 169
msd &
rmsd
x2f
Variance &
standard
deviation
Example 2 : Method B
Mean square deviation
The number of children per family, x, for a random selection of 100 families, is given
by the following table:
No. of children, x
0
1
2
3
4
5
6
>6
Frequency, f
9
24
35
19
8
3
2
0
 Divide the ‘sum of squares’ Sxx by n :
x = 2.1
n = f
x
f
xf
x2f
1
0
9
0
0
2
1
24
24
24
3
2
35
70
140
4
3
19
57
171
5
4
8
32
128
6
5
3
15
75
7
6
2
12
72
Total
21
100
210
610
Mean square deviation
=
S xx
n
=
169
100
= 1.69
x2f
=
 x 2 f  nx 2
n
Example 2 : Method B
Root mean square deviation
The number of children per family, x, for a random selection of 100 families, is given
by the following table:
No. of children, x
0
1
2
3
4
5
6
>6
Frequency, f
9
24
35
19
8
3
2
0
 Divide the ‘sum of squares’ Sxx by n and take the square root :
x = 2.1
n = f
x
f
xf
x2f
1
0
9
0
0
2
1
24
24
24
3
2
35
70
140
4
3
19
57
171
5
4
8
32
128
6
5
3
15
75
7
6
2
12
72
Total
21
100
210
610
Root mean square
deviation
=
S xx
n
=
169
100
=
 x 2 f  nx 2
n
= 1.3
x2f
RETURN
Example 2 : Method B
Variance
The number of children per family, x, for a random selection of 100 families, is given
by the following table:
No. of children, x
0
1
2
3
4
5
6
>6
Frequency, f
9
24
35
19
8
3
2
0
 Divide the ‘sum of squares’ Sxx by n – 1 :
x = 2.1
n = f
x
f
xf
x2f
1
0
9
0
0
Variance
2
1
24
24
24
3
2
35
70
140
=
S xx
n 1
4
3
19
57
171
5
4
8
32
128
=
169
99
6
5
3
15
75
7
6
2
12
72
Total
21
100
210
610
=
 x 2 f  nx 2
n 1
= 1.71 (to 3 s.f.)
x2f
Example 2 : Method B
Standard deviation
The number of children per family, x, for a random selection of 100 families, is given
by the following table:
No. of children, x
0
1
2
3
4
5
6
>6
Frequency, f
9
24
35
19
8
3
2
0
 Divide the ‘sum of squares’ Sxx by n – 1 and take the square root :
x = 2.1
n = f
x
f
xf
x2f
1
0
9
0
0
2
1
24
24
24
3
2
35
70
140
4
3
19
57
171
5
4
8
32
128
6
5
3
15
75
7
6
2
12
72
Total
21
100
210
610
Standard deviation s
=
S xx
=
n 1
=
169
99
 x 2 f  nx 2
n 1
= 1.31 (to 3 s.f.)
x2f
RETURN
Various forms of ‘sum of squares’ Sxx
Sxx
Definition
Alternative
version
Raw data
 ( x  x )2
 x 2  nx 2
Frequency
distribution
 ( x  x )2 f
 x 2 f  nx 2
Mean square  S xx
n
deviation
S xx
Variance 
n 1
Root mean square 
deviation (rmsd)
Standard deviation s 
S xx
n
S xx
n 1
RETURN
Notes on using the presentation
The presentation covers calculations using
Raw data (Example 1) or a
Frequency distribution (Example 2).
The ‘sum of squares’ Sxx is evaluated using the
Definition formula (Method A) or the
Alternative formula (Method B).
For each example and each method the ‘sum of squares’ Sxx is used to
calculate the
mean square deviation and root mean square deviation or the
variance and standard deviation
Use the links in the presentation to choose the appropriate example and
method, together with the desired calculations.
RETURN