Standard Error of estimate

Standard error of prediction

Y on X X on Y

  y  yp    x  xp 
2 2

E yx  Exy 
n n

y p  predict valueof y x p  predict value of x

Example : Find the standard error of estimate of y on x by following data

X 1 2 3 4 5
Y 2 5 3 8 7
Solution :

Line of regression of y on x is given by

y = a + bx …(1)

a and b find by formula

Σy = na + b Σx …(2)

Σxy = aΣx + bΣx2 …(3)

xy y p  1.1  1.3x y y 

2
x y x2 p

1 2 1 2 2.4 0.16
2 5 4 10 3.7 1.69
3 3 9 9 5 4
4 8 16 32 6.3 2.89
5 7 25 35 7.6 0.36
 x  15  y  25  x 2
 55  xy  88  y  y  p
2
 9.10

n=5
Putting all value from the table in equation (2) and (3) we get
25 = 5a + 15b
88 = 15a + 55b
On solving a = 1.1 and b = 1.3
Put in equation (1) the line of regression y on x is
y = 1.1 + 1.3x …(4)
 y  y 
2
Now find yp by equation (4) and calculate p

the standard error of estimate of y on x

  y  yp 
2
9.10
E yx    1.349
n 5
Standard Deviation
If  2 is the variance, then , is called the standard deviation, is given by

 X  X   x     x 
2 2

(1) raw data     

n n  n 
(2) when the ungrouped data is given

  f  X  X  
  fx     fx 
2
2 2

  
 
N N  N 

  fd     fd 
2 2

  
N  N 
(3) when the grouped data is given in class interval

  fu     fu 
2 2
X A
 2
   i u
N  N  h

Find standard deviation for following data

Example 1: 57, 64, 43, 67, 49, 59, 44, 47, 61, 59 Mean

57  64  43  67  49  59  44  47  61  59 550
Solution : Mean X    55
10 10

X X  X  X  55 X  X 
2

57 2 4
64 9 81
43 -12 144
67 12 144
49 -6 36
59 4 16
44 -11 121
47 -8 64
61 6 36
59 4 16
Total 662

 X  X 
2
662
standard deviation     8.13
n 10
Example 2 : 8, 9, 15, 23, 5, 11, 19, 8, 10, 12
Solution :

X X2
8 64
9 81
15 225
23 529
5 25
11 121

19 361

8 64

10 100

12 144

120 1714
Total


 x     x 
2

standard deviation  
n  n 

1714  120 
    27.4  5.23
10  10 
Example 3:

Solution :

X f fx X  X  X 8 X  X  f X  X 
2 2

1 3 3 -7 49 147
3 3 9 -5 25 75
5 4 20 -3 9 36
7 14 98 -1 1 14
9 7 63 1 1 7
11 4 44 3 9 36

13 3 39 5 25 75

15 4 60 7 49 196

Total 42 336 585

X
 fx  336  8
N 42

  f  X  X  
2

   585  3.732
N 42
Example 4:

X 10 11 12 13 14
f 3 12 18 12 3
Solution :

X f fx fx 2
10 3 30 300
11 12 132 1452
12 18 216 2592
13 12 156 2028
14 3 42 588
Total 48 576 6960

  fx     fx 
2 2

standard deviation    
N  N 

2
6960  576 
    1 1
48  48 
Example 5:

Class 4-8 8-12 12-16 16-20

Interval
f 3 6 4 7
Solution :

Class f Mid-Value fx X  X  X  13 X  X  f X  X 
2 2

Interval X
4-8 3 6 18 -7 49 147
8-12 6 10 60 -3 9 54
12-16 4 14 56 1 1 4
16-20 7 18 126 5 25 175
Total 20 260 380

X
 fx  260  13
N 20

  f  X  X  
2

   380  4.36
N 20
Example 6 :

Class Interval 0-30 30-60 60-90 90-120 120-150 150-180 180-210

f 9 17 43 82 81 44 24

Solution :

Class f Mid-value X  105 fu f u2

Interval u
X 30
0-30 9 15 -3 -27 81
30-60 17 45 -2 -34 68
60-90 43 75 -1 -43 43
90-120 82 105 0 0 0
120-150 81 135 1 81 81
150-180 44 165 2 88 176

180-210 24 195 3 72 216

Total 300 137 665

  fu     fu 
2 2

standard deviation      i
N  N 
665  137 
    30
300  300 
 42.5
 Exercise Find standard deviation for following data

Q.1

Q.2