Page 302 - DMTH404_STATISTICS

Statistics

Notes The values of different terms, given in the formula, are calculated from the following table :

X Y X Y X 2 Y 2
i i i i i i
12 14 168 144 196
9 8 72 81 64
8 6 48 64 36
10 9 90 100 81
11 11 121 121 121
13 12 156 169 144
7 3 21 49 9
70 63 676 728 651

Here n = 7 (no. of pairs of observations)

7 676 70 63



r XY   0.949
7 728    2 7 651    2
63
70


Example 2: Calculate the Karl Pearson's coefficient of correlation between X and Y from
the following data:
2 2


No. of pairs of observations n = 8,  X  X  = 184,  Y  Y  =148,
i
i
  X  X  Y  Y  = 164, X =11 and Y =10

i
i
Solution.
  X X  Y  Y 
r  i i
Using the formula, XY 2 2 , we get
  X  X    Y  Y 
i
i
164
r   0.99
XY
184 148
Example 3:
(a) The covariance between the length and weight of five items is 6 and their standard
deviations are 2.45 and 2.61 respectively. Find the coefficient of correlation between length
and weight.

(b) The Karl Pearson's coefficient of correlation and covariance between two variables X and
Y is – 0.85 and – 15 respectively. If variance of Y is 9, find the standard deviation of X.
Solution.

(a) Substituting the given values in formula (1) for correlation, we get
6
r XY   0.94
2.45 2.61


294 LOVELY PROFESSIONAL UNIVERSITY