Question

Starts in inches and weighed 7 pounds for a sample National Football League quarterbacks quarterback

X. 77. 76. 77. 72. 76. 77

Y. 248. 225. 241. 209. 225. 243

Compute correlation coefficient.

Answer 1

( X)	( Y)	X^2	Y^2	X*Y
77	248	5929	61504	19096
76	225	5776	50625	17100
77	241	5929	58081	18557
72	209	5184	43681	15048
76	225	5776	50625	17100
77	243	5929	59049	18711

calculation procedure for correlation

sum of (x) = 455

sum of (y) = 1391

sum of (x^2) = 34523

sum of (y^2) = 323565

sum of (x*y) = 105612

to caluclate value of r( x,y) = covariance ( x,y ) / sd (x) * sd (y)

covariance ( x,y ) = [ sum (x*y - N *(sum (x/N) * (sum (y/N) ]/n-1

= 105612 - [ 6 * (455/6) * (1391/6) ]/6- 1

= 21.3056

and now to calculate r( x,y) = 21.3056/ (SQRT(1/6*105612-(1/6*455)^2) ) * ( SQRT(1/6*105612-(1/6*1391)^2)

=21.3056 / (1.7717*13.4464)

=0.8943

value of correlation is =0.8943

coeffcient of determination = r^2 = 0.7998

properties of correlation

1. If r = 1 Corrlation is called Perfect Positive Corrlelation

2. If r = -1 Correlation is called Perfect Negative Correlation

3. If r = 0 Correlation is called Zero Correlation

& with above we conclude that correlation ( r ) is = 0.8942> 0 ,perfect positive correlation