Question

The following data lists the grades of 6 students selected at random: Mathematics grade: (70, 92, 80, 74, 65, 85) English grade: (69, 88, 75, 80, 78, 90)

a). Find the regression line.

b). Compute and interpret the correlation coefficient.

Answer 1

x	y	x^2	y^2	xy
70	69	4900	4761	4830
92	88	8464	7744	8096
80	75	6400	5625	6000
74	80	5476	6400	5920
65	78	4225	6084	5070
85	90	7225	8100	7650

sum x	sum y	sum x^2	sum y^2	sum xy	n
466.0000	480.0000	36690.0000	38714.0000	37566.0000	6.0000

A)

Slope:

b1={n*sum(xy)- sum(x)*sum(y)}/{n*sum(x^2 )- [sum(x)]^2 }

b1={6*37566 - 466*480}/{6*36690 - [466]^2 }

b1=0.5751

Intercept:

={sum(y)*sum(x^2 )- sum(x)*sum(xy)}/{n*sum(X^2 )- [sum(X)]^2 }

={480*36690 - 466*37566}/{6*36690 - [466]^2 }

=35.3365        

Regression equation:

y = bo + b1*x

y = 35.3365 + 0.5751*x          

           

B)

Correlation Coefficient

r={n*sum(XY)- sum(X)*sum(Y)}/{ sqrt(n*sum(x^2 )-[sum(x)]^2 )* sqrt(n*sum(y^2 )-[sum(y)]^2 )}

r={6*37566-466*480}/{ sqrt(6*36690-[466]^2 )* sqrt(6*38714-[480]^2 )}

r=0.7237

The value of correlation Coefficient being 0.72 indicates that there is a strong positive linear relationship between the two variables.