Answer to Question #196063 in Statistics and Probability for Ben

Question

Relationship between X and Y. To test this relationship, a research team decides to conduct a study using a sample of tertiary students.

 x= 7,9,9,2,8,5,7,6,8,10

y= 7,6,6,3,9,7,6,7,6,9

 

 

a.    Use the data supplied in a table to manually calculate the correlation coefficient (r). Show all your calculations. (14 Marks)

 

b.    What does the ‘correlation coefficient’ that you calculated tell you about the direction and strength (magnitude) of the relationship between information and readiness? (3 Marks)

 

c.    Calculate and interpret the coefficient of determination for this relationship. (3 marks)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Answer 1

a.

$r=\frac{cov_{xy}}{\sigma_x\sigma_y}$

$\mu_x=\frac{7+9+9+2+8+5+7+6+8+10}{10}=7.1$

$\mu_y=\frac{7+6+6+3+9+7+6+7+6+9}{10}=6.6$

$\sigma_x=\sqrt{\frac{\sum (x_i-\mu_x)^2}{N}}=\sqrt{\frac{0.1^2+1.9^2+1.9^2+5.1^2+0.9^2+2.1^2+0.1^2+1.1^2+0.9^2+2.9^2}{10}}=2.21$

$\sigma_y=\sqrt{\frac{\sum (y_i-\mu_y)^2}{N}}=\sqrt{\frac{0.4^2+0.6^2+0.6^2+3.6^2+2.4^2+0.4^2+0.6^2+0.4^2+0.6^2+2.4^2}{10}}=1.62$

$cov_{xy}=\frac{\sum (x_i-\mu_x)(y_i-\mu_y)}{N-1}=$

$=\frac{-0.1\cdot0.4-1.9\cdot0.6-1.9\cdot0.6+5.1\cdot3.6+0.9\cdot2.4-2.1\cdot0.4+0.1\cdot0.6-1.1\cdot0.4-0.9\cdot0.6+2.9\cdot2.4}{9}=2.6$

$r=\frac{2.6}{2.21\cdot1.62}=0.726$

b.

The correlation coefficient tells us that relationship between two variables is not strong (r<0.8), and that two variables move in the same direction (r>0).

c.

The coefficient of determination:

$R=r^2=0.726^2=0.527$

This value suggests that 52.7% of the dependent variable is predicted by the independent variable.