Question

Survey tests on seIf-concept and on leadership skill were administered to student-leaders. Both tests use a 10-point Likert scale with 10 indicating the highest scores for each test. Scores for students on the tests follow:

Student | A | B | C | D | E | F | G |

Self -Concept | 7.1 | 5.6 | 6.8 | 7.8 | 8.3 | 5.4 | 6.3 |

Leader | 3.4 | 6.0 | 7.8 | 8.8 | 7.0 | 6.5 | 8.3 |

-ship Skill

1. Compute the coeficient of correlation r.

2. Interpret the results in terms of strength and direction of correlation.

3. Find the regression line that will predict the leadership skill if the self-concept score is known

Answer 1

Solution:

(1): Let X denotes the random variable of Self -Concept, and Y denotes leadership skill.

X = {7.1 , 5.6 , 6.8 , 7.8 , 8.3 , 5.4 , 6.3}

Y = {3.4 , 6.0 , 7.8 , 8.8 , 7.0 , 6.5 , 8.3}

Mean of X values$=M_X=\dfrac{47.3}{7}=6.757$

Mean of Y values$=M_Y=\dfrac{47.8}{7}=6.829$

Now, $\Sigma(X-M_X)^2 =SS_X=6.977$

$\Sigma(Y-M_Y)^2 =SS_Y=19.574$

And, sum of products, $\Sigma(X-M_X)(Y-M_Y) =SP=1.919$

We know that coefficient of correlation, $r=\dfrac{\Sigma(X-M_X)(Y-M_Y)}{\sqrt{SS_X\times SS_Y}}$

$\Rightarrow r=\dfrac{1.919}{\sqrt{6.977\times19.574}}\approx0.1642$

(2): We know that the nearer the value is to zero, the weaker the relationship between the variables. Although technically we get a positive correlation here (0.1642), the relationship between these variables is weak.

(3):

$b=\dfrac{SP}{SS_X}=\dfrac{1.919}{6.977}=0.27498$

$a=M_Y-bM_X=6.829-0.27498(6.757)=4.9705$

Regression equation: $\hat{y}=bX+a$

Putting values,

$\hat y=0.27498X + 4.9705$