Answer to Question #349774 in Statistics and Probability for erika miguel

Question

Survey tests on leadership skills and self-concept were administered to student-leaders.

Both tests use a 10-point Likert Scale, with 10 indicating the highest scores for each test.

Scores for the student-leaders on the tests follow:

Student Code A B C D E F G H I J

Self-concept 9.5 9.2 6.3 4.1 5.4 8.3 7.8 6.8 5.6 7.1

Leadership Skill 9.2 8.8 7.3 3.4 6.0 7.8 8.8 7.0 6.5 8.3

a. Compute the correlation coefficient r.

b. Interpret the results in terms of (a) strength and (b) direction of correlation.

c. Find the regression line that will predict the leadership skill if the self-

concept score is known.

d. Predict the leadership skill of a student leader whose self-concept skill is

1.5.

Answer 1

In order to compute the regression coefficients, the following table needs to be used:

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & X & Y & XY & X^2 & Y^2 \\ \hline & 9.5 & 9.2 & 87.4 & 90.25 & 84.64 \\ \hdashline & 9.2 & 8.8 & 80.96 & 84.64 & 77.44 \\ \hdashline & 6.3 & 7.3 & 45.99 & 39.69 & 53.29 \\ \hdashline & 4.1 & 3.4 & 13.94 & 16.81 & 11.56 \\ \hdashline & 5.4 & 6.0 & 32.4 & 29.16 & 36 \\ \hdashline & 8.3 & 7.8 & 64.74 & 68.89 & 60.84 \\ \hdashline & 7.8 & 8.8 & 68.64 & 60.84 & 77.44 \\ \hdashline & 6.8 & 7.0 & 47.6 & 46.24 & 49 \\ \hdashline & 5.6 & 6.5 & 36.4 & 31.36 & 42.25 \\ \hdashline & 7.1 & 8.3 & 58.93 & 50.41 & 68.89 \\ \hdashline Sum= & 70.1 & 73.1 & 537 & 518.29 & 561.35 \\ \hdashline \end{array}$$

$$\bar{X}=\dfrac{1}{n}\sum _{i}X_i=\dfrac{70.1}{10}=7.01$$

$$\bar{Y}=\dfrac{1}{n}\sum _{i}Y_i=\dfrac{73.1}{10}=7.31$$

$$SS_{XX}=\sum_iX_i^2-\dfrac{1}{n}(\sum _{i}X_i)^2$$ $$=518.29-\dfrac{70.1^2}{10}=26.889$$

$$SS_{YY}=\sum_iY_i^2-\dfrac{1}{n}(\sum _{i}Y_i)^2$$ $$=561.35-\dfrac{(73.1)^2}{10}=26.989$$

$$SS_{XY}=\sum_iX_iY_i-\dfrac{1}{n}(\sum _{i}X_i)(\sum _{i}Y_i)$$ $$=537-\dfrac{70.1(73.1)}{10}=24.569$$

a.

$$r=\dfrac{SS_{XY}}{\sqrt{SS_{XX}SS_{YY}}}=\dfrac{24.569}{\sqrt{26.889(26.989)}}$$

$$=0.912025$$

b. There is a strong positive correlation between the self-concept skill and leadership skill of the students.

c.

$$m=slope=\dfrac{SS_{XY}}{SS_{XX}}=\dfrac{24.569}{26.889}=0.9137$$ $$n=\bar{Y}-m\bar{X}=7.31-\dfrac{24.569}{26.889}(7.01)=0.9048$$

The regression equation is:

$$y=0.9048+0.9137x$$

d.

$x=1.5$

$$y=0.9048+0.9137(1.5)=2.275$$