Answer to Question #175556 in Statistics and Probability for Janine

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

The coefficient of correlation is calculated by the formula:

$$r=\frac{\sum(x_i-\bar x)(y_i-\bar y)}{\sqrt{\sum(x_i-\bar x)^2\sum(y_i-\bar y)^2}}$$

where x is denoted for self-concept and y is denoted for leader

$\bar x =\frac{\sum x_i}{n}=\frac{7.1+5.6+6.8+7.8+8.3+5.4 +6.3}{7}=6.76$

$\bar y =\frac{\sum y_i}{n}=\frac{3.4 +6.0+ 7.8 +8.8 +7.0 +6.5+8.3 }{7}=6.83$

$r=\frac{(7.1-6.76)(3.4-6.83)+(5.6-6.76)(6.0-6.83)+...+(6.3-6.76)(8.3-6.83)}{\sqrt{((7.1-6.76)^2+(5.6-6.76)^2+...+(6.3-6.76)^2)((3.4-6.83)^2+(6.0-6.83)^2+...+(8.3-6.83)^2)}}=0.16$

The correlation is positive and weak.

The regression line is calculated by the formula:

$\hat y=ax+b$

where $a=r\frac{s_y}{s_x}$ and $b=\bar y-a\bar x$

$s_y=\sqrt{\frac{1}{n}\sum(y_i-\bar y)^2}=1.67$

$s_x=\sqrt{\frac{1}{n}\sum(x_i-\bar x)^2}=1$

$a=r\frac{s_y}{s_x} =0.16\cdot\frac{1.67}{1}=0.27$

$b=\bar y-a\bar x=6.83-0.27\cdot6.76=5$

Therefore the regression line is calculated by the formula:

$\hat y = 0.27x+5$