Answer to Question #349773 in Statistics and Probability for erika miguel

Question

a. r =

=

=

=

=

=

r =

b. There is a _____________________between the self-concept skill and leadership skill of the students.

x y xy x² y²

9.5 9.2

9.2 8.8

6.3 7.3

4.1 3.4

5.4 6.0

8.3 7.8

7.8 8.8

6.8 7.0

5.6 6.5

7.1 8.3

$\sum$= $\sum$= $\sum$= $\sum$= $\sum$=

c)

= a + bx a= ( $\sum$ y)($\sum$ x²) - ($\sum$ x)($\sum$ xy) / n($\sum$ x²) -$\sum$ x²

^{= = = a=}

b= ($\sum$ xy) - ($\sum$ x) ($\sum$ y) / n($\sum$x² ) -($\sum$ x)²

d)= _____________

= ____________(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.

$x=1.5$

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