Question #174983

B. 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.


1
Expert's answer
2021-03-30T07:33:52-0400

1. Let us first determine xi\sum x_i , xi2\sum x^2_i , yi\sum y_i ,yi2\sum y^2_i , xiyi\sum x_iy_i , where the xix_i are the first coordinates of the ordered pairs and yiy_i are the second coordinates of the ordered pairs.


xi=47.3xi2=326.59yi=47.8yi2=345.98xiyi=324.91\sum x_i = 47.3 \\ \sum x^2_i = 326.59 \\ \sum y_i = 47.8 \\ \sum y^2_i = 345.98 \\ \sum x_iy_i = 324.91

n represents the sample size and thus n is equal to the number of ordered pairs.

n = 7

We can then determine the covariance using the formula

sxy=xiyi(xi)(yi)nn1=324.91(47.3)(47.8)771=0.319s_{xy} = \frac{\sum x_iy_i - \frac{(\sum x_i)(\sum y_i)}{n}}{n -1} \\ = \frac{ 324.91 - \frac{(47.3)(47.8)}{7}}{7 -1} \\ = 0.319

Let us next determine the sample variance s2s^2 using the formula:

s2=xi2(xi)2nn1sx2=326.59(47.3)2771=1.1628sy2=345.98(47.8)2771=3.2623s^2 = \frac{\sum x^2_i - \frac{(\sum x_i)^2}{n}}{n-1} \\ s^2_x = \frac{ 326.59 - \frac{(47.3)^2}{7}}{7-1} = 1.1628 \\ s^2_y = \frac{ 345.98 - \frac{(47.8)^2}{7}}{7-1} = 3.2623

The sample standard deviation is the square root of the population sample:

sx=sx2=1.1628=1.0783sy=sy2=3.2623=1.8061s_x = \sqrt{s^2_x} = \sqrt{1.1628} = 1.0783 \\ s_y = \sqrt{s^2_y} = \sqrt{3.2623} = 1.8061

We can then determine the correlation coefficient r using the formula:

r=sxysxsyr=0.3191.0783×1.8061=0.1637r = \frac{s_{xy}}{s_x s_y} \\ r = \frac{0.319}{1.0783 \times 1.8061} = 0.1637

2. The relationship between the variables is weak, positive, linear relationship.

3. Next, we can determine the slope b using the formula:

b=rsysxb=0.1637×1.80611.0783=0.2742b = r\frac{s_y}{s_x} \\ b = 0.1637 \times \frac{1.8061}{1.0783} = 0.2742

Next, we can determine the y-intercept a using the formula a=yˉbxˉa = \bar{y} -b \bar{x} , where the sample mean is the sum of all values divided by the number of values.

a=yˉbxˉ=yinbxina=frac47.870.274247.37=6.82851.8528=4.9757a = \bar{y} -b \bar{x} = \frac{\sum y_i}{n} -b \frac{\sum x_i}{n} \\ a = frac{47.8}{7} -0.2742 \frac{47.3}{7} \\ = 6.8285 -1.8528 = 4.9757

Finally, we then obtain the regression line:

y = a + bx = 4.9757 + 0.2742x

Where x represents the self-concept and y represents the leadership.


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