Answer to Question #279077 in Statistics and Probability for yaya

Question

Eleven secretaries at a university in Kuala Lumpur were asked to take a special three-day intensive course to improve their keyboarding skills. At the beginning and again at the end of the course, they were given a particular two-page letter and asked to type it flawlessly. The data from the typing tests are shown in the following Table 3.

Table 3

Secretary

Number of years of experience x

Improvement

(words per minute)

5

140

3

120

c

2

80

4

100

5

130

4

90

4

110

5

120

6

130

5

130

4

100

a) Compute the correlation coefficient between the number of years of experience and the improvement.        (5 Marks)

b) Develop the regression equation for the improvement based on number of years of experience. (5 Marks)

Answer 1

a) Correlation coefficient

Following formula is used to compute the correlation coefficient between the number of years of experience and the improvement:

$r=\frac{n(\Sigma xy)-(\Sigma x)(\Sigma y)}{\sqrt{[n\Sigma x^{2}-(\Sigma x)^2][n\Sigma y^{2}-(\Sigma y)^2]}}$

The table below shows the calculation:

$r=\frac{11(5500)-(47)(1250)}{\sqrt{[11(213)-(47)^2][11(14500)-(1250)^2]}}$

$r=0.75$

b) Regression equation

The regression equation for the improvement based on number of years of experience is expressed as:

$y=a+bx$

where;

$b=\frac{n(\Sigma xy)-(\Sigma x)(\Sigma y)}{[n\Sigma x^{2}-(\Sigma x)^2]}$

$b=\frac{11(5500)-(47)(1250)}{[11(213)-(47)^2]}=13.06$

$a=\bar{y}-(b\bar{x})$

$a=113.64-(13.06\times4.27)=57.87$

Then, regression equation is:

$y=57.87+13.06x$