The president of a company was interested in determining whether there is a corr

Question

The president of a company was interested in determining whether there is a correlation between sales made by different sales teams and hours spent on employee training. These figures are shown.

Sales
(in thousands)

Training
Hours

Compute the correlation coefficient for the data. (Round your answer to 4 decimal places, the tolerance is +/-0.0001.)
The correlation coefficient is

What is your interpretation of this value? (Do not round your intermediate computations to answer this question.)
There is

linear association between sales and training hours.

Using the data, what would you expect sales to be if training was increased to eighteen hours? Use the linear regression model. (Round your answer to 2 decimal places, the tolerance is +/-0.01.)
Sales =

(in thousands)

Sales
(in thousands)

Training
Hours

12 8 30 11 26 11 44 15 9 5

Explanation / Answer

calculation procedure for correlation

sum of (x) = x = 121

sum of (y) = y = 50

sum of (x^2)= x^2 = 3737

sum of (y^2)= y^2 = 556

sum of (x*y)= x*y = 1417

to caluclate value of r( x,y) = covariance ( x,y ) / sd (x) * sd (y)

covariance ( x,y ) = [ x*y - N *(x/N) * (y/N) ]/n-1

= 1417 - [ 5 * (121/5) * (50/5) ]/5- 1

= 41.4

and now to calculate r( x,y) = 41.4/ (SQRT(1/5*1417-(1/5*121)^2) ) * ( SQRT(1/5*1417-(1/5*50)^2)

=41.4 / (12.7185*3.3466)

=0.9726

value of correlation is =0.9726

coeffcient of determination = r^2 = 0.946

properties of correlation

1. If r = 1 Corrlation is called Perfect Positive Corrlelation

2. If r = -1 Correlation is called Perfect Negative Correlation

3. If r = 0 Correlation is called Zero Correlation

& with above we conclude that correlation ( r ) is = 0.9726> 0 ,strongpositive correlation

Line of Regression Y on X i.e Y = bo + b1 X

calculation procedure for regression

mean of X = X / n = 24.2

mean of Y = Y / n = 10

(Xi - Mean)^2 = 808.8

(Yi - Mean)^2 = 56

(Xi-Mean)*(Yi-Mean) = 207

b1 = (Xi-Mean)*(Yi-Mean) / (Xi - Mean)^2

= 207 / 808.8

= 0.25593472

bo = Y / n - b1 * X / n

bo = 10 - 0.25593472*24.2 = 3.80637982

value of regression equation is, Y = bo + b1 X

Y'=3.80637982+0.25593472* X

expect sales to be if training was increased to eighteen hours,

Y'=3.80637982+0.25593472* (8)

=5.85385758

( X) ( Y) X^2 Y^2 X*Y 12 8 144 64 96 30 11 900 121 330 26 11 676 121 286 44 15 1936 225 660 9 5 81 25 45

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