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
14	6
33	12
20	11
44	15
8	5

1. 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 a strong positive linear association between sales and training hours.
   
2. 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)

Answer 1

The table given below ,

Training Hours (X)	Sales(Y)	X^2	Y^2	XY
6	14	36	196	84
12	33	144	1089	396
11	20	121	400	220
15	44	225	1936	660
5	8	25	64	40
X = 49	Σ Υ = 119	$\sum X^2=551$	$\sum Y^2=3685$	$\sum XY=1400$

From table ,

$\bar{X}=\frac{\sum X}{n}=\frac{49}{5}=9.8$

$\bar{Y}=\frac{\sum Y}{n}=\frac{119}{5}=23.8$

$\sigma^2_{X}=\frac{\sum X^2}{n}-\bar{X}^2=\frac{551}{5}-9.8^2=14.16$

$\sigma^2_{Y}=\frac{\sum Y^2}{n}-\bar{Y}^2=\frac{3685}{5}-23.8^2=170.56$

$Cov(X,Y)=\frac{\sum XY}{n}-\bar{X}*\bar{Y}=\frac{1400}{5}-9.8*23.8=46.76$

Therefore ,

a) The correlation coefficient for the data is ,

$r=\frac{Cov(X,Y)}{\sqrt{\sigma_{X}^2*\sigma_{Y}^2}}=\frac{46.76}{\sqrt{14.16*170.56}}=0.9515$

Interpretation : Here , the value of the correlation coefficient is positive and close to 1.

Therefore , there is strong positive linear association between training hours and sales.

b) The regression coefficient is ,

$b=\frac{cov(X,Y)}{\sigma_{X}^2}=\frac{46.76}{14.16}=3.3023$

The intercept of the regression equation is ,

$a=\bar{Y}-b*\bar{X}=23.8-3.3023*9.8=-8.5621$

Therefore , the linear regression equation is ,

$\hat{y}=a+bx=-8.5621+3.3023x$

Now , for x=8 hours

$\therefore \hat{y}=-8.5621+3.3023x=-8.5612+3.3023*18=50.88$

Therefore , if the training was increased to 18 hours , then the expected sale is 50.88 in thosands.