Question

A manager wishes to determine the relationship between the number of miles traveled​ (in hundreds of​ miles) by her sales representatives and their amount of sales​ (in thousands of​ dollars) per month. Find the equation of the regression line for the given data. What would be the predicted sales if the sales representative traveled 0​ miles? Is this​ reasonable? Why or why​ not? Round the predicted sales to the nearest dollar. Round the regression line values to the nearest hundredth.

Answer 1

Answer:

$y=3.53 x +37.92$  

And for 0 miles if we use the model we got:

$y(0)= 3.53*0 +37.92 = 37.92$

But for this case. No is not reasonable for a representative to travel o miles and have a positive amount of sales.

Step-by-step explanation:

For this case we assume the following data:

Miles traveled (X): 2,3,10,7,8,15,3,1,11

Sales (Y): 31,33,78,62,65,61,48,55,120

We want to found a linear model like this one : $y = mx+b$. Where:

$m=\frac{S_{xy}}{S_{xx}}$  

Where:  

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}$  

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}$  

With these we can find the sums:  

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=582-\frac{60^2}{9}=182$  

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}=4329-\frac{60*553}{9}=642.33$  

And the slope would be:  

$m=\frac{642.33}{182}=3.529$  

Nowe we can find the means for x and y like this:  

$\bar x= \frac{\sum x_i}{n}=\frac{60}{9}=6.667$  

$\bar y= \frac{\sum y_i}{n}=\frac{553}{9}=61.444$  

And we can find the intercept using this:  

$b=\bar y -m \bar x=61.444-(3.529*6.667)=37.917$  

So the line would be given by:  

$y=3.53 x +37.92$  

And for 0 miles if we use the model we got:

$y(0)= 3.53*0 +37.92 = 37.92$

But for this case. No is not reasonable for a representative to travel o miles and have a positive amount of sales.