Question

The owners of an e-business have been successful in selling fashion products but are now venturing into another domain. Knowing that the impact of advertising on profit cannot be overemphasized, they are interested in determining the right amount to allocate to advertising for the new business. Based on a monthly report from the fashion e-business, a regression analysis of monthly profit (in thousands of dollars) on advertising spending (in hundreds of dollars) produced the following results:

slope	yy-intercept	rr
1.28	2.488	0.7379

where yy = profit (in $1000s)
           xx = advertising spending (in $100s)

a. State the least-squares regression line for the data.
ŷ = ŷ =

++

xx

b. Interpret the value of the slope as it relates to this problem.

For every $1 increase in advertising spending, there is a $1.213 increase in profit.

For every $100 increase in advertising spending, there is a $1,213 increase in profit.

For every $100 increase in advertising spending, there is a $121.3 increase in profit.

For every $1,000 increase in advertising spending, there is a $121.3 increase in profit.

c. Compute and interpret the coefficient of determination.
R2=R2=

Round to 4 decimal places

d. Predict the monthly profit for a month when advertising is $2,300.

Round to the nearest cent

e. If the expected profit in a particular month is $43,448, about how much should be set aside for advertising that month?

Round to the nearest cent

Answer 1

Y= monthly profit (in thousands of dollars) on x= advertising spending (in hundreds of dollars)

a) The regression equation is

y = 1.213 ( x ) + 2.488

b)  Interpret the value of the slope

The slope tells that the in one unit increase in variable related to slope, how much unit will increase the response variable.

Here For every $1 increase in advertising spending, there is a $1.213 increase in profit.

Answer:- For every $1 increase in advertising spending, there is a $1.213 increase in profit.

c) Compute and interpret the coefficient of determination.

The coefficient of determination.is R² = (r) ^2 = ( 0.7379) ^2 = 0.5445

54.45% of the variation in the response variable ( i.e expected profit) is explained by the linear regression model.

d) Predict the monthly profit for a month when advertising is $2,300. so in model x = advertising spending (in $100s)

so x = 2300 / 100 = 23

x= 23

y = 1.213 ( x ) + 2.488

y = 1.213 ( 23) + 2.488

= 27.90+2.488

y=$ 30.387

as y = profit (in $1000s) = 30.387 * 1000 = $30387

$30387 monthly profit for a month when advertising is $2,300.

e)

If the expected profit in a particular month is $43,448, about how much should be set aside for advertising that month?

y = 43448 , x= ?

y = profit (in $1000s) = 43448 / 1000 = 43.448

y = 1.213 ( x ) + 2.488

43.448 = 1.213 ( x ) + 2.488

43.448 -2.488 = 1.213(x)

40.96 =1.213(x)

x = 40.96 / 1.213

x = 33.7675

So x = advertising spending (in $100s) = 33.7675*100 = 3376.75

expected profit in a particular month is $43,448, about $ 3376.75 should be set aside for advertising that month.