Question

In trying to evaluate the effectiveness of its advertising campaign, a firm completed the following information:
Year : 1 2 3 4 5 6 7 8
Advertising Expenditure (000) : 12 15 15 23 24 38 42 48
Sales (00000) : 5.0 5.6 5.8 7.0 7.2 8.8 9.2 9.5
i) Is the advertising expenditure and sales of this firm related?
ii) Given an advertising expenditures of 60,000, what will the sales turn out to be?
iii) Calculate the 3 yearly moving average of the sales.
iv) Find the mean and standard deviation of a normal distribution when 8% of the items are over 64 and 31% are under 45.

Answer 1

Below is the output results of the regression.

i) r=0.9878 implies that there is a strong relationship between expenditure and sales.

ii)expenditure =60000, is equivalent to 60 in (000)

the relationship between sales and expenditure is;

sales(00000)=3.871+0.125 expenditure(000)

=3.871+0.125(60)

=11.3817

For an expenditure of 60000, the sales associated to it is 1138170.

iii) 3-year moving average

the first point 3 year moving average is;

(5+5.6+5.8)/3=5.466667

the second point is;

(5.6+5.8+7.0)/3=6.133333

the third point is;

(5.8+7.0+7.2)/3=6.666667

the fourth point is;

(7.0+7.2+8.8)/3=7.666667

the fifth point is;

(7.2+8.8+9.2)/3=8.4

the sixth point is;

(8.8+9.2+9.5)/3= 9.166667

so the 3 point averages are

5.466667 , 6.133333 , 6.666667 , 7.666667 , 8.4 , 9.166667

iv) P(z<45)=0.31 and P(z>64)=0.08

the corresponding z values from the table is

$\phi^{-1}(0.31)=-0.5$

$\phi^{-1} (0.08)=1.41$

$z=\frac{x-\mu}{\sigma}$

-0.5=$\frac{45-\mu}{\sigma}$

$-0.5\sigma=45-\mu$

$-0.5\sigma+\mu=45$ .....(i)

$1.41=\frac{64-\mu}{\sigma}$

$1.41\sigma=64-\mu$

$1.41\sigma+\mu=64$ ...(ii)

solving equations (i) and (ii) simultaneously gives;

$\sigma\approx 10$

$\mu \approx 50$

The mean and standard deviation are 50 and 10 respectively.