BUS 207: Assignment 6 a) The independent variable for this data

BUS 207: Assignment 6
a) The independent variable for this data is ‘Advertising Expenses’.
b) The dependent variable for this data is ‘Sales’.
c)
$0 $500 $1,000 $1,500 $2,000 $2,500 $3,000 $3,500 $4,000
$0
$20,000
$40,000
$60,000
$80,000
$100,000
$120,000
$140,000 Sales Versus Advertising expenses
Advertising Expenditure
Sales
d)
$0 $500 $1,000 $1,500 $2,000 $2,500 $3,000 $3,500 $4,000
$0
$20,000
$40,000
$60,000
$80,000
$100,000
$120,000
$140,000
f(x) = 35.1667158221807 x + 6291.6576831365
R² = 0.885842411774408
Sales Versus Advertising expenses
Advertising Expenditure
Sales
e) Correlation coefficient , r = 0.9412 ; Coefficient of determination, R2 = 0.8858 ;
Slope of the regression line, b = 35.17 ; y-intercept of regression line, a = 6291.66
f) The value of the slope shows that there is a $35.17 increase in sales for every dollar
increase in advertising expenses.
g) The coefficient of correlation value, 0.9412, indicate that the two variables,
advertising expenses and sales, have strong and positive statistical relationship.
h) The coefficient of determination value, 0.8858, indicate that advertising expenses
explain 88.58% of the variation in sales.
i) Yes, at 0.01 level of significance, we can conclude that there is a positive
correlation between the advertising expenditure an sales for Woodbon. This is
because the ANOVA significance level of the linear regression analyses is less that
0.01.
ANOVA
df SS MS F Significance
F

Regression 1 17713801540.06 17713801540.06 201.76 0.00
Residual 26 2282759140.05 87798428.46
Total 27 19996560680.11
j) Yes, the amount spent on adverting can be use to effectively predict the sales for
Woodbon within a given year. Statistically, advertising expenses an sales have been
found to be strongly related with the amount spent on advertising explains 88.58% of
the change in sales. Therefore, the linear regression equation,
will effectively predict the sales for
Woodbon.
k) If the company spends $2,600 on advertising, the predicted annual sales is:
l) The standard error of the equation is ±9370.08, therefore the range of the predicted
sales in (k) is $88,363.58 to $107,103.74
m) I believe it is reasonable to use this regression equation to predict the sales for a
year in which $50,000 is spent on advertising because the equation is not limited to
the amount spend on advertising.