Question

2. The owner of Showtime Movie Theaters would like to estimate weekly gross revenue as a function of advertising expenditures. Historical data for a sample of eight weeks are as below: TV Advertising $1000s 5.0 2.0 4.0 2.5 3.0 3.5 2.5 3.0 Week Weekly gross revenue $1000s 96 90 95 92 95 94 94 94 Newspaper Advertising $1000s 2.0 2.5 3.3 2.3 4.2 2.5 4 6 Following are the regression results for the data using Excel. In this problem, you will be interpreting the regression results. (For Practice, you may want to see if you can replicate these results using the data above in Excel.) (8 Points) SUMMARY OUTPUT ression Statistics Multiple R 0.958663 0.919036 0.88665 0.642587 Adjusted R Square Standard Error Observations ANOVA nificance F MS Regression Residual Total 2 23.4354 7177 28.3777684 0.00186524 5 2.064592 0.412918 25.5 990% Coefficientsandard Em t Stat P-value Lower 95% , 95% Lower 99.0% 83.23009 1.573869 52.88248 4.5717E-08 79.1843328 87.27585 76.88402704 89.57615634 Television Advertising ( 2.290184 0.304065 7.531899 0.00065323 1.5085608 3.0106 1064151855 3.516215387 Advertising 1.300989 0.320702 4.056697 0.0097608 0.4765994 2.125379 0.007874404 2.594103793 Ne a. Write the estimated regression equation. b. Interpret the coefficient of TV Advertising. (Hint: Pay attention to the units of measurement) c. Interpret the coefficient of Newspaper Advertising. d. What is the estimate of the weekly gross revenue for a week when S3500 is spent on TV advertising and $1800 is spent on newspaper advertising? e. Which variables have a statistically significant relationship with the dependent variable at the 190 level? f. What is the 99% confidence interval estimate for the coefficient of TV Advertising? g. What would be average weekly gross revenues if nothing is spent on TV and newspaper advertising? h. At what level (probability) would the relationship between the dependent variable and the Newspaper Advertising will NOT be statistically significant?

Answer 1

Y be Weekly gross revenue ( in $1000's )

X_1 be TV advertising ( in $1000's )

X_2 be Newspaper advertising ( in $1000's )

a)

Y = 83.23009 + 2.290184*X_1 + 1.300383*X_2

b)

2.290184 is the coefficient for ( $1000's) X_1

2.290184*1000 = $2290.184 increase in TV advertisement will increase weekly sales by $1000

c)

1.300383 is the coefficient for ( $1000's) X_2

1.300383*1000 = $1300.383 increase in TV advertisement will increase weekly sales by $1000

d)

spent on TV = $3500 = 0.35 ($1000's) , X_1 = 0.35

spent on newspaper = $1800 = 0.18 ($1000's) , X_2 = 0.18

Y = 83.23009 + 2.290184*X_1 + 1.300383*X_2 = Y = 83.23009 + 2.290184*0.35 + 1.300383*0.18 = 83.23009 + 0.8015644 + 0.23406894 = 84.262346 ($1000's)

Weekly Sales is $84262.346

e)

since, p value for both the variables is less than 0.01, therefore, both are significant at 1 % level of significance

f)

99% confidence interval for coefficient for TV advertising is ( 1.064151855, 3.516215387‬ )

g)

If nothing is spent on TV and newspaper advertising, X_1 = X_2 = 0

Y = 83.23009 + 2.290184*X_1 + 1.300383*X_2 = Y = 83.23009 + 2.290184*0 + 1.300383*0 = 83.23009

weekly gross revenue = $83230.09

h)

p value for newspaper advertising is :

p = 0.0097608

it will not be statistical siginificant if it is greater than alpha

at alpha = 0.009 it will not be statistical siginificant

level of significance = 0.9%