You are an owner of Carrefour supermarket. You have made feature advertisings fo
ID: 3374933 • Letter: Y
Question
You are an owner of Carrefour supermarket. You have made feature advertisings for last three years. You want to know the effectiveness of this feature advertising on store traffic(numbers of shoppers) in different week. In data set, you have: average numbers of shoppers, average numbers of feature advertising, and average price each week.
With the tables bellow it was done a REGRESSION model in Excel and you should interpret the results obtained from the equation based on the questions.
Q1. Consider a regression model (Model I) that has feature advertising as a single independent variable with intercept. Estimate your model and interpret your estimation results.
Q2. Update above regression model (Model II) by adding an additional independent variable. average price in order to capture the effect of price promotion activities such as coupon during week. Estimate your model and interpret your estimation results. Do you think which model makes more sense between Model I and Model II? Why?
PS: It was done a ANOVA REGRESSION model in Excel WITH THAT DATA and you should interpret the results obtained from the equation based on the questions. IF YOU DON'T KNOW HOW TO interpret ANOVA regression results , PLEASE DON'T ANSWER.
THANK YOU.
Regression Statistics Multiple R 0,53969388 R Square 0,29126948 Adjusted R Square 0,28666733 Standard Error 205,827509 Observations 156 ANOVA df SS MS F Significance F Regression 1 2681275,289 2681275,3 63,28992304 3,59619E-13 Residual 154 6524204,403 42364,964 Total 155 9205479,692 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95,0% Upper 95,0% Intercept 609,046483 19,25202773 31,635446 2,91178E-69 571,0143324 647,0786342 571,0143324 647,0786342 feature 9,48205612 1,191887424 7,9554964 3,59619E-13 7,127496744 11,83661549 7,127496744 11,83661549Explanation / Answer
Model: shoppers = 609.046483 + 9.48205612*feature
Ho: the model is not significant. V/s h1: the model is significant. with (F=63.28992304, P<5%), The null hypothesis is rejected at 5% level of significance and conclude that model is significant.
With one unit increase in an average number of feature advertising, there is 9.48205612 units increase in the average number of Shoppers. This value is significant with (t=7.9554964, p<5%).
The coefficient of determination, R square is 29%. There is only 29% variation in Shoppers which is explained by feature advertising. This percentage is very less and hence the fitted model is not set to be a good fit to the data.
Model: shoppers = 960.3924697 + 7.806303253*feature-65.10617381*price
Ho: the model is not significant. V/s h1: the model is significant. with (F=32.8081625, P<5%), The null hypothesis is rejected at 5% level of significance and conclude that model is significant.
With one unit increase in an average number of feature advertising, there is 7.806303253 units increase in the average number of Shoppers. This value is significant with (t=4.616424443, p<5%).
With one unit increase in an average price each week, there is 65.10617381 units decrease in the average number of Shoppers. This value is NOT significant with (t=-1.392860929, p>5%).
The coefficient of determination, R square is 30%. There is only 30% variation in Shoppers which is explained by feature advertising AND price. This percentage is very less and hence the fitted model is not set to be a good fit to the data.
There is not a much significant difference in the value of adjusted R squared between the two models. Hence I can say that model one is better as compared to model 2. Since model one has only one independent variable which saves time and money in calculations.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.