Colossal Genius
Running head: BENCHMARK CASE STUDY
BENCHMARK CASE STUDY 5
Benchmark Case Study Data
Belinda Smith
Grand Canyon University BUS-660
April 23, 2019
First Evaluation
To understand the factors that contribute to customer satisfaction and spending, we fit linear regression model and check for the significance of the independent variables to our model. A linear regression model has dependent and independent variables. In this case, the dependent variable is customer satisfaction and spending while the independent variables include Dine in (1)/Take out (2), satisfaction with service, satisfaction with food, and driving distance to restaurant.
Let;
Y = Customer Satisfaction and spending
X1 = Dine in (1)/Take out (2)
X2 = Satisfaction with service
X3 = Satisfaction with food
X4 = Driving distance to restaurant
The linear regression model is of the form,
Where Bi, I = 1,2,3,4 are the coefficients and is the error term. The method of the least squares is used to minimize the error term.
This model consists of two components mainly the explained components and the unexplained components. The explained components are the dependent variables while the unexplained components are the predictor variables. Multiple regressions with overall satisfaction as a dependent variable are carried out in excel. The following table contains the results from excel worksheet.
To pick the best fitted model, we check whether the independent variables have a significant effect on the dependent variable. We can check this using the p-value or using the t statistic. In our case we test the hypothesis to test significance of each model.
For Dine in (1)/Take out (2), we test the hypothesis: H0: B1 = 0 (The variable has no significant effect on the model or is not a significant predictor) vs. H1: B1≠ 0 (The variable is a significant predictor). If the p-value is less than 0.05, we reject the null hypothesis and conclude that the predictor is a significant predictor. For Dine in (1)/Take out (2) variable, the p-value is 0.84532 which is greater than 0.05. Thus, we fail to reject the null hypothesis and conclude that Dine in (1)/Take out (2) is not a significant predictor of customer satisfaction and spending.
For Satisfaction with service, we test Hypothesis H0: B2 = 0 vs. H1: B2 ≠ 0. From the table, the p-value is 1.4E-10 which is less than 0.05. Thus, we reject the null hypothesis and conclude that satisfaction with service is a significant predictor of customer satisfaction and spending.
For satisfaction with food, we carry out the same procedure. We test the hypothesis H0: B3 = 0 vs. H1: B3 ≠ 0. The p-value is 0.00019 which is less than 0.05. We reject the null hypothesis that the variable is not a significant predictor and conclude that satisfaction with food is a significant predictor.
For driving distance to the restaurant, we test the hypothesis H0: B4 = 0 vs. H1: B4 ≠ 0. The p-value is 0.51115 which is great than 0.05. We therefore fail to reject the null hypothesis and conclude that for driving distance to the restaurant is not a significant predictor.
Y = 0.70943 + 0.58037 X2 + 0.26681 X3
Interpretation of the equation.
Satisfaction with service
A unit increase in the level of satisfaction with the food (X2) increases the customer satisfaction and spending by (1-0.58037) * 100% = 41.96%. This is to say that when the customer’s satisfaction with the food is increased with a unit, the general customer satisfaction increases by 41.96%.
Satisfaction with food
A unit increase in the level of satisfaction with the food increases the general customer satisfaction and spending by (1- 0.26681) *100% = 73.319%
B0 is the intercept. The intercept is a constant and does not affect in any significant. This means that the intercept has no significant effect on the model.
The coefficient is allowing that the R-squared value which is 0.863. It means that 86.3% of variability in values of overall satisfaction is explained by this model. This is with the assumption that every independent variable in the model explains the variation in the independent variable. The Adjusted R for this model is 0.8472. This means that 84.72% of the variation in the general satisfaction and disburse are relevant facts by the predictor. This is righter since it assumes that the percentage of variation is only relevant fact by the variables that only affect the dependent variable.
So, based on the analysis, Michael and Tony should focus on increasing the satisfaction and spending level of customers through better food and service
Second Evaluation
The error values for 4 month moving average, weighted moving average and exponential smoothing with alpha of 0.9 are calculated in excel. We see that the MAD values are 54.1 for 4 month moving average, 42.4 for weighted moving average and 41.21 for exponential smoothing. We see that the MAPE values are 5.98% for 4 month moving average, 4.8% for weighted moving average and 4.83% for exponential smoothing.
The error values are lower for exponential smoothing with MSE value and Bias also lower than the other two methods. So, we say that exponential smoothing is the best forecasting method for this scenario. The forecast for December based on exponential smoothing forecast is 931.88 or 932 customers. The business owner should use exponential smoothing with NSE value and Bias.
Third Evaluation
Based on solver solution in excel, the staff requirement for each shift is shown as
|
10 AM - 1 PM |
3 |
|
1:00 PM - 4:00 PM |
4 |
|
4:00 PM - 7:00 PM |
6 |
|
7:00 PM - 10:00 PM |
7 |
|
10:00 PM - 1:00 AM |
4 |