Interpret the forecasting model for the selected product.
Questions | Responses | |
How many espresso beverages are made each hour? | 30, on average | |
How many ounces of espresso beans are used for each beverage? | 1.5 ounces | |
How many hours per day is the coffee shop open? | 6:00 a.m.–8:00 p.m., Sunday–Saturday | |
How many days each month is the business open? |
|
You have the following six months of data to work with for pounds of espresso beans (y) used each month and monthly advertising expenditures (x):
Month | Lbs/Beans(y) | Advertising Dollars (x) |
---|---|---|
1 | 987 | $1,050 |
2 | 1,412 | $1,500 |
3 | 1,020 | $1,000 |
4 | 1,140 | $1,250 |
5 | 1,322 | $1,500 |
6 | 1,399 | $1,500 |
7 | 1253 | $1,350 |
From the above table data, using Simple linear regression model to show the forecast, here software used - Minitab, steps -
Go to Stat> Regression > Fit Regression Model, then let me put Y column data as Response and X Column data as Predictors, then click OK, we will get below result,
From the Above result, Regression Equation is shown, so knowing the Advertising dollars, we can calculate Lbs/ Beans,
As P value for Advertising dollars is Zero, which is less than 0.05 or alpha level, so we can conclude that Data for Advertising Dollars is statistically significant,
Next, we will look at the R Square Value, as R square value is 95.18%, thus it sounds good, and variability of all the response data, here Lbs/Beans data are almost around the mean
Next, we will calculate How many pounds of espresso beans will the company need, on average, each day, here i have used Linear forecasting to get Lbs/Bean data for upto 12 Month, using others forecasting techniques, result will come different,
A | B | C | D | E | F- Intercept | G - Slope |
2 | Month | Lbs/Beans(y) | Advertising Dollars (x) | Linear Forecasting | 1065.57 | 38.36 |
3 | 1 | 987 | $1,050 | 1103.93 | ||
4 | 2 | 1,412 | $1,500 | 1142.29 | ||
5 | 3 | 1,020 | $1,000 | 1180.64 | ||
6 | 4 | 1,140 | $1,250 | 1219.00 | ||
7 | 5 | 1,322 | $1,500 | 1257.36 | ||
8 | 6 | 1,399 | $1,500 | 1295.71 | ||
9 | 7 | 1253 | $1,350 | 1334.07 | ||
10 | 8 | 1372.43 | ||||
11 | 9 | 1410.79 | ||||
12 | 10 | 1449.14 | ||||
13 | 11 | 1487.50 | ||||
14 | 12 | 1525.86 | ||||
15 | Total | 15778.71 | ||||
16 | Forecast data- pounds of espresso beans per day | 43.35 = 43 |
Formula
A | B | C | D | E | F- Intercept | G - Slope |
2 | Month | Lbs/Beans(y) | Advertising Dollars (x) | Linear Forecasting | =INTERCEPT(C3:C9,B3:B9) | =SLOPE(C3:C9,B3:B9) |
3 | 1 | 987 | 1050 | =$F$2+($G$2*B3) | ||
4 | 2 | 1412 | 1500 | =$F$2+($G$2*B4) | ||
5 | 3 | 1020 | 1000 | =$F$2+($G$2*B5) | ||
6 | 4 | 1140 | 1250 | =$F$2+($G$2*B6) | ||
7 | 5 | 1322 | 1500 | =$F$2+($G$2*B7) | ||
8 | 6 | 1399 | 1500 | =$F$2+($G$2*B8) | ||
9 | 7 | 1253 | 1350 | =$F$2+($G$2*B9) | ||
10 | 8 | =$F$2+($G$2*B10) | ||||
11 | 9 | =$F$2+($G$2*B11) | ||||
12 | 10 | =$F$2+($G$2*B12) | ||||
13 | 11 | =$F$2+($G$2*B13) | ||||
14 | 12 | =$F$2+($G$2*B14) | ||||
15 | Total | =SUM(E3:E14) | ||||
16 | Forecast data - pounds of espresso beans per day | =E15/364 |
Interpret the forecasting model for the selected product. Use a simple linear regression model to show...
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