question archive Looking at the following data set for a menu items sales volume for five consecutive Fridays in a row: Friday (1): 22 Friday (2): 26 Friday (3): 41 Friday (4): 24 Friday (5): 26 What would be the most logical forecast for that same menu items to prep and sell on the following Friday, or Friday (6)? A: 36 B: 41 C: 22 D: 42 E: 27
Looking at the following data set for a menu items sales volume for five consecutive Fridays in a row: Friday (1): 22 Friday (2): 26 Friday (3): 41 Friday (4): 24 Friday (5): 26 What would be the most logical forecast for that same menu items to prep and sell on the following Friday, or Friday (6)? A: 36 B: 41 C: 22 D: 42 E: 27
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Looking at the following data set for a menu items sales volume for five consecutive Fridays in a row:
Friday (1): 22
Friday (2): 26
Friday (3): 41
Friday (4): 24
Friday (5): 26
What would be the most logical forecast for that same menu items to prep and sell on the following Friday, or Friday (6)?
A: 36
B: 41
C: 22
D: 42
E: 27
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Based on given information, first we will try to look a trend or pattern in data by linear trend equation:
| Friday, time Period (t) | Sales Volume (y) | y*t | t^2 | |
| 1 | 22 | 22 | 1 | |
| 2 | 26 | 52 | 4 | |
| 3 | 41 | 123 | 9 | |
| 4 | 24 | 96 | 16 | |
| 5 | 26 | 130 | 25 | |
| Sum | 15 | 139 | 423 | 55 |
| n = | 5 | |||
| Slop b = (n * Σty – Σt * Σy) / {n * Σ(t^2) - (Σt)^2} = | 0.6000 | |||
| Intercept a = (Σy – b * Σt) / n = | 26.00 | |||
| Regression Equation | Y = a + bt = | 26 + 0.60*t | ||
| Friday (6) = | 29.60 |
As no option is matching with this, therefore the most logical forecast for that same menu items to prep and sell on the following Friday, Friday (6) will be 27 because forecast 41, 42 and 36 are outliers and 22 will be too low.
Therefore, correct answer is option E 27
