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Historical demand for a product is: a. Using a weighted moving average with weights of 0.40 (June), 0.20 (May), and 0.40 (April), find the July forecast. (Round your answer to 1 decimal place.) July forecast b. Using a simple three-month moving average, find the July forecast. (Round your answer to 1 decimal place.) July forecast c. Using single exponential smoothing with alpha = 0.20 and a June forecast = 12, find the July forecast. (Round your answer to 1 decimal place.) July forecast d. Using simple linear regression analysis, calculate the regression equation for the preceding demand data. (Do not round intermediate calculations. Round your intercept value to 1 decimal place and slope value to 2 decimal places.) Y = + t e. Using the regression equation in d, calculate the forecast for July. (Do not round intermediate calculations. Round your answer to 1 decimal place.) July forecast
Expert Answer
Answer:- Weighted moving average ; –
Forecast for July = (14*0.4 + 15*0.2+13*0.4)/(0.4+0.2+0.4) = (5.6+3+5.2)/1
Forecast for July =13.8
Answer:- Simple three-month moving average:-
Forecast for July =(14+15+13)/2=14
Answer:-
| Month | Demand | ||
| Jan | 13 | ||
| Feb | 10 | ||
| March | 14 | ||
| April | 13 | ||
| May | 15 | ||
| June | 14 | ||
| Month (X) | Number of Accident(Y) | X*Y | X*X |
| 1 | 13 | 13 | 1 |
| 2 | 10 | 20 | 4 |
| 3 | 14 | 42 | 9 |
| 4 | 13 | 52 | 16 |
| 5 | 15 | 75 | 25 |
| 6 | 14 | 84 | 36 |
| ∑X = 21 | ∑Y = 79 | ∑XY = 286 | ∑X^2 =91 |
| Slope(b) = (N∑XY – (∑X)(∑Y)) / (N∑X2 – (∑X)2) | |||
| Slope B = (6*286 – 21*79)/(6*91-21^2) | |||
| B = 0.543 | |||
| Intercept(a) = (∑Y – b(∑X)) / N | |||
| a=(79-0.543*21)/6 | |||
| a=11.27 | |||
| Regression Equation(y) = a + bx | |||
| y=11.27+0.543*x | |||
| Forecast for July | |||
| y=11.27+0.543*7 | |||
| y=15.071 | |||
