MGSC 272 Final Review 1 W10 Solved

MGSC 272 Final Review 1 W10 Solved - MGSC 272 Final Review...

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MGSC 272 Final Review W09 Question 1 The following data represents quarterly sales in thousands of dollars of sports equipment at a mountain resort. Y = original data, T = Trend, S = Seasonal index. The trend line for the data is given by T t = 152 + 3.6 t . (1) Find the seasonally adjusted value for the 1 st quarter of 2004. 100(Y/S) = 100(272/143.1) = 190.08
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Interpret the seasonally adjusted value. In the first quarter the seasonal index of 143.1 means that the seasonal effect is to increase the value to 43.1% above the trend line. Thus the actual data (272) is inflated and removing the seasonal effect results in the true underlying value (consisting of trend, cyclical and random effects) of 190.08. (2) Calculate a forecast for the first quarter of 2006 based on trend and seasonal effects. F = T x S = 213.75 x 143.1/100 = 305.9 Determine the magnitude of the error term if this forecast is compared to the actual value. Error = Y – F = 315 – 305.9 = 9.1 (3) Calculate the deseasonalized percent of trend for the third quarter of 2005. Y/(TS) = 270/(206.48 x 1.3051) = 0.7942 Which time series components are contained in this value? Cyclical and Random (4) The trend line for the data is given by T t = 152 + 3.6 t . Calculate a forecast for the first quarter of 2008. t =25. T 25 = 152 +3.6(25) = 242 F 25 = T 25 x S 1 = 242 x 1.431 = 346.3
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(5) Calculate the random component for the 2 nd quarter of 2002. Year Q Y T S Y/(TS)=CR 3MA = C CR/C=R 2002 1 205 155.53 143.1 .9224 2 96 159.17 61.89 .9745 .9336 1.0405 3 194 162.81 130.51 .9130 4 102 166.45 64.51 Question 2 In the following spreadsheet, Excel column D (variable Y) contains time series data and column E (Forecast F) contains a forecast based on a time series model.
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MAD = 172/20 = 8.6 MSE = 2025/20 = 101.25 MAPE = [(18/205) + (3/96) + (18/194) + (5/102) + (13/230) + (3/105) + (14/245) + (3/120) + (8/272) + (7/110) + (5/255) + (12/114) + (11/296) + (4/130) + (1/270) + (4/140) + (9/315) + (15/150) + (7/295) + (13/158)] = 0.9807 x 100 = 98.07% (2) Explain why the mean error t
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This note was uploaded on 02/28/2012 for the course MANAGEMENT MGSC 272 taught by Professor Smith during the Spring '12 term at McGill.

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MGSC 272 Final Review 1 W10 Solved - MGSC 272 Final Review...

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