In: Statistics and Probability
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Consider the following time series data:
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| - Select your answer -Graph (i)Graph (ii)Graph (iii)Graph (iv)Item 1 | ||||||||||||||||||||||||||
| What type of pattern exists in the data? | ||||||||||||||||||||||||||
| - Select your answer -Positive trend patternHorizontal patternVertical patternNegative trend patternItem 2 | ||||||||||||||||||||||||||
| (b) | Develop a three-month moving average for this time series. Compute MSE and a forecast for month 8. | |||||||||||||||||||||||||
| If required, round your answers to two decimal places. Do not round intermediate calculation. | ||||||||||||||||||||||||||
| MSE: | ||||||||||||||||||||||||||
| The forecast for month 8: | ||||||||||||||||||||||||||
| (c) | Use α = 0.2 to compute the exponential smoothing values for the time series. Compute MSE and a forecast for month 8. | |||||||||||||||||||||||||
| If required, round your answers to two decimal places. Do not round intermediate calculation. | ||||||||||||||||||||||||||
| MSE: | ||||||||||||||||||||||||||
| The forecast for month 8: | ||||||||||||||||||||||||||
| (d) | Compare the three-month moving average forecast with the exponential smoothing forecast using α = 0.2. Which appears to provide the better forecast based on MSE? | |||||||||||||||||||||||||
| - Select your answer -3-month moving averageexponential smoothingItem 7 | ||||||||||||||||||||||||||
| (e) | Use trial and error to find a value of the exponential smoothing coefficient α that results in the smallest MSE. | |||||||||||||||||||||||||
| If required, round your answer to two decimal places. | ||||||||||||||||||||||||||
| α = |
1. The graphs are not seen but here is a picture of what the graph should look like. Please match it correctly. with the ones given

Here the x-axis has the months and the y-axis has the values.
3-month moving average
Here first we find the total of previous 'n years and then it is divided by 'n'.
For e.g.: For month = 4 with n = 3
Total =( 1st + 2nd + 3rd ) / 3
| Month | Value | Total | Avg (forecast) | Error (value -forecast) | Error^2 |
| 1 | 23 | ||||
| 2 | 13 | ||||
| 3 | 21 | ||||
| 4 | 13 | 57 | 19 | -6 | 36 |
| 5 | 19 | 47 | 15.667 | 3.333 | 11.111 |
| 6 | 21 | 53 | 17.667 | 3.333 | 11.111 |
| 7 | 17 | 53 | 17.667 | -0.667 | 0.444 |
| 8 | 57 | 19 | |||
| Total | 58.667 | ||||
| MSE | 14.667 |
MSE =
.....here it is divide by 4 because we only have 4 pairs of
actual and forecast
Exponential smoothing with smoothing constant = 0.2

| Month | Value | Forecast | Error | Error ^2 |
| 1 | 23 | 23 | ||
| 2 | 13 | 23 | -10 | 100 |
| 3 | 21 | 21 | 0 | 0 |
| 4 | 13 | 21 | -8 | 64 |
| 5 | 19 | 19.4 | -0.4 | 0.16 |
| 6 | 21 | 19.32 | 1.68 | 2.822 |
| 7 | 17 | 19.656 | -2.656 | 7.054 |
| 8 | 19.1248 | |||
| Total | 174.037 | |||
| MSE | 29.006 |
Here MSE =
.here it is divide by 6 because we only have 4
pairs of actual and forecast
Since we haven't been given the forecast for motnh = 1, we take it same as the actual.
| (a) | What type of pattern exists in the data? |
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- Select your answer -Positive trend patternHorizontal patternVertical patternNegative trend pattern It is not positive or negative since the points are not sloping upward or downward. |
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| (b) | Develop a three-month moving average for this time series. Compute MSE and a forecast for month 8. |
| If required, round your answers to two decimal places. Do not round intermediate calculation. | |
| MSE: 14.667 | |
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The forecast for month 8: 19 (ans explained above) |
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| (c) | Use α = 0.2 to compute the exponential smoothing values for the time series. Compute MSE and a forecast for month 8. |
| If required, round your answers to two decimal places. Do not round intermediate calculation. | |
| MSE: 29.006 | |
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The forecast for month 8: 19.1248 |
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| (d) | Compare the three-month moving average forecast with the exponential smoothing forecast using α = 0.2. Which appears to provide the better forecast based on MSE? |
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- Select your answer -3-month moving average exponential smoothingItem 7 This is because MSE represents error lack of accuracy. So the smaller the MSE more the accuracy. |
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| (e) | Use trial and error to find a value of the exponential smoothing coefficient α that results in the smallest MSE. |
| If required, round your answer to two decimal places. | |
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α = 0.3 Since if we look at less than 0.3 or greater than 0.3 we see that MSE starts increasing so the smallest is givne at 0.3. |
At α = 0.1
| Month | Value | Forecast | Error | Error ^2 |
| 1 | 23 | 23 | ||
| 2 | 13 | 23 | -10 | 100 |
| 3 | 21 | 22 | -1 | 1 |
| 4 | 13 | 21.9 | -8.9 | 79.21 |
| 5 | 19 | 21.01 | -2.01 | 4.0401 |
| 6 | 21 | 20.809 | 0.191 | 0.036 |
| 7 | 17 | 20.8281 | -3.8281 | 14.654 |
| 8 | 20.44529 | |||
| Total | 198.941 | |||
| MSE | 33.157 |
α =0.2
MSE = 29.006
α =0.3
| Month | Value | Forecast | Error | Error ^2 |
| 1 | 23 | 23 | ||
| 2 | 13 | 23 | -10 | 100 |
| 3 | 21 | 20 | 1 | 1 |
| 4 | 13 | 20.3 | -7.3 | 53.29 |
| 5 | 19 | 18.11 | 0.89 | 0.7921 |
| 6 | 21 | 18.377 | 2.623 | 6.880 |
| 7 | 17 | 19.1639 | -2.1639 | 4.682 |
| 8 | 18.51473 | |||
| Total | 166.645 | |||
| MSE | 27.774 |
α =0.4
| Month | Value | Forecast | Error | Error ^2 |
| 1 | 23 | 23 | ||
| 2 | 13 | 23 | -10 | 100 |
| 3 | 21 | 19 | 2 | 4 |
| 4 | 13 | 19.8 | -6.8 | 46.24 |
| 5 | 19 | 17.08 | 1.92 | 3.6864 |
| 6 | 21 | 17.848 | 3.152 | 9.935 |
| 7 | 17 | 19.1088 | -2.1088 | 4.447 |
| 8 | 18.26528 | |||
| Total | 168.309 | |||
| MSE | 28.051 |