For a smoothing constant of 0.2
Time period – 1 2 3 4 5 6 7 8 9 10
Actual value – 46 55 39 42 63 54 55 61 52
Forecast – 58 55.6 55.48 52.18 50.15 52.72 52.97 53.38 54.90
Forecast error - -12 -.6 -16.48 – 10.12 12.85 1.28 2.03 7.62 -2.9
The mean square error is 84.12
The mean forecast for period 11 is 54.38
For a smoothing constant of 0.8
Time period – 1 2 3 4 5 6 7 8 9 10
Actual value – 46 55 39 42 63 54 55 61 52
Forecast – 58 48.40 53.68 41.94 41.99 58.80 54.96 54.99 59.80
Forecast error - -12 6.60 -14.68 0.06 21.01 -4.80 0.04 6.01 -7.80The mean square error is 107.17
The mean forecast for period 11 is 53.56
Based on the MSE, smoothing constant of .2 offers a better model since the mean forecast is much better compared to the 53.56 of the smoothing constant of 0.8.
To find the standard error, divide the standard deviation by the square root of the sample size
standard deviation = 7.5
sample size = 25......sqrt 25 = 5
standard error = 7.5 / 5 = 1.5 <==
B but i would also ask a parent or teacher for help
Answer
a nd b
Step-by-step explanation:
