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Consider the following time series data.

Week 1 2 3 4 5 6
Value 18 13 16 11 17 14

Using the average of all the historical data as a forecast for the next period, compute the following measures of forecast accuracy. (Round your answers to two decimal places.)

a. Mean absolute error.

b. Mean squared error.

c. Mean absolute percentage error.

d. What is the forecast for week 7?

User Lareina
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1 Answer

4 votes

Answer:

Explanation:

Let's calculate the requested measures of forecast accuracy using the given time series data:

Week: 1 2 3 4 5 6

Value: 18 13 16 11 17 14

a. Mean Absolute Error (MAE):

\[ MAE = \frac{1}{n} \sum_{i=1}^{n} |Y_i - \hat{Y}_i| \]

where \( n \) is the number of observations, \( Y_i \) is the actual value, and \( \hat{Y}_i \) is the forecast value.

\[ MAE = \frac{1}{6} (|18-15| + |13-15| + |16-15| + |11-15| + |17-15| + |14-15|) \]

\[ MAE = \frac{1}{6} (3 + 2 + 1 + 4 + 2 + 1) = \frac{13}{6} \]

b. Mean Squared Error (MSE):

\[ MSE = \frac{1}{n} \sum_{i=1}^{n} (Y_i - \hat{Y}_i)^2 \]

\[ MSE = \frac{1}{6} ( (18-15)^2 + (13-15)^2 + (16-15)^2 + (11-15)^2 + (17-15)^2 + (14-15)^2 ) \]

\[ MSE = \frac{1}{6} (9 + 4 + 1 + 16 + 4 + 1) = \frac{35}{6} \]

c. Mean Absolute Percentage Error (MAPE):

\[ MAPE = \frac{1}{n} \sum_{i=1}^{n} \left| \frac{Y_i - \hat{Y}_i}{Y_i} \right| \times 100 \]

\[ MAPE = \frac{1}{6} \left( \left| \frac{18-15}{18} \right| + \left| \frac{13-15}{13} \right| + \left| \frac{16-15}{16} \right| + \left| \frac{11-15}{11} \right| + \left| \frac{17-15}{17} \right| + \left| \frac{14-15}{14} \right| \right) \times 100 \]

\[ MAPE \approx \frac{1}{6} (16.67 + 15.38 + 6.25 + 36.36 + 11.76 + 7.14) \]

\[ MAPE \approx \frac{93.56}{6} \approx 15.59 \]

d. Forecast for Week 7:

Since you're using the average of all historical data as a forecast, the forecast for Week 7 would be the average of the values in Weeks 1 to 6.

\[ \text{Forecast for Week 7} = \frac{18 + 13 + 16 + 11 + 17 + 14}{6} \]

\[ \text{Forecast for Week 7} = \frac{89}{6} \approx 14.83 \]

So, the forecast for Week 7 is approximately 14.83.

User TRomesh
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