Answer:
a) T = 91.7 degrees
b) T = 90 degrees
c) MAD = 1.9
d) MSE = 5.05
Explanation:
Given:
- Daily high temperatures in St. Louis for the last week were as follows:
95, 92, 93, 92, 95, 90, 90
Find:
a) Forecast the high temperature today, using a 3-day moving average.
b) Forecast the high temperature today, using a 2-day moving average.
c) Calculate the mean absolute deviation based on a 2-day moving average, covering all days in which you can have a forecast and an actual temperature.
d) The mean squared error for the 2-day moving average
Solution:
a)
- The set of 3 day moving average is as follows:
4. (95 + 92 + 93) ÷ 3 = 93.33⁰C
5. (92 + 93 + 92) ÷ 3 = 92.33⁰C
6. (93 + 92 + 95) ÷ 3 = 93.33⁰C
7. (92 + 95 + 90) ÷ 3 = 92.33⁰C
8. (95 + 90 + 90) ÷ 3 = 91.667⁰C
- Now use these points on excel sheet to forecast the temperature for today. The line of best fit is given:
T = 91.7 degrees
b)
- The set of 2 day moving average is as follows:
3. (95 + 92) ÷ 2 = 93.5⁰C
4. (95 + 93) ÷ 2 = 92.5⁰C
5. (93 + 92) ÷ 2 = 92.5⁰C
6. (92 + 95) ÷ 2 = 93.5⁰C
7. (95 + 90) ÷ 2 = 92.5⁰C
8. (90 + 90) ÷ 2 = 90⁰C
- Now use these points on excel sheet to forecast the temperature for today. The line of best fit is given:
T = 90 degrees
c)
Error Error^2
3. 93.5⁰C 0.5 0.25
4. 92.5⁰C 0.5 0.25
5. 92.5⁰C 2.5 6.25
6. 93.5⁰C 3.5 12.25
7. 92.5⁰C 2.5 6.25
8. 90⁰C
- The mean absolute deviation as follows:
MAD = Sum of all errors / 5
MAD = (0.5+0.5+2.5+3.5+2.5) / 5
MAD = 1.9
d)
- The mean squared error deviation as follows:
MSE = Sum of all error^2 / 5
MSE = (0.25+0.25+6.25+12.25+6.25) / 5
MSE = 5.05