Daily high temperatures in St. Louis for the last week were as follows: 95, 92, 93, 92, 95, 90, 90 (yesterday). a) The high temperature for today using a 3-day moving average = 91.7 degrees (round your response to one decimal place). b) The high temperature for today using a 2-day moving average = 90 degrees (round your response to one decimal place). c) The mean absolute deviation based on a 2-day moving average = 1.9 degrees (round your response to one decimal place). d) The mean squared error for the 2-day moving average = nothing degrees squared (round your response to one decimal place).

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