Daily high temperatures in St. Louis for the last week were as​ follows: 95​, 92​, 93​, 92​, 95​, 90​, 90 ​(yesterday). ​a) The high tempera

Question

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).

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3 weeks 2022-01-06T21:43:43+00:00 1 Answer 0 views 0

a) T = 91.7 degrees

b) T = 90 degrees

d) MSE = 5.05

Step-by-step 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