Question

Daily high temperatures in St. Louis for the last week were as​ follows: 95​, 94​, 95​, 95​, 95​, 90​, 90 ​(yesterday).

​a) The high temperature for today using a​ 3-day moving average ​= nothing degrees ​(round your response to one decimal​ place).

​b) The high temperature for today using a​ 2-day moving average ​= nothing degrees ​(round your response to one decimal​ place).

​c) The mean absolute deviation based on a​ 2-day moving average​ = nothing 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). ​

e) The mean absolute percent error​ (MAPE) for the​ 2-day moving average​ = nothing​% ​(round your response to one decimal place​).

Answer 1

	Temperature	3-Day MA	2-Day MA	MAD	MSE	MAPE
	95
	94		=(95+94)/2	=ABS((94-94.5))	=(94-93.43)^2	=(ABS((94-93.43)/94))*100
	95	=(95+94+95)/3	=(94+95)/2	=ABS((95-94.5))	=(95-93.43)^2	=(ABS((95-93.43)/95))*100
	95	=(94+95+95)/3	=(95+95)/2	=ABS((95-95))	=(95-93.43)^2	=(ABS((95-93.43)/95))*100
	95	=(95+95+95)/3	=(95+95)/2	=ABS((95-95))	=(95-93.43)^2	=(ABS((95-93.43)/95))*100
	90	=(95+95+90)/3	=(95+90)/2	=ABS((90-92.5))	=(90-93.43)^2	=(ABS((90-93.43)/90))*100
	90	=(95+90+90)/3	=(90+90)/2	=ABS((90-91.67))	=(90-93.43)^2	=(ABS((90-93.43)/90))*100
Average	=SUM(95+94+95+95+95+90+90)/7 = 93.4	=SUM(91.67+93.33+95)/3 = 93.3	=SUM(94.5+94.5+95+95+92.5+90)/6 = 93.6	=SUM(0.5+0.5+0+0+2.5+1.67)/6 =0.9	=SUM(0.32+2.46+2.46+2.46+11.76+11.76)/6 =5.2	=(SUM(0.61+1.65+1.65+1.65+3.81+3.81)/6) = 2.2

a) SUM(91.67+93.33+95)/3 = 93.4

b) SUM(94.5+94.5+95+95+92.5+90)/6 = 93.6

c) SUM(0.5+0.5+0+0+2.5+1.67)/6 = 0.9

d) SUM(0.32+2.46+2.46+2.46+11.76+11.76)/6 = 5.2

e) SUM(0.61+1.65+1.65+1.65+3.81+3.81)/6) = 2.2