Answer to Question #253026 in Statistics and Probability for Joseph

Question

 

The number of fishing rods selling each day is given below. Perform analyses of the time series to determine which model should be used for forecasting.

 

a. 3 day moving average analysis

b. 4 day moving average analysis

c. 3 day weighted moving average analysis with weights w1=0.2, w2=0.3 and w3=0.5 with w1 on the oldest data

d. exponential smoothing analysis with a = 0.3.

e. Which model provides a better fit of the data?

f. Forecast day 13 sales of fishing rods using the model chosen in part (e).

Day

Rods sold

1 60

2 70

3 110

4 80

5 70

6 85

7 115

8 105

9 65

10 75

11 95

12 85

Answer 1

1. Analysis of methods:

1.1 3 day moving average

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} day & sales & Model & Error & Cummulated \space Err\\ \hline 1 & 60 & 60 &0& 0 \\ \hline 2 & 70 & 70 &0& 0 \\ \hline 3 & 110 & \frac{60+70+110}{3}=80 &110-80=30& 0+30=30 \\ \hline 4 & 80 & \frac{70+110+80}{3}=88.6 &86.6-80=6.6& 30+6.6=36.6 \\ \hline 5 & 70 & \frac{110+80+70}{3}=88.6 &86.6-70=16.6& 36.6 +16.6=53.2\\ \hline 6 & 85 & \frac{80+70+85}{3}=78.3 &85-78.3=6.6& 53.2+6.6=60\\ \hline 7 & 115 & \frac{70+85+115}{3}=90 &115-90=25& 60+25=85\\ \hline 8 & 105 & \frac{85+115+105}{3}=101.6 &105-101.6=3.4& 85+3.4=88.4\\ \hline 9 & 65 & \frac{115+105+65}{3}=95 &95-65=30& 88.4+30=118.4\\ \hline 10 & 75 & \frac{105+65+75}{3}=81.6 &81.6-75=6.6&118.4+6.6=125\\ \hline 11 & 95 & \frac{65+75+95}{3}=78.3 &95-78.3=16.6&125+6.6=141.6\\ \hline 12 & 85 & \frac{75+95+85}{3}=85 &85-85=0&141.6+0=141.6\\ \hline \end{array}$

MEAN ERROR =$141.6/12=11.8$

1.1  4 day moving average

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} day & sales & Model & Error & Cummulated \space Err\\ \hline 1 & 60 & 60 &0& 0 \\ \hline 2 & 70 & 70 &0& 0 \\ \hline 3 & 110 & 110 &0& 0+30=30 \\ \hline 4 & 80 & \frac{60+70+110+80}{4}=80 &80-80=0& 0+0=0 \\ \hline 5 & 70 & \frac{70+110+80+70}{4}=82.5 &82.5-70=12.5& 0 +12.5=12.5\\ \hline 6 & 85 & \frac{110+80+70+85}{4}=86.25 &86.25-85=1.25& 12.5.+1.25=13.75\\ \hline 7 & 115 & \frac{80+70+85+115}{4}=87.5 &115-87.5=1.27.5& 13.75+27.5=41.25\\ \hline 8 & 105 & \frac{70+85+115+105}{4}=93.75 &105-93.75=11.25.5& 41.25+11.25=52.5\\ \hline 9 & 65 & \frac{85+115+105+65}{4}=92.5 &92.5-65=27.5& 52.5+27.5=80\\ \hline 10 & 75 & \frac{115+105+65+75}{4}=90 &90-75=15& 80+15=95\\ \hline 11 & 95 & \frac{105+65+75+95}{4}=85 &95-85=10& 95+10=105\\ \hline 12 & 85 & \frac{65+75+95+85}{4}=80 &85-80=5& 105+5=110\\ \hline \end{array}$

MEAN ERROR =$110/12=9.17$

Thus method of 4 day moving average more exactly correspond to data than method of  3 day moving average.

1,3 3 day weighted moving average analysis with weights w1=0.2, w2=0.3 and w3=0.5 with w1 on the oldest data

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} day & sales & Model & Error & Cummulated \space Err\\ \hline 1 & 60 & 60 &0& 0 \\ \hline 2 & 70 & 70 &0& 0 \\ \hline 3 & 110 & 60\cdot 0.2+70\cdot0.3+110\cdot 0.5=88 &110-88=22& 0+22=22 \\ \hline 4 & 80 & 70\cdot 0.2+110\cdot0.3+80\cdot 0.5=87 &87-80=7& 22+7=29 \\ \hline 5 & 70 & 110\cdot 0.2+80\cdot 0.3+70\cdot 0.5=81 &81-70=11& 29+11=40 \\ \hline 6 & 85 & 80\cdot 0.2+70\cdot 0.3+85\cdot 0.5=79.5 &85-79.5=5.5& 40 +5.5=45.5\\ \hline 7 & 115 & 70\cdot 0.2+85\cdot 0.3+115\cdot 0.5=97 &115-97=18& 45.5+18=63.5\\ \hline 8 & 105 & 85\cdot 0.2+115\cdot 0.3+105\cdot 0.5=104 &105-104=1& 63.5=1=64.5\\ \hline 9 & 65 & 115\cdot 0.2+105\cdot 0.3+65\cdot 0.5=87 &87-65=22&64.5+22=86.5\\ \hline 10 & 75 & 105\cdot 0.2+65\cdot 0.3+75\cdot 0.5=78 &78-75=3&86.5+3=89.5\\ \hline 11 & 95 & 65\cdot 0.2+75\cdot 0.3+95\cdot 0.5=83 &95-83=12&89.5+12=101.5\\ \hline 12 & 85 & 75\cdot 0.2+95\cdot 0.3+85\cdot 0.5=88 &86-85=1&101.5+1=102.5\\ \hline \end{array}$

MEAN ERROR =$102.5/12=8.5$

Thus last merhod is a best from three considered.

1.4 exponential smoothing analysis with a = 0.3

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} day & sales & Model & Error & Cummulated \space Err\\ \hline 1 & 60 & 60 &0& 0 \\ \hline 2 & 70 & 0.7\cdot 60 +0.3\cdot 70=63&70-63=7& 0+7=7 \\ \hline 3 & 110 & 0.7\cdot 63 +0.3\cdot 110=77.1&110-77.1=32.9& 7+32.9=39.9 \\ \hline 4 & 80 & 0.7\cdot 77.1 +0.3\cdot 80=77.97&80-77.97=2.03& 39.9 +2.03=41.93\\ \hline 5 & 70 & 0.7\cdot 77.97 +0.3\cdot 70=75.579&75.579-70=5.579& 41.93+5.579=47.509\\ \hline 6 & 85 & 0.7\cdot 75.579 +0.3\cdot 85=78.4053&85-78.4053=6.5947&47.509+6.5947=54.1037\\ \hline 7 & 115 & 0.7\cdot 78.4053 +0.3\cdot 115=89.383&115-89.383=25.616&54.1037+25.616=79.720\\ \hline 8 & 105 & 0.7\cdot 89.383 +0.3\cdot 105=94.069&105-94.069=10.931&79.720+9.931=90.651\\ \hline 9 & 65 & 0.7\cdot 94.069 +0.3\cdot 65=85.348&85.348-65=20.348&90.651+20.348=110.999\\ \hline 10 & 75 & 0.7\cdot 85.348 +0.3\cdot 75=82.244&82.244-75=7.244&110.999+7.244=118.243\\ \hline 11 & 95 & 0.7\cdot 82.244 +0.3\cdot 95=86.07&95-86.07=9.929&118.243+8,929=127.17\\ \hline 12 & 85 & 0.7\cdot 86.07 +0.3\cdot 85=85.749&85.749-85=0.749&127.17+0.749=127.92\\ \hline \end{array}$

MEAN ERROR =$127.9/12=10.7$

Thus the method of exponential smoothing analysis with a = 0.3 gives the bigger mean error than best method of day weighted moving average analysis with weights w1=0.2, w2=0.3 and w3=0.5 with w1 on the oldest data and finally we take lass method as a best at whole.

f. Forecast day 13 sales of fishing rods using the model chosen in part (e)

we apply /the best method of 'day weighted moving average analysis with weights w1=0.2, w2=0.3 and w3=0.5 with w1 on the oldest data ',

use data of last three days and calculate forecast value:

$75\cdot 0.2+95\cdot 0.3+85\cdot 0.5=88$

Thus in 13 day we expect 88 sales.