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Statistics Tutor and Assignment Illustrations and Solutions

Illustration 1. Find the 4 yearly moving averages from the following data (i) by centering

the averages, and (ii) by centering the totals:

Year : Prodn. (in tones):

1995 75

1996 85

1997 98

1998 90

1999 95

2000 108

2001 124

2002 140

2003 150

2004 160

Solution. (i) Computation of the 4 yearly moving averages by centering the

averages

Year (1)

Production (2)

4 yearly moving totals (3)

4 yearly moving average

(4)

Moving total of moving

averages in twos(5)

4 yearly moving average

centred (col. 5 + 2) (6)

1995 1996 1997 1998 1999 2000 2001 2002 2003 2004

75 85 98 90 95 108 124 140 150 160

- 348 368 391 417 467 522 574 - -

- 87 92 97.75 104.25 116.75 130.50 143.50 - -

- - 179.00 189.75 202.00 221.00 247.25 274.00 - -

- - 89.50 94.87 101.00 110.50 123.63 137.00 - -

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(ii) Computation of the 4 yearly Moving Averages by centering the totals

Year (1)

Production (2)

4 yearly moving totals (3)

Centering of the two adjacent totals (4)

4 yearly moving average centred

(col. 5 + 2) (6)

1995 1996 1997 1998 1999 2000 2001 2002 2003 2004

75 85 98 90 95 108 124 140 150 160

- - 348 368 391 417 467 522 574 - -

- - 716 759 808 884 989 1096 - -

- - 89.5 94.87 101.00 110.50 123.63 137.00 - -

Weighted Moving Averages

Under this method, weights are assigned rationally to different items of the groups in a

moving manner. Each item is multiplied by its respective weight, and the moving average of

the group is obtained by dividing the weighted total of the group by the total of the weights.

Thus, the weighted moving average of a group is obtained by

MA(w) = + +

++ =

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Illustration 2. From the following data, calculate the 3 yearly weighted moving averages,

the weights being 1, 2 and 3 respectively.

Year : Value :

1998 2

1999 6

200 3

2001 5

2002 7

2003 4

2004 2

Solution. Computation of the 3 yearly weighted moving averages (Weights being 1, 2 and

3 respectively)

Year Value X

3 Yearly weighted moving totals ( )

3 yearly weighted moving averages i.e. ( / )

1998 1999 2000 2001 2002 2003 2004

2 6 3 5 7 4 2

- (2 1 + 6 2 + 3 3) = 23 (6 1 + 3 2 + 5 3) = 27 (3 1 + 5 2 + 7 3) = 34

(5 1 + 7 2 + 4 3) = 31 (7 1 + 4 2 + 2 3) = 21 -

- 23/6 = 3.83 27/6 = 4.50 34/6 = 5.67 31/6 = 5.17 21/6 = 3.50 -

Determination of Trend Values through Moving Average Method

The moving average method discussed above can be used as a simple device of reducing

the fluctuations, and of obtaining the trend values in a time series with a fair degree of

accuracy. Under this method, the moving averages calculated represent the trend values for

the middle point of the period of moving averages. The moving averages when plotted on a

graph paper gives us a trend line in the form of a smoothed curve by reducing the

fluctuations in her time series.

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Illustration 3. Using the straight line method of least square, compute the trend values,

and draw the line of the best fit for the following series.

Day : Sales :

1 20

2 30

3 40

4 20

5 20

6 60

7 80

Also, show the curve for the original data on the same graph paper.

Solution (a) Computation of the trend values by the straight line method of least

square.

Days

X (1)

Sales

Y (2)

XY

(3)

X2

(4)

Trend

values Yc = 7.14 + 8.93X

Deviations

of items from trend values (Y Yc) (6)

1

2 2 4 5 6 7

20

30 40 20 50 60 80

20

60 120 80 250 360 560

1

4 9 16 25 36 49

16.07

25.00 33.93 42.86 51.79 60.72 69.65

3.93

5.00 6.07 -22.86 -1379 -0.72 10.35

Total 28 300 1450 140 N = 7 0.00

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Working

The trend value shown in the 5th column above have been found as under :

By the formula of straight line equation we have,

Yc = a + bX

Where, a and b are the two constants, the values of which are obtained by solving

simultaneously the following two normal equations (since 0).

= Na + b

= a + b 2

Substituting the respective values in the above we get,

300 = 7a + 28b

1450 = 28a + 140b

Multiplying the eqn (i) by 4 under the eqn (iii), and subtracting the same from the eqn (ii)

we get,

28a + 140b = 1450

= 28 + 112 = 1200

28 = 250

b = 250

28 = 8.93 approx.

Putting the value of b in the eqn (i) by 4 under the eqn(i) we get,

7a + 28(8.93) = 300

or 7a = 300 250 = 50

a = 50

7 = 7.14

Thus a = 7.14 and b =8.93

Putting the above values of a and b in the linear equation Yc = a + bX we get,

Yc = 7.14 + 8.93 X

Where, X = value of the time variable

Computation of the Trend values

Substituting the values of X successively in the linear equation, Yc= 7.14 + 8.93 X, we

compute the trend values as under:

When X = 1, Yc = 7.14 + 8.93 (1) = 16.07

When X = 2, Yc = 7.14 + 8.93 (2) = 25.00

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When X = 3, Yc = 7.14 + 8.93 (3) = 33.93

When X = 4, Yc = 7.14 + 8.9 (4) = 42.86

When X = 5, Yc = 7.14 + 8.93(6) = 51.79

When X = 6, Yc = 7.14 + 8.93 (6) = 60.72

When X = 7, Yc = 7.14 + 8.93 (7) = 69.65

Aliter

The above trend values could have been obtained by simply adding 8.93 (value of b i.e. rate

of change of the slope) successively to 7.14 (the value of the trend origin, a), as follows :

When X = 1, Yc = 7.14 + 8.93 (1) = 16.07

When X = 2, Yc = 7.14 + 8.93 (2) = 25.00

When X = 3, Yc = 7.14 + 8.93 (3) = 33.93

When X = 4, Yc = 7.14 + 8.9 (4) = 42.86

When X = 5, Yc = 7.14 + 8.93(6) = 51.79

When X = 6, Yc = 7.14 + 8.93 (6) = 60.72

When X = 7, Yc = 7.14 + 8.93 (7) = 69.65

Note. From the last column of the table given, it may be observed that the sum of the

deviation of the original values from their corresponding trend values is nearly zero. The

slight difference is due to the error in approximation.

(b) Graphic representation of the trend values, and the original data values

Days

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Illustration 4. Taking the deviations of the time variable, compute the trend values for the

following data by the method of the least square :

Days : Sales (in $) :

1 20

2 30

3 40

4 20

5 50

6 60

7 80

Solution. Computation of the Trend Values Taking the deviations of the time

variable by the method of the least

The trend values of Y are given by Yc = a + bX

Where, a =

[ = a + b 2 and = 0]

= 300

7 = 42.86 approx.

And b =

2 [ = a + b 2 and = 0]

= 250

8 = 8.93 approx.

Putting the values of a and b in the above, we get the required trend line equation as:

Yc = 42.86 + 8.93X

Where, Yc represents the computed trend value of Y, and X the deviation of the time

variable.

Using the above trend equation, the various trend values will be computed as under:

Computation of the Trend Values

When X = -3, Yc = 42.86 + 8.93 (-3) = 16.07

When X = -2, Yc = 42.86 + 8.93 (-2) = 25.00

Days

Sales Y

Time dvn. From mid value 4

XY X2 Trend values =42.86 + 8.93X

1 2 2 4

5 6 7

20 30 40 20

50 60 80

-3 -2 -1 0

1 2 3

-60 -60 -40 0

50 120 240

9 4 1 0

1 4 9

16.7 25.00 33.93 42.86

51.79 60.72 69.65

Total 300 0 140 N = 7 0.00

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When X = -1, Yc = 42.86 + 8.93 (-1) = 33.93

When X = 0, Yc = 42.86 + 8.93 (0) = 42.86

When X = 1, Yc = 42.86 + 8.93 (1) = 51.79

When X = 2, Yc = 42.86 + 8.93 (2) = 60.72

When X = 3, Yc = 42.86 + 8.93 (3) = 69.65

Aliter

The above trend values could be obtained by simply adding 8.93 (the value of b i.e. rate of

change of the slope) successively to 42.86 (the value of the trend origin at t = 4) for each

time period succeeding the time of the origin, and by deducting 8.93 successively from

42.86 for each item period preceding the time of the origin as follow:

When X is at the origin 4, Yc = 42.86

When X is at 5, Yc = 42.86 + 8.93 = 51.79

When X is at 6, Yc = 51.79 + 8.93 = 60.72

When X is at 7, Yc = 60.72 + 8.93 = 69.65

When X is at 3, Yc = 42.86 8.93 = 33.93

When X is at 2, Yc = 33.93 8.93 = 25.00

When X is at 1, Yc = 25 8.93 = 16.07

From the above it must be seen that the trend values, thus obtained on the basis of the

time deviations, are the same as they were obtained on the basis of the original data in the

illustration 8 before, where the values of the constants a and b were determined through

the lengthy procedure of simultaneous equations.

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