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Statistics Homework Sample Questions and Answers:

Question 1. Calculate the 3 yearly, and 5 yearly moving averages for the following

time series:

Year Prodn. (in quintals) :

1994 500

1995 540

1996 550

1997 530

1998 520

1999 560

2000 600

2001 640

2002 620

2003 610

2004 640

Solution. Computation of the 3 yearly, and 5 yearly moving averages

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Year Prodn. 3 Yearly Moving totals Moving av.

5 Yearly Moving totals Moving av.

1994 1995 1996 1997 1998 1999 2000

2001 2002 2003 2004

500 540 550 530 520 560 600

640 620 610 640

- - 1590 530 1620 540 1600

533 1610 537 1680 560 1800 600

1860 620 1870 623 1870 623 -

-

- - - - 2640 528 2700 540 2760 552 2850 570 2940 588

3030 606 3110 622 - - - -

Note. For 3 yearly calculations, the first group consists of 1994, 1995 and 1995; the second

group of 1995, 1996 and 1997 and so on. In the similar manner, groupings have been

made in case of 5 yearly calculations viz : 1994 + 1995 + 1996 + 1997 + 1998 ;

1995 + 1996 + 1997 + 1998 + 1999 and so on.

Question 2. 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

- -

(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) = 𝑋1𝑊1 + 𝑋2𝑊2 + 𝑋3𝑊3

𝑊1 +𝑊2 +𝑊2 =

𝑋𝑊

𝑊

Question 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

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 𝑋 ≠ 0).

𝑌 = Na + b 𝑋

𝑋𝑌 = a 𝑋 + b 𝑋2

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

= − 28𝑎 + 112𝑏 = 1200

28𝑏 = 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

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

Alter

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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