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About Statistics: Complexity of statistics subject is well

known. Statistics involves solving complex problems having

multi-dimensional data using computational methods.

Information technology has played a vital role in handling and

simplifying such complex methods and scenarios but students

face a lot of difficulties in understanding the right application

of statistical concepts and implementing them using statistical

softwares. Our Statistics Assignment and Homework help

service has been strategized and simplified to help students in

learning statistics problem solving. We use latest and genuine

tools & softwares to make statistics easy. We deliver step by step help with statistics

assignments which is self-explanatory to make students understand the method of solving

problems without any inconvenience.

Sample Statistics Assignment and Homework Help Service Questions:

Depreciation Sample Questions

Question-1: Find the trend line equation and obtain the trend values for the following data

using the method of the least square. Also, forecast the earning for 2006.

Year :

Earning in ’000 $

1997

38

1998

40

1999

65

2000

72

2001

69

2002

60

2003

87

2004

95

Solution. Here, the number of items being 8 (i.e. even), the time deviation X will be taken

as 𝒕−𝒎𝒊𝒅 𝒑𝒐𝒊𝒏𝒕 𝒐𝒇 𝒕𝒊𝒎𝒆𝟏

𝟐 𝒐𝒇 𝒕𝒊𝒎𝒆 𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍

to avoide the decimal numbers. Thus, the working will run as under:

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(a) Determination of the Trend Line Equation and the Trend Values

Working

The trend line equation is given by Y = a + bX

Where, a = 𝑌

𝑁 [ ∵ 𝑋𝑌 = Na + b 𝑋, and 𝑋=0]

= 526

8 = 65.75

And b = 𝑋𝑌

𝑋2 [ ∵ 𝑋𝑌 = a 𝑋 + b 𝑋2, and 𝑋=0]

= 616

168 = 3.67 approx.

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

as Yc = 65.75 + 3.67 X

Where ,trend origin is 2000.5,

Y unit = annual earning, and X unit = time deviation

Putting the respective values of X in the above equation, we get the different trend values

as under:

Trend Values

For 1997 When X = -7, Yc = 65.75 + 3.67 (-7) = 40.06

Year t

Earnings Y

Time dvn. i.e. 𝒕−𝟐𝟎𝟎𝟎.𝟓

𝟏/𝟐 ×𝟏

X

XY X2 Trend values Yc=65.75 + 3.67X

1997 1998 1999 2000 2000.5 (mid time)

38 40 65 72 - 69 60 87 95

-7 -5 -3 -1 0 1 3 5 7

-266 -200 -195 -72 0 69

180 435 665

49 25 9 1 0 1 9 25 49

40.06 47.40 54.74 62.08 A = 65.75 69.42 76.76 84.10 91.44

Total 526 0 616 168 N = 8

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1998 When X = -5, Yc = 65.75 + 3.67 (-5) = 47.40

1999 When X =-3, Yc = 65.75 + 3.67 (-3) = 54.74

2000 When X = -1, Yc = 65.75 + 3.67 (-1) = 62.08

2001 When X =1, Yc = 65.75 + 3.67 (1) = 69.42

2002 When X = 3, Yc = 65.75 + 3.67 (3) = 76.76

2003 When X = 5, Yc = 65.75 + 3.67 (5) = 84.10

2004 When X = 7, Yc = 65.75 + 3.67 (7) = 91.44

(b) Forecasting of earnings for 2006

For 2006, X = 𝒕−𝒎𝒊𝒅 𝒑𝒐𝒊𝒏𝒕 𝒐𝒇 𝒕𝒊𝒎𝒆𝟏

𝟐 𝒐𝒇 𝒕𝒊𝒎𝒆 𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍

= 2006−2000 .5

1

2×1

= 11

Thus, Yc = 65.75 + 3.67 (11) = 106.12

Hence, the earnings for 2005 is expected to be

= $ 106.12 × 100 = $ 106120

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Question-2: Obtain the straight line trend equation for the following data by the method of

the least square.

Year : Sales in ’000

$

1995 140

1997 144

1998 160

1999 152

2000 168

2001 176

2004 180

Also, estimate the sales for 2002

Solution. (a) Determination of the straight line trend equation by the method of least

square

Year t

Sales Y

Time dvn. i.e. t-1999 X

XY 𝑋2

1995 1997 1998 1999 2000 2001

2004

140 144 160 152 168 176

180

-4 -2 -1 0 1 2

5

-560 -288 -1 0 1 2

5

-16 4 1 0 1 4

25

Total 13994 1120 1 412 51 N = 7

Note. *In the above case, the average of the time variable is given by 𝑋 = 𝑡

𝑁 =

13994

7 =

1999 approx.

Hence, 1999 has been taken as the year of origin in the above table.

Working

The trend line equation is given by Yc = a + bX

Here, since 𝑋 ≠0, the value of the two constants a, and b are to be found out by solving

simultaneously the following two normal equations:

𝑌 = Na + b 𝑋

𝑋𝑌 = a 𝑋 + b 𝑋2

Substituting the respective values in the above we get

1120 = 7a + b

412 = a + 51b

Multiplying the eqn (ii) by 7 under the eqn (iii) and getting the same deducted from the

equation (i) we get

7a + b =1120

= − 7𝑎+357𝑏=2884

−356𝑏= −1764

b = 1764

356 = 4.96

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Putting the above value of b in the equation (i) we get,

7a + 4.96 = 1120

7a = 1120 – 4.96 = 1115.04

or a = 1115.04/7 = 159.29

Putting the above values of a and b in the format of the equation we get the straight line for

trend as under:

Yc = 159.29 + 4.96X

Where, the year of working origin = 1999,

Y unit = annual sales (in ’000 $) and

X unit = time deviations.

(b)Estimation of the Sale for 2002

For 2002, X = 2002 – 1999 =3

Thus when, X = 3, Yc = 159.29 + 4.96 (3)= 159.29 + 14.88 = 174.17

Hence, the sales for 2002 are expected to be 174.17 × 103 = $ 174170.

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Question-3: Production figures of a Textile industry are as follows:

Year Production (in units) :

1998 12

1999 10

2000 14

2001 11

2002 13

2003 15

2004 16

For the above data.

(i) Determine the straight line equation by change of the origin under the least square

method.

(ii) Find the trend values, and show the trend line on a graph paper, and

(iii) Estimate the production for 2005 and 2007.

Solution (i). Determination of the straight line equation by change of the origin under the

least square method.

Year T

Prodn. Y

Successive values of

time variable X

XY 𝑿𝟐 Trend values

T

1998 1999 2000 2001 2002

2003 2004

12 10 14 11 13

15 16

1 2 3 4 5

6 7

12 20 42 44 65

90 112

1 4 9 16 25

36 49

10.745 11.50 12.25 13.00 13.75

14.50 15.25

Total 𝑌 = 91 𝑋 =28 𝑋𝑌 = 385 𝑋2 = 140 N = 7

Note. The successive values of time variable X, have been taken as a matter of change of

the origin to reduce their magnitude for the sake of convenience in calculations.

Working

The straight line equation is given by Yc = a + bX

Here, since 𝑥 ≠ 0, we are to work out the values of the two constants, a and b by

simultaneous solution of the following two normal equations:

𝑌 = Na + b 𝑋

𝑋𝑌 = a 𝑋 + b 𝑋2

Substituting the respective values obtained from the above table in the above equation we

get,

91 = 7a + 28b

385 = 28a + 140b

Multiplying the eqn. (i) by 4 under the equation (iii) and subtracting the same from the eqn.

(ii) we get,

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28a + 140b = 385

= − 28𝑎+112𝑏=364

28𝑏=21

∴ b = 21

28 = .75

Putting the above value of b in the eqn. (i) we get,

7a + 28 (.75) = 91

= 7a = 91 – 21 = 70

∴ a = 70/7 = 10

Putting the above values of a and b in the relevant equation we get the straight line

equation naturalized as under:

Yc = 10 + 0.75 X

Where, X, represents successive values of the time variable, Y, the annual production and

the year of origin is 1997 the previous most year.

(ii) Calculation of the Trend values & Their Graphic Representation

1998 When X = 1, Yc = 10 + 0.75 (1) = 10.75

1999 When X =2, Yc = 10 + 0.75 (2) = 11.50

2000 When X = 3, Yc = 10 + 0.75 (3) = 12.25

2001 When X =4, Yc = 10 + 0.75 (4) = 13.00

2002 When X = 5, Yc = 10 + 0.75 (5) = 13.75

2003 When X = 6, Yc = 10 + 0.75 (6) = 14.50

2004 When X = 7, Yc = 10 + 0.75 (7)= 15.25

Graphic Representation of the Trend Line & the Original Data

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(i) Estimation the Production Figure for 2004 and 2006.

Since the last successive value of X for 2004 is 7, the successive values of X for 2005 and

2007 are 8 and 10 respectively.

Thus, for 2005, when X = 8, Yc = 10 + 0.75 (8) = 16.00

And for 2007, when X = 10, Yc = 10 + 0.75 (10) = 17.50

Hence, the estimate figures of production for 2005 and 2007 are 16000 units and 17500

units respectively.

Test of Suitability of Straight line Method

If the differences between the successive observations of a series are found to be constant,

or nearly so, the straight line model is considered to be a suitable measure for

representation of trend components, otherwise not. This fact can be determined by the

method of First Differences illustrated as under:

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Question-4. State by using the method of First Differences, if the straight line model is

suitable for finding the trend values of the following time series:

Year : Sales :

1997 30

1998 50

1999 72

2000 90

2001 107

2002 129

2003 147

2004 170

Solution. Determination of Suitability of the Straight line model by the method of

First Differences

Year T

Sales Y

First Differences

1997 1998 1999 2000 2001 2002 2003 2004

30 50 72 90 107 129 147 170

50-30 = 20 72 – 50 = 22 90 – 72 = 18 107 – 90 = 17 129 – 107 = 22 147 – 129 = 18 170 – 147 = 23

From the above table, it must be seen that the first differences in the successive

observations are almost constant by 20 or nearly so. Hence, the straight line model is quite

suitable for representing the trend components of the given series.

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Vi. Parabolic Method of the least square

This method of least square is used only when the trend of a series is not linear, but

curvilinear. Under this method, a curve of parabolic type is fitted to the data to obtain their

trend values and to obtain such a curve, an equation of power series is determined in the

following model.

Yc = a + bX + cX2 + dX3 + ….+ mXn

It may be noted that the above equation can be carried to any power of X according to the

nature of the series. If the equation is carried only up to the second power of X, (i.e. X2) it

is called the parabola of second degree, and if it is carried up to the 3rd power of X, (i.e.

X3) it is called the Parabola of 3rd degree. However, in actual practice the parabolic curve of

second degree is obtained in most of the cases to study the non-linear of a time series. For

this, the following equation is used.

Yc = a + bX +cX2

Where, Yc represents the computed trend value of the Y variable ‘a’ the intercept of Y, b’

the slope of the curve at the origin of X and c’, the rate of change in the slope.

In the above equation, a, b, and c are the three constants, the value of which are

determined by solving simultaneously the following three normal equations:

𝑌 = 𝑁𝑎 + 𝑏 𝑋 + 𝑐 𝑋3

It may be noted here that the first normal equation has been derived by multiplying each

set of the observed relationship by the respective coefficients of a, and getting them all

totaled ; the second normal equation has been derived by multiplying each set of the

observed relationship by the respective coefficients of b, and getting them all totaled ; and

the third equation has been derived by multiplying each set of the observed relationship by

the respective coefficients of c and getting them all totaled.

Further, it may be noted that by taking the time deviations from the midpoint of the time

variable, if 𝑋 and 𝑋3 could be made zero, the above three normal equations can be

reduced to the simplified forms to find the values of the relevant constants as follow :

From the above, it must be noticed that the value of b can be directly obtained as b = 𝑿𝒀

𝑿𝟐,

and the values of the other two constants a and c can be obtained by solving simultaneously

the rst of the following two normal equations:

a = 𝒀− 𝒄 𝑿𝟐

𝑿𝟒

c = 𝑿𝟐𝒀−𝒂 𝑿𝟐

𝑿𝟒

Once, the values of the three constants a, b and c are determined in the above manner, the

trend line equation can be fitted to obtain the trend values of the given time series by

simply substituting the respective values of X therein.

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