Help With Statistics Assignment

Help with Statistics Assignment

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Sample Statistics Assignment Questions and Answer:

Depreciation Sample Question

Question-1. 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]

=

= 42.86 approx.

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


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

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.


Question-2. 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

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

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


Working

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

Where, a =

[ = Na + b , and =0]

=

= 65.75

And b =

[ = a + b 2, and =0]

=

= 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:

Year t

Earnings Y

Time dvn. i.e. .

/

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


Trend Values

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

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 =

= .

= 11

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

Hence, the earnings for 2005 is expected to be = $ 106.12 100 = $106120


Question-3. 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 2

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 =

=

=

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

simultaneously the following two normal equations:

= Na + b

= a + b 2

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

= +=

=

b = 1764

356 = 4.96


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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