# Help With Statistics Assignment

of 8

## Content

Tutorhelpdesk

Help with Statistics Assignment

Tutorhelpdesk
David Luke

Phone: (617) 807 0926
Web: http://www.tutorhelpdesk.com
Email: [email protected]
Blog: http://tutorhelpdesk.blogspot.com

the field of econometrics, applied statistics, quantitative methods,
mathematical statistics, business statistics and operations research.
We follow a strategic approach in solving and presenting the
assignment. With the comprehensive use of tools, graphs,
histograms, pie charts, diagrams, tables you can be rest assured of
based assignments are concerned we provide analysis and
explanation in a word file along with necessary code file and output
file. We also assist students in their online statistics quizzes, exams and homework.
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
Days

Sales
Y

Time dvn.
From mid
value 4

XY

X2

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

Trend
values
=42.86
+ 8.93X
16.7
25.00
33.93
42.86
51.79
60.72
69.65

Total

300

0

140

N=7

0.00

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

π
π΅
πππ
π

[β΅

ππ = a π + b π 2 and

π = 0]

= 42.86 approx.

And

b=
=

ππ
π2
250
8

[β΅

ππ = a π + b π 2 and

π = 0]

= 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
π−πππ πππππ ππ ππππ
π
ππ ππππ ππππππππ
π

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

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

Year
t

Earnings
Y

Time
dvn.
i.e.

XY

X2

Trend
values
Yc=65.75 +
3.67X

266
200
195
-72
0
69
180
435
665
616

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

168

N=8

π−ππππ.π
π/π ×π

1997
1998
1999
2000
2000.5 (mid
time)

38
40
65
72
69
60
87
95

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

Total

526

0

Working
The trend line equation is given by Y = a + bX
Where, a =
=
And

πππ
π

b=
=

πππ
πππ

π
π΅

[β΅

ππ = Na + b π, and

π=0]

= 65.75
πΏπ
πΏπ

[β΅

ππ = a π + b π 2 , and

π=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:

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 =

π−πππ πππππ ππ ππππ
π
ππ ππππ ππππππππ
π

=

ππππ−ππππ.π
π
×π
π

= 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 :
1995
1997
Sales in ’000 140
144
\$:
Also, estimate the sales for 2002

1998
160

1999
152

2000
168

2001
176

2004
180

Solution
(a) Determination of the straight line trend equation by the method of least square
Year
t
1995
1997
1998
1999
2000
2001
2004
Total 13994

Sales
Y
140
144
160
152
168
176
180
1120

Time dvn.
i.e. t-1999 X
-4
-2
-1
0
1
2
5
1

π2

XY
-560
-288
-1
0
1
2
5
412

-16
4
1
0
1
4
25
51

Note. *In the above case, the average of the time variable is given by π =

N=7
π
π΅

=

πππππ
π

=

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 π ≠0, the value of the two constants a, and b are to be found out by solving
simultaneously the following two normal equations:
π = Na + b π
ππ = a π + b π 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

=

− ππ+ππππ=ππππ
−ππππ= −ππππ

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.

## Recommended

#### Statistics Homework Help

Or use your account on DocShare.tips

Hide