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Help with Statistics Assignment

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David Luke

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Email: [email protected]

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About Help with Statistics Assignment:

We have excellent statistics assignment tutors to help you in

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

expecting a higher grade in your assignments. As far as software

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΅

πππ

π

[β΅

ππ = a π + b π 2 and

π = 0]

= 42.86 approx.

Copyright © 2010-2015 Tutorhelpdesk.com

And

b=

=

ππ

π2

250

8

[β΅

ππ = 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.

Copyright © 2010-2015 Tutorhelpdesk.com

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−πππ πππππ ππ ππππ

π

ππ ππππ ππππππππ

π

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

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

Copyright © 2010-2015 Tutorhelpdesk.com

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−ππππ.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

πππ

π

b=

=

πππ

πππ

π

π΅

[β΅

ππ = Na + b π, and

π=0]

= 65.75

πΏπ

πΏπ

[β΅

ππ = 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:

Copyright © 2010-2015 Tutorhelpdesk.com

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−πππ πππππ ππ ππππ

π

ππ ππππ ππππππππ

π

=

ππππ−ππππ.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

Copyright © 2010-2015 Tutorhelpdesk.com

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

π2

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΅

=

πππππ

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

ππ = 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

=

− ππ+ππππ=ππππ

−ππππ= −ππππ

b=

1764

356

= 4.96

Copyright © 2010-2015 Tutorhelpdesk.com

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.

Copyright © 2010-2015 Tutorhelpdesk.com

Help with Statistics Assignment

Tutorhelpdesk

David Luke

Contact Us:

Phone: (617) 807 0926

Web: http://www.tutorhelpdesk.com

Email: [email protected]

Facebook: https://www.facebook.com/Tutorhelpdesk

Twitter: http://twitter.com/tutorhelpdesk

Blog: http://tutorhelpdesk.blogspot.com

About Help with Statistics Assignment:

We have excellent statistics assignment tutors to help you in

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

expecting a higher grade in your assignments. As far as software

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΅

πππ

π

[β΅

ππ = a π + b π 2 and

π = 0]

= 42.86 approx.

Copyright © 2010-2015 Tutorhelpdesk.com

And

b=

=

ππ

π2

250

8

[β΅

ππ = 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.

Copyright © 2010-2015 Tutorhelpdesk.com

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−πππ πππππ ππ ππππ

π

ππ ππππ ππππππππ

π

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

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

Copyright © 2010-2015 Tutorhelpdesk.com

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−ππππ.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

πππ

π

b=

=

πππ

πππ

π

π΅

[β΅

ππ = Na + b π, and

π=0]

= 65.75

πΏπ

πΏπ

[β΅

ππ = 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:

Copyright © 2010-2015 Tutorhelpdesk.com

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−πππ πππππ ππ ππππ

π

ππ ππππ ππππππππ

π

=

ππππ−ππππ.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

Copyright © 2010-2015 Tutorhelpdesk.com

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

π2

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΅

=

πππππ

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

ππ = 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

=

− ππ+ππππ=ππππ

−ππππ= −ππππ

b=

1764

356

= 4.96

Copyright © 2010-2015 Tutorhelpdesk.com

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.

Copyright © 2010-2015 Tutorhelpdesk.com