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Statistics Tutor and Assignment Illustrations and Solutions
Illustration 1. 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
-
-

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(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) =
𝑿 𝟏 𝑾 𝟏+ 𝑿 𝟐 𝑾 𝟐+ 𝑿 𝟑 𝑾 𝟑
𝑾 𝟏+𝑾 𝟐+𝑾 𝟐
=
𝑿𝑾
𝑾

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Illustration 2. From the following data, calculate the 3 yearly weighted moving averages,
the weights being 1, 2 and 3 respectively.
Year :
Value :
1998
2
1999
6
200
3
2001
5
2002
7
2003
4
2004
2
Solution. Computation of the 3 yearly weighted moving averages (Weights being 1, 2 and
3 respectively)
Year Value
X
3 Yearly weighted
moving totals ( 𝑿𝑾)
3 yearly weighted
moving averages
i.e. ( 𝑿𝑾/ 𝑾)
1998
1999
2000
2001
2002
2003
2004
2
6
3
5
7
4
2
-
(2 × 1 + 6 × 2 + 3 ×
3) = 23
(6 × 1 + 3 × 2 + 5 ×
3) = 27
(3 ×1 + 5 × 2 + 7 ×
3) = 34
(5 × 1 + 7 × 2 + 4 ×
3) = 31
(7 × 1 + 4 × 2 + 2 ×
3) = 21
-
-
23/6 = 3.83
27/6 = 4.50
34/6 = 5.67
31/6 = 5.17
21/6 = 3.50
-
Determination of Trend Values through Moving Average Method
The moving average method discussed above can be used as a simple device of reducing
the fluctuations, and of obtaining the trend values in a time series with a fair degree of
accuracy. Under this method, the moving averages calculated represent the trend values for
the middle point of the period of moving averages. The moving averages when plotted on a
graph paper gives us a trend line in the form of a smoothed curve by reducing the
fluctuations in her time series.

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Illustration 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

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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

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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
Aliter
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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Illustration 4. 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 =
𝒀
𝑵
[∵ 𝑋𝑌 = a 𝑋 + b 𝑋2
and 𝑋 = 0]
=
300
7
= 42.86 approx.
And b =
𝑋𝑌
𝑋2
[∵ 𝑋𝑌 = a 𝑋 + b 𝑋2
and 𝑋 = 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
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

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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.

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