Data Analysis and Forecasting: A Practical Approach

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DATA ANALYSIS AND FORECASTING
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Contents
Answer 1......................................................................................................................................................3
Answer 2......................................................................................................................................................4
Answer 3......................................................................................................................................................5
Answer 4....................................................................................................................................................11
References.................................................................................................................................................16
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Answer 1
Weather report of London, Great Britain form 1st may 2018 to 10th may 2018 is shown below which
includes four parameters, namely – temperature, humidity, windspeed and pressure. This data was
gathered with the help of online weather reports.
Date Temperature (°F) Windspeed (mph) Humidity (%) Pressure (Hg)
01-May 49.3 13.1 62 29.7
02-May 49.8 14.7 78.6 29.7
03-May 50.3 6.6 67.5 30
04-May 55.9 4.8 64.1 30.2
05-May 55.2 6.3 64.5 30.2
06-May 57.7 5.3 67.3 30.1
07-May 60.4 5.2 70.9 30
08-May 65.2 6.6 64 29.7
09-May 60.5 8.6 67.6 29.7
10-May 55.1 8.3 64.8 29.9
In above table, temperature is in degree Fahrenheit, windspeed is in miles per hour, humidity is in
percentage and pressure is in mercury (Weather Underground et al, 2019).
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Answer 2
Data presented in answer 1 can be graphically visualized with the help of charts. For better representation
of the relationships of all the parameters with the number of days, a column chart and a line chart are
prepared.
01-May 02-May 03-May 04-May 05-May 06-May 07-May 08-May 09-May 10-May
0
10
20
30
40
50
60
70
80
90
Column chart
Temperature (°C) Windspeed (mph) Humdity (%) Pressure (Hg)
01-May 02-May 03-May 04-May 05-May 06-May 07-May 08-May 09-May 10-May
0
10
20
30
40
50
60
70
80
90
Line Chart
Temperature (°C) Windspeed (mph) Humdity (%) Pressure (Hg)
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Answer 3
(i) Mean
It the ratio of sum of all numbers in a set to the total numbers in a set.
Mean = ∑ of all numbers∈a set
Total number ∈a set
For different variables, different values of mean will be calculated.
For temperature,
Mean = 49.3 + 49.8 + 50.3 + 55.9 + 55.2 + 57.7 + 60.4 + 65.2 + 60.5 + 55.1
10
= 559.4
10 = 55.94
For windspeed,
Mean = 13.1 + 14.7 + 6.6 + 4.8 + 6.3 + 5.3 + 5.2 + 6.6 + 8.6 + 8.3
10
= 79.5
10 = 7.95
For humidity,
Mean = 62 + 78.6 + 67.5 + 64.1 + 64.5 + 67.3 + 70.9 + 64 + 67.6 + 64.8
10
= 671.3
10 = 67.13
For pressure,
Mean = 29.7 + 29.7 + 30 + 30.2 + 30.2 + 30.1 + 30 + 29.7 + 29.7 + 29.9
10
= 299.2
10 = 29.92
(ii) Median
For temperature data,
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Ascending order is 49.3, 49.8, 50.3, 55.1, 55.2, 55.9, 57.7, 60.4, 60.5, 65.2
Median = 55.2 + 55.9
2 = 55.55
(Data36 et al, 2019).
For windspeed data,
Ascending order is 4.8, 5.2, 5.3, 6.3, 6.6, 6.6, 8.3, 8.6, 13.1, 14.7
Median = 6.6 + 6.6
2 = 6.6
For humidity data,
Ascending order is 62, 64, 64.1, 64.5, 64.8, 67.3, 67.5, 67.6, 70.9, 78.6
Median = 67.3 + 64.8
2 = 66.05
For pressure data,
Ascending order is 29.7, 29.7, 29.7, 29.7, 29.9, 30, 30, 30.1, 30.2, 30.2
Median = 29.9 + 30
2 = 29.95
(iii) Mode
For temperature data,
Temperat
ure (°C)
Frequen
cy
49.3 1
49.8 1
50.3 1
55.1 1
55.2 1
55.9 1
57.7 1
60.4 1
60.5 1
65.2 1
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As all numbers are occurring with same frequency, so there will be NO mode.
For windspeed data,
Windspeed (mph) Frequenc
y
4.8 1
5.2 1
5.3 1
6.3 1
6.6 2
8.3 1
8.6 1
13.1 1
14.7 1
Here 6.6 is the most occurred number, so mode is 6.6.
For humidity data,
Humidity (%) Frequenc
y
62 1
64 1
64.1 1
64.5 1
64.8 1
67.3 1
67.5 1
67.6 1
70.9 1
78.6 1
As all numbers are occurring with same frequency, so there will be NO mode.
For pressure data,
Pressure (Hg) Frequenc
y
29.7 4
29.9 1
7

30 2
30.1 1
30.2 2
Here 29.7 is the most occurred number, so mode is 29.7
(iv) Range:
For temperature data,
Highest number = 65.2
Lowest number = 49.3
Range = 65.2 – 49.3 = 15.9
For windspeed data,
Highest number = 14.7
Lowest number = 4.8
Range = 14.7 – 4.8 = 9.9
For humidity data,
Highest number = 78.6
Lowest number = 62
Range = 78.6 – 62 = 16.6
For pressure data,
Highest number = 30.2
Lowest number = 29.7
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Range = 30.2 – 29.7 = 0.5
(v) Standard deviation:
Standard deviation = √ ∑ ¿ ¿ ¿ ¿
Where x is the number from the set
x bar is the mean of the set
and n is the total count of numbers
For temperature data,
consider temperature as x
x x - x bar (x-x bar) ^2
49.3 -6.64 44.0896
49.8 -6.14 37.6996
50.3 -5.64 31.8096
55.9 -0.04 0.0016
55.2 -0.74 0.5476
57.7 1.76 3.0976
60.4 4.46 19.8916
65.2 9.26 85.7476
60.5 4.56 20.7936
55.1 -0.84 0.7056
x bar = 55.94 244.384
Standard deviation = √ 244.384
9 = 5.2109
For windspeed data,
Consider windspeed as x
x x - x bar (x-x bar) ^2
13.1 5.15 26.5225
14.7 6.75 45.5625
6.6 -1.35 1.8225
4.8 -3.15 9.9225
6.3 -1.65 2.7225
9

5.3 -2.65 7.0225
5.2 -2.75 7.5625
6.6 -1.35 1.8225
8.6 0.65 0.4225
8.3 0.35 0.1225
x bar = 7.95 103.505
Standard deviation = √ 103.505
9 = 3.391247
For humidity data,
Consider humidity as x
x x - x bar (x-x bar) ^2
62 -5.13 26.3169
78.6 11.47 131.5609
67.5 0.37 0.1369
64.1 -3.03 9.1809
64.5 -2.63 6.9169
67.3 0.17 0.0289
70.9 3.77 14.2129
64 -3.13 9.7969
67.6 0.47 0.2209
64.8 -2.33 5.4289
x bar = 67.13 203.801
Standard deviation = √ 203.801
9 = 4.75863
For pressure data,
Consider pressure as x,
x x - x bar (x-x bar) ^2
29.7 -0.22 0.0484
29.7 -0.22 0.0484
30 0.08 0.0064
30.2 0.28 0.0784
30.2 0.28 0.0784
30.1 0.18 0.0324
30 0.08 0.0064
10

29.7 -0.22 0.0484
29.7 -0.22 0.0484
29.9 -0.02 0.0004
x bar = 29.92 0.396
Standard deviation = √ 0.396
9 = 0.209762
(Zady et al, 2019).
Answer 4
For linear forecasting model, equation can be written as
y = mx + c
where m and c are constants
m and c can be solved using formulas:
m = ∑ (x-x bar)(y-y bar)
∑ (x-x bar)2
and c = y ¯-m*x bar
Where x is the number from the set
x bar is the mean of the set
For temperature data,
Consider number of days as x and temperature as y.
x y x-x
bar
y-y bar (x-x bar) (y-y
bar)
(x-x bar)
^2
1 49.3 -5 -6.64 30 20.25
2 49.8 -4 -6.14 21 12.25
3 50.3 -3 -5.64 14 6.25
4 55.9 -2 -0.04 0 2.25
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5 55.2 -1 -0.74 0 0.25
6 57.7 1 1.76 1 0.25
7 60.4 2 4.46 7 2.25
8 65.2 3 9.26 23 6.25
9 60.5 4 4.56 16 12.25
10 55.1 5 -0.84 -4 20.25
x bar =
5.5
y bar =
55.94
Total 109 82.5
So, m = 109
82.5 = 1.319
And c = 55.94 – (1.319 x 5.5) = 48.687
Putting both constants in liner equation,
y = 1.319x + 48.687
using this equation, we can forecast temperature on future days.
For day 15th,
Put x = 15 in liner equation,
We get y = (1.319 * 15) + 48.687 = 68.468
For day 23rd,
Put x = 23 in liner equation,
We get y = (1.319 * 23) + 48.687 = 79.024
For windspeed data,
Consider number of days as x and windspeed as y.
x y x-x
bar
y-y bar (x-x bar) (y-y
bar)
(x-x bar)
^2
1 13.1 -5 5.15 -23 20.25
2 14.7 -4 6.75 -24 12.25
3 6.6 -3 -1.35 3 6.25
4 4.8 -2 -3.15 5 2.25
5 6.3 -1 -1.65 1 0.25
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