Introduction
I’m going to investigate how the condition of a car e.g. bodywork, mileage, age and any special features, affects the price. In my opinion the age of the car will have the biggest effect on its price, as with most cars, the older it becomes the more its price will depreciate. I’m going to draw a scatter graph to find the relationship between the price and the age.
I will investigate the first 30 cars on my datasheet except number 15 because it doesn’t have a price number. I have a secondary data sheet showing different information which may help me in my investigation.
In my opinion not only the age of the car will affect the car price, there are some other factors, which might also affect the car price, such as the colour, mileage, engine size and number of owners. The second factor I will investigate is the relationship between the price and the mileage, but this time I will choose 50 random numbers from my data sheet.
Now I’m going to draw two scatter graphs (Age VS % depreciation) and (Mileage VS % depreciation)
On my second graph some cars fall way above my line of best fit e.g. car no. 38 Volvo which is 11 years old but has a low mileage, car no.157 Daewoo which is 11 years old but has a low mileage, car no. 201 which is also 11 years old but has a low mileage and car ...
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Now I’m going to draw two scatter graphs (Age VS % depreciation) and (Mileage VS % depreciation)
On my second graph some cars fall way above my line of best fit e.g. car no. 38 Volvo which is 11 years old but has a low mileage, car no.157 Daewoo which is 11 years old but has a low mileage, car no. 201 which is also 11 years old but has a low mileage and car no.190 which is 10 years old but has a low mileage.
d = depreciation
a = age
m = gradient
c = intercept
d = ma + c m= h/b m=9.5, c=8,
- I worked the equation for my first graph which is: d= (9.5 x a) + 8
d = ma=c m= h/b m= 0.005, c= 24
- My equation for my second graph is: d= (0.0005 x m) + 24
% d= ((9.5 x a) + 8)
% d= ((0.0005 x m) + 24)
% depreciation = X ((9.5 x a) + 8) + Y ((0.0005 x m) + 24)
X and Y are constants
Car no – 11 is a Nissan Micra aged 11, its mileage 28000 and its
% deprecation is 88.
- When X = 1
- And Y= 0
1 ((9.5x 11) + 8) + 0((0.0005x 28000) + 8)
1(104.5+8) + 0(14+24)
= 112.5 = 0
112.5+0= TOO BIG
- When X = 0.9
- And Y = 0.1
0.9((9.5x11) + 8) + 0.1((0.0005x28000) +24)
0.9(104.5+8) + 0.1(14+24)
= 101.25 = 3.8
101.25+3.8=105.05 TOO BIG
- When X = 0.6
- And Y = 0.4
0.6((9.5x11) +8) + 0.4((0.0005x28000) +24)
0.6(104.5+8) + 0.4(14+24)
=67.5 =15.2
67.5+15.2= 82.7 too small
- When X = 0.7
- And Y = 0.3
0.7((9.5x11) +8) + 0.3((0.0005x28000) + 24)
0.7(104.5+8) + 0.3(14+24)
=78.75 = 11.4
78.75+11.4= 90.15 TOO BIG
- When X = 0.65
- And Y =0.35
0.65((9.5x11) +8) +
0.65(104.5+8)
= 73.125
73.125+13.3= 86.425 too small
- When X = 0.67
- And Y = 0.33
0.67((9.5x11) +8) +
0.67(104.5+8)
= 75.375
75.375+12.54=87.915
87.915~88
Therefore I have concluded that X = 0.67and Y = 0.33
I’m going to pick another 4 cars to find the value of X and Y as the previous formula.
- Car No – 196 is a Ford Explorer aged 6, it’s mileage 10000 and its % depreciation is 63.7
% depreciation = X ((9.5x6) +8) + Y ((0.0005x10000) +24)
- When X = 1
- And Y = 0
1((9.5x6) +8) + 0((0.0005x10000) +24)
1(57+8) 0(5+10000)
=65 = 0
65+0=65 TOO BIG
- When X = 0.9
- And Y = 0.1
0.9((9.5x6) +8) + 0.1((0.0005x10000) +24)
0.9(57+8) 0.1(5+24)
=58.5 = 2.9
58.5+2.9=61.4 too small
- When X = 0.95
- And Y = 0.05
0.95((9.5x6) +8) + 0.05((0.0005x10000) +24)
0.95(57+8) 0.05(5+24)
=61.57 = 1.45
61.4+1.45=62.85 too small
- When X = 0.96
- And Y = 0.04
0.96((9.5x6) +8) + 0.04((0.0005x10000) +24)
0.96(57+8) 0.04(5+24)
=62.4 =1.16
62.4+1.16=63.56 too small
- When X = 0.965
- And Y = 0.035
0.965((9.5x6) +8) + 0.035((0.0005x10000) +24)
0.965(57+8) 0.035(5+24)
=62.725 =1.015
62.725+1.015=63.74
63.74~63.7
Therefore I have concluded that X = 0.965 and Y = 0.035
- Car No – 59 is an Audi Coupe 88-96 aged 9, its mileage 27000 and its % depreciation is 70.
% depreciation = X ((9.5x9) +8) + Y ((0.0005x27000) +24)
- When X = 1
- And Y = 0
1((9.5x9) +8) + 0((0.0005x27000) +24)
1(85.5+8) 0(13.5+24)
=93.5 = 0
93.5+0=93.5 TOO BIG
- When X = 0.8
- And Y = 0.2
0.8((9.5x9) +8) + 0.2((0.0005x27000) +24)
0.8(85.5+8) 0.2(13.5+24)
= 74.8 =7.5
74.8+7.5=82.3 TOO BIG
- When X = 0.6
- And Y = 0.4
0.6((9.5x9) +8) + 0.4((0.0005x27000) +24)
0.6(85.5+8) 0.4(13.5+24)
= 56.1 =15
56.1+15=71.1 TOO BIG
- When X = 0.58
- And Y = 0.42
0.58((9.5x9) +8) + 0.42((0.0005x27000) +24)
0.58(85.5+8) 0.42(13.5+24)
= 54.23 =15.75
54.23+15.75=69.98
69.98~70
Therefore I have concluded that X = 0.58 and Y = 0.42
- Car No – 89 is a Mercedes CLK 97 – 03 aged 9, its mileage 3700 and its % depreciation is 65.
% depreciation = X ((9.5x9) +8) + Y ((0.0005x37000) +24)
- When X= 1
- And Y = 0
1((9.5x9) +8) + 0((0.0005x37000) +24)
1(85.5+8) 0(18.5+24)
=93.5 = 0
93.5+0=93.5 TOO BIG
- When X= 0.7
- And Y = 0.3
0.7((9.5x9) +8) + 0.3((0.0005x37000) +24)
0.7(85.5+8) 0.3(18.5+24)
=65.45 =12.75
65.45+12.75=78.2 TOO BIG
- When X= 0.5
- And Y = 0.5
0.5((9.5x9) +8) + 0.5((0.0005x37000) +24)
0.5(85.5+8) 0.5(18.5+24)
=46.75 =21.25
46.75+21.25=68 TOO BIG
- When X= 0.45
- And Y = 0.55
0.45((9.5x9) +8) + 0.55((0.0005x37000) +24)
0.45(85.5+8) 0.55(18.5+24)
=42.075 =23.375
42.075+23.375=65.45 TOO BIG
- When X= 0.445
- And Y = 0.555
0.445((9.5x9) +8) + 0.555((0.0005x37000) +24)
0.445(85.5+8) 0.555(18.5+24)
=41.6075 =23.5875
41.6075+23.5875=65.195
65.195~65
Therefore I have concluded that X = 0.445 and Y = 0.555
- Car No – 173 is a Mazda Premacy aged 7, it’s mileage 18000and its % depreciation is 69.
% depreciation = X ((9.5x7) +8) + Y ((0.0005x18000) +24)
- When X = 1
- And Y= 0
1((9.5x7) +8) + 0((0.0005x18000) +24)
1(66.5+8) 0(9+24)
=74.5 =0
74.5+0=74.5 TOO BIG
- When X = 0.8
- And Y= 0.2
0.8((9.5x7) +8) + 0.2((0.0005x18000) +24)
0.8(66.5+8) 0.2(9+24)
=59.6 =6.6
59.6+6.6=66.2 too small
- When X = 0.9
- And Y= 0.1
0.9((9.5x7) +8) + 0.1((0.0005x18000) +24)
0.9(66.5+8) 0.1(9+24)
=67.05 =3.3
67.05+3.3=70.35 TOO BIG
- When X = 0.88
- And Y= 0.12
0.88((9.5x7) +8) + 0.12((0.0005x18000) +24)
0.88(66.5+8) 0.12(9+24)
=65.56 =3.96
65.56+3.96=69.52 TOO BIG
- When X = 0.87
- And Y= 0.13
0.87((9.5x7) +8) + 0.13((0.0005x18000) +24)
0.87(66.5+8) 0.13(9+24)
=64.815 =4.29
64.815+4.29=69.105
69.105~69
Therefore I have concluded that X = 0.87and Y= 0.13
I will now work out the average value of X and Y by:
- Adding all the Xs together and dividing by 5
- Adding all the Ys together and dividing by 5
- X = 0.67and Y = 0.33
- X = 0.965 and Y = 0.035
- X = 0.58 and Y = 0.42
- X = 0.445 and Y = 0.555
- X = 0.87and Y= 0.13
- 0.67 + 0.965+ 0.58 + 0.445 + 0.87 = 3.53
3.53/5 = 0.706 ~ 0.71
So X= 0.71
- 0.33 + 0.035 + 0.42 + 0.555 + 0.13= 1.47
1.47/5 = 0.294~ 0.29
So Y=0.29
