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Used car prices.
Extracts from this document...
Introduction
Used car prices
Introduction
During this coursework I will be working from the form the data that has been given to me. It is established on figuring out and representing the data in different forms. The table below is the given data that is to be interpreted:
Car  Make  Model  Price when new  Second hand price  Age (Years)  Mileage  Engine Size  
1  Ford  Orion  16000  7999  1  7000  1.8  
2  Mercedes  A140 Classic  14425  10999  1  14000  1.4  
3  Vauxhall  Vectra  18580  7999  2  20000  2.5  
4  Vauxhall  Astra  14325  6595  4  30000  1.6  
5  Nissan  Micra  7995  3999  3  37000  1  
6  Renault  Megane  13610  4999  4  33000  1.6  
7  Mitsubishi  Carisma GDI  14875  5999  2  24000  1.8  
8  Rover  623 Gsi  22980  6999  4  30000  2.3  
9  Renault  Megane  13175  6999  3  41000  1.6  
10  Vauxhall  Tigra  13510  7499  4  27000  1.4  
11  Fiat  Bravo  10351  3495  5  51000  1.4  
12  Vauxhall  Vectra  18140  6499  4  49000  2.5  
13  BMW  525i SE  28210  5995  8  55000  2.5  
14  Vauxhall  Corsa  8900  4995  2  24000  1.6  
15  Fiat  Punto  8601  3995  4  31000  1.2  
16  Rover  820 SLi  21586  3795  6  51000  2  
17  Mitsubishi  Carisma  15800  5999  2  33000  1.8  
18  Fiat  Cinquecento  6009  1995  6  20000  0.9  
19  Rover  416i  13586  3795  6  49000  1.6  
20  Nissan  Micra  6295  1795  8  47000  1.2  
21  Daewoo  Lanos  11225  5999  3  42000  1.6  
22  Rover  114 Sli  8595  2495  6  33000  1.4  
23  Ford  Escort  8785  1595  7  68000  1.3  
24  Fiat  Uno  6864  1495  8  51000  1  
25  Rover  Metro  6645  895  7  43000  1.1  
26  Vauxhall  Nova  5599  1000  10  75000  1.4  
27  Toyota  Corrolla  13800  7495  2  25000  2  
28  Vauxhall  Cavalier  10150  850  10  73000  1.6  
29  Volkswagen  Golf  400  15  1.4  
30  Volkswagen  Golf  9524  3695  7  49000  1.4  
31  Seat  Ibiza  5995  795  7  45000  0.9  
32  Rover  214i  9565  1700  8  55000  1.4  
33  Ford  Fiesta  7310  1050  8  90000  1.1  
34  Fiat  Tempra  10423  1295  6  81000  1.6  
35  Ford  Fiesta  7875  1495  11  74000  1.8  
36  Hyundai  Sonnata  11598  1195  9  65000  2  
37  Renault  Clio  6795  1995  8  47000  1.2  
38  Citroen  Debut  5715  1495  7  50000  0.95  
39  Renault  Clio  7403  1495  9  98000  1.2  
40  Fiat  Tipo  8272  1500  7  32000  1.4  
41  BMW  316i  13650  6995  6  71000  1.8  
42  Citroen  AX Diesel  7680  1295  6  1.5  
43  Ford  Fiesta LX  8748  1995  7  60000  1.1  
44  Nissan  Micra  5340  1595  9  40000  1.1  
45  Ford  Escort Duet  9105  2300  7  64000  1.4  
46  Nissan  Sunny  7799  2595  7  41000  1.4  
47  Vauxhall  Astra  13740  2900  6  58000  1.6  
48  Hyundai  Accent  6899  2800  6  49000  1.3  
49  Daewoo  Nubira  13850  6895  4  14730  2  
50  Daewoo  Lanos  9525  4395  3  32400  1.4  
51  Volkswagen  Golf  12999  3595  6  58000  1.8  
52  Ford  Escort  12125  4295  3  29000  1.4  
53  Ford  Escort  11800  4700  5  34000  1.8  
54  Bentley  TurboR  170841  37995  8  55000  2  
55  Fiat  Punto  7864  4500  3  13000  1.2  
56  Lexus  LS400  39728  6250  7  4  
57  Ford  Fiesta  8680  3200  4  27000  1.8  
58  Nissan  Almera  12590  4300  4  17000  1.6  
59  Rover  623GSi  24086  2975  5  96000  2.3  
60  Rover  620Si  17795  3400  5  66000  2  
61  Fiat  Bravo  10954  6795  1  3000  1.2  
62  Peugot  406LX  13975  5795  3  53000  1.8  
63  Volkswagen  Golf GTi  16139  6995  5  35000  2  
64  Ford  Focus  14505  8800  2  7200  1.6  
65  Ford  Puma  13230  8250  3  34000  1.4  
66  Peugot  206  9125  7500  1  18000  1.1  
67  Peugot  406  ...read more.
Middle
Midinterval (m) f*m 4 ≤ x < 6  0 0 5 0 6 ≤ x < 8 III 3 3 7 21 8 ≤ x < 10 IIII 5 8 9 45 10 ≤ x < 12 I 1 9 11 11 12 ≤ x < 14 III 3 12 13 39 14 ≤ x < 16 II 2 14 15 30 16 ≤ x < 18 II 2 16 17 34 I shall firstly find out the mode that is also known as the modal. The mode is the highest frequency. By looking at the table above, in the frequency column, it can be easily seen that 5 is the highest frequency. Therefore mode is equal to £8k ≤ x < £10k. This shows me that this is the most frequent price. It is the central tendency from all the Ford cars. Secondly I shall calculate the range. This will show me the price between the lowest and the highest price of this car make. The range can be calculated as the following: Range = Highest Price – Lowest Price Range = 18 – 4 = £14k The mean can be found by dividing the number of cars by means of the number of makes. This can be written as the following: Mean = ∑ Frequency × Midinterval ∑ Frequency Mean = 21 + 45 + 11 + 39 + 30 + 34 ÷ 16 Mean = £11.25kThe median represents the middle value out of collection of numbers. Finding the median is very straightforward. The frequency column that was shown above can find the median. The frequency should always arrange from smallest to largest. If there were a large amount of data that is present in the frequency column it would be very time consuming to put the column in ascending order. Therefore an easier step can be taken into deliberation. This has been shown below with the formula that is used at all times to calculate the median: Median = n + 1 ÷ 2 Where n is equal to highest cumulative frequency Median = n + 1 ÷ 2 Median = 16 + 1 ÷ 2 Median = 17 ÷ 2 Median = 8.5 Now when this median is found this number is taken and the cumulative frequency column is looked at. It can be noticed that 8.5 lies between 8 and 9. It can now be said that the median is equal to 10 ≤ x < 12. Therefore, Median = 10 ≤ x < 12 Lower Quartile, Upper Quartile and Inter Quartile Range – Ford To make my estimations accurate I will find out the lower quartile, upper quartile and inter quartile range. Median = n + 1 ÷ 2 Median = 16 + 1 ÷ 2 Median = 8.5 I now looked at 8.5 on the yaxis (cumulative frequency) of the graph on the next page and dropped a perpendicular line until it touched the plotted line. Therefore Median is the value that is obtained on the xaxis. This has been shown on the next page. Hence, Median is equal to £10.5kTo find the lower quartile (LQ) I will use the following formula: ¼ (n + 1) Where ‘n’ again is known as the highest cumulative frequency. ¼ (n + 1) ¼ (16 + 1) ¼ (17) 17 ÷ 4 = 4.25 I now looked at 4.25 on the yaxis (cumulative frequency) of the graph on the next page and dropped a perpendicular line until it touched the plotted line. Therefore LQ is the value that is obtained on the xaxis. This has been shown on the next page It is now known that the lower quartile is equal to £8.4k. This is the value of onequarter way into the division. The same steps are to be considered while finding out the upper quartile, but this time the following formula was used: Upper Quartile = ¾ (n + 1) ¾ (16 + 1) ¾ (17) 17 х 3 ÷ 4 51 ÷ 4 12.75 The same steps were taken into deliberation. This is the part of three quarters of the way into the division. The value that I obtained on the xaxis is £15k. UQ = £14.5k Now with the known upper and lower quartile I can determine calculating the interquartile range. The following formula shows how to calculate the interquartile range: Inter quartile range (IQR) = Upper quartile – Lower quartile IQR = 14.5 – 8.4 I.Q.R = £6.1k Mean, mode, median and range – VauxhallI now have shown the mean, mode, median and range of a specific make of car. I have taken Vauxhall into consideration. The mean, mode and median will be found by using the prices of Vauxhall cars when new. From the following table the mean, mode, median and range can be found:
I shall firstly find out the mode that is also known as the modal. The mode is the highest frequency. By looking at the table above, in the frequency column, it can be seen that 3 is the highest frequency because there are two numbers that are 3. This is known as a bimodal. This basically means that there are two highest frequencies in the Vauxhall model. Therefore mode is equal to £12k ≤ x < £14k and £18k ≤ x < £20k. This shows me that this is the most frequent price. It is the central tendency from all the Vauxhall cars. Secondly I shall calculate the range. The range can be calculated as the following: Range = Highest Price – Lowest Price Range = 20 –4 = £16k The mean can be found by dividing the number of cars by means of the number of makes. This can be written as the following: Mean = ∑ Frequency × Midinterval ∑ Frequency Mean = 5 + 14 + 18 + 11 + 39 + 15 + 0 + 57 ÷ 19 Mean = £8.37kThe median represents the middle value out of collection of numbers. Finding the median is very straightforward. The frequency column that was shown above can find the median. The frequency should always arrange from smallest to largest. If there were a large amount of data that is present in the frequency column it would be very time consuming to put the column in ascending order. Therefore an easier step can be taken into deliberation. This has been shown below with the formula that is used at all times to calculate the median: Median = n + 1 ÷ 2 Where n is equal to highest cumulative frequency Median = n + 1 ÷ 2 Median = 13 + 1 ÷ 2 Median = 14 ÷ 2 Median = 7 Now when this median is found this number is taken and the cumulative frequency column is looked at. It can be noticed that 7 lies between 6 and 18. It can now be said that the median is equal to 12 ≤ x < 14. Therefore, Median = £12k ≤ x < £14k Lower Quartile, Upper Quartile and Inter Quartile Range – Vauxhall To make my estimations accurate I will find out the lower quartile, upper quartile and inter quartile range. Median = n + 1 ÷ 2 Median = 13 + 1 ÷ 2 Median = 7 I now looked at 7 on the yaxis (cumulative frequency) of the graph on the next page and dropped a perpendicular line until it touched the plotted line. Therefore Median is the value that is obtained on the xaxis. This has been shown on the next page. Hence, Median is equal to £13kTo find the lower quartile (LQ) I will use the following formula: ¼ (n + 1) Where ‘n’ again is known as the highest cumulative frequency. ¼ (n + 1) ¼ (13 + 1) ¼ (14) 14 ÷ 4 = 3.5 I now looked at 3.5 on the yaxis (cumulative frequency) of the graph on the next page and dropped a perpendicular line until it touched the plotted line. Therefore LQ is the value that is obtained on the xaxis. This has been shown on the next page It is now known that the lower quartile is equal to £8.5k. This is the value of onequarter way into the division. The same steps are to be considered while finding out the upper quartile, but this time the following formula was used: Upper Quartile = ¾ (n + 1) ¾ (13 + 1) ¾ (14) 14 х 3 ÷ 4 42 ÷ 4 10.5 The same steps were taken into deliberation. This is the part of three quarters of the way into the division. The value that I obtained on the xaxis is £18.5k. UQ = £18.5k Now with the known upper and lower quartile I can determine calculating the interquartile range. The following formula shows how to calculate the interquartile range: Inter quartile range (IQR) = Upper quartile – Lower quartile IQR = 18.5 – 8.5 I.Q.R = £10k Mean, mode, median and range – FiatThe third make of car I shall investigate is Fiat. I have shown a table below, with all the necessary details that I will be using during the calculations of discovering the mean, mode, median and range.
I shall firstly find out the mode that is also known as the modal. The mode is the highest frequency. By looking at the table above, in the frequency column, it can be seen that 4 is the highest frequency because there are two numbers that are 3. This is known as a bimodal. This basically means that there are two highest frequencies in the Fiat model. Therefore mode is equal to £6k ≤ x < £8k and £10k ≤ x < £12k. This shows me that this is the most frequent price. It is the central tendency from all the Fiat cars. Secondly I shall calculate the range. This will show me the price between the lowest and the highest price of this car make. The range can be calculated as the following: Range = Highest Price – Lowest Price Range = 12 – 4 = £8k The mean can be found by dividing the number of cars by means of the number of makes. This can be written as the following: Mean = ∑ Frequency × Midinterval ∑ Frequency Mean = 0 + 28 + 18 + 44 ÷ 10 Mean = £9kThe median represents the middle value out of collection of numbers. Finding the median is very straightforward. The frequency column that was shown above can find the median. The frequency should always arrange from smallest to largest. If there were a large amount of data that is present in the frequency column it would be very time consuming to put the column in ascending order. Therefore an easier step can be taken into deliberation. This has been shown below with the formula that is used at all times to calculate the median: Median = n + 1 ÷ 2 Where n is equal to highest cumulative frequency. The added one is done so that the figure can be rounded off to give a most accurate result that could be obtained. Median = n + 1 ÷ 2 Median = 10 + 1 ÷ 2 Median = 11 ÷ 2 Median = 5.5 Now when this median is found this number is taken and the cumulative frequency column is looked at. It can be noticed that 5.5 lies between 4 and 6. It can now be said that the median is equal to 8 ≤ x < 10. Therefore, Median = 8 ≤ x < 10 Lower Quartile, Upper Quartile and Inter Quartile Range – Fiat To make my estimations accurate I will find out the lower quartile, upper quartile and inter quartile range. Median = n + 1 ÷ 2 Median = 10 + 1 ÷ 2 Median = 5.5 I now looked at 5.5 on the yaxis (cumulative frequency) of the graph on the next page and dropped a perpendicular line until it touched the plotted line. Therefore Median is the value that is obtained on the xaxis. This has been shown on the next page. Hence, Median is equal to £8.4kTo find the lower quartile (LQ) I will use the following formula: ¼ (n + 1) Where ‘n’ again is known as the highest cumulative frequency. ¼ (n + 1) ¼ (10 + 1) ¼ (11) 17 ÷ 4 = 2.75 I now looked at 2.75 on the yaxis (cumulative frequency) of the graph on the next page and dropped a perpendicular line until it touched the plotted line. Therefore LQ is the value that is obtained on the xaxis. This has been shown on the next page. It is now known that the lower quartile is equal to £7.4k. This is the value of onequarter way into the division. The same steps are to be considered while finding out the upper quartile, but this time the following formula was used: Upper Quartile = ¾ (n + 1) ¾ (10 + 1) ¾ (11) 11 х 3 ÷ 4 33 ÷ 4 8.25 The same steps were taken into deliberation. This is the part of three quarters of the way into the division. The value that I obtained on the xaxis is £11.4k. UQ = £11.4k Now with the known upper and lower quartile I can determine calculating the interquartile range. The following formula shows how to calculate the interquartile range: Inter quartile range (IQR) = Upper quartile – Lower quartile IQR = 11.4 – 7.4 I.Q.R = £4k Box whisker diagrams In this part pf my coursework I shall produce box whisker diagrams for each of these makes that I have calculated the mean, mode, median, range, upper quartile, lower quartile and inter quartile range. After making the box whisker diagrams I shall compare each of the diagrams produced. This will also tell me how much tendency there is from the central point. The box whisker diagram shall look like the diagram below: I will be producing all the box whisker diagrams on the same scale this will allow me to compare them easily. I have shown the box whisker diagrams for three makes below: Conclusion on box whisker diagrams By looking at the previous page, which shows the three box whisker diagrams for Ford, Vauxhall and Fiat, it can be seen that the Vauxhall is the widest drawn diagram and Fiat is the smallest. Once again the median shows the middle price of the make of car. By visual observation id can be seen that Vauxhall once again has the largest median and the Fiat has the smallest median. This tells me that the central tendency of Vauxhall is higher than any of the other two makes. If the middle price of the Vauxhall make of cars is taken into consideration that would be more expensive than the other makes, this will not be suitable for a normal working class person. I would prefer buying the Fiat make of cars, which has a not so expensive central tendency. The Ford is the average car between these two makes. It can also be seen that Vauxhall has cars that are of a higher range than the other two makes that I have been investigating. This is a good feature, as the person that will buy the vat will obviously buy a car, which is reliable to his/her pocket, and this may vary to very rich and not so rich. There I mean to say that there is a wide choice of choosing a very expensive car or which will be reasonable and affordable, by working class people. I have also notices that comparison between the medians to the upper quartile also varies. Vauxhall again has the more dispersion. This is due to the contributing factors that raise the price of the second hand car. There can be many contributing factors that raise the used car price. Some of them have been mentioned below:
I would prefer a sales man that has to travel up and down different states most probably drives the Vauxhall make of. A normal working class person that travels to and from work every weekday would most probably drive the Fiat. And the Ford is a mixture of both. Second hand price and new price I know shall begin the second part of the coursework. In this part of the coursework I will be comparing the current price (second hand price) and the original cost (price when new) of the each make of car. I am investigating this, as I will be able to judge which make of car decrease in value the most and which preserves most value. I will be scrutinizing five make of cars. The following are the make of cars that I shall be investigating:
PeugeotI shall begin with Peugeots; the data for this make of car has been shown on the table below:
The above data has been represented on the following bar chart. This bar chart has especially been created as it clearly shows the comparisons between the cost of each car when new and the second hand price for a certain years of age. By looking at the bar chart it is obviously known that the price when new is higher than the second hand price. It clearly represents a comparison between them both. The age is a contributing factor to the depreciation of the price. Many other things can also act as a contributing factor such as door number and engine size. FiatThe next make of car that I will be investigating will be Fiat. The same steps will be taken into consideration as for the Peugeot. The data for this make of car has been shown on the table below:
Many factors can also be a contributing factor here that shows more depreciation in the second hand price. These can be for instance the age of the car as the more older the car is the more depreciation will be caused or the larger the engine size the more the price will be when new and then it would reduce in price much more faster as the person that would buy it would not buy a car with a rather large engine as this will cost him/her more money on fuel, therefore the owner would have to bring the price down so that it could be sold. Therefore depreciation can be caused by many contributing factors. The following bar chart shows that the cost of buying a car brand new is always more than the current price of the car: RoverThe data for this make of car has been shown on the table below:
The above data has been epitomized in the bar graph below: Looking at the graph it is easily noticed that the decrease in the original cost of the car has reduces rapidly. For example Car number 7 has decreased from its original price to a very low second hand price. I have also noticed that car number 4 is an anomaly, this is because its second hand price is more than car number 2 which is 7 years old whereas car number 4 is 8 years old. Ignoring car number 4 that has been taken as an anomaly, I have noticed that the more the age of the car the more the car depreciates. Ford The same steps are taken into deliberation. The data for this make of car has been shown on the table below:
