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• Level: GCSE
• Subject: Maths
• Word count: 8793

Used second hand cars

Extracts from this document...

Introduction

Introduction:

I have been provided with a set of data, which altogether includes the factors that affects the second hand car prices of the listed hundred cars. Such as *Age, *Mileage, *Owners, *Insurance group and *MPG to name a few. You will see a spread sheet of this information on the following pages.

My task is to relate the prices of the second hands to number of variables. Also so how and state why they are important elements. I must then interpret my results and finally come to conclusion from them.

Many problems of statistical nature are made clearer when they are put in the forms of hypothesis. A hypothesis is generally considered to be a statement which may be true, but for which no proof has yet been found. I am going to choose o less than three factors, which I feel has an affect on the prices of the second hand cars. To start my project I have come up with three responsible questions, which I believe, could help me with my investigation. The questions I have decided to experiment are:

1. Does the age leave a huge impact on the second hand value?
2. What about make? Would you rather spend more money buying a cool car rather than buying a new car which is cheaper and less stylish?
3. How many people care about the mileage? Is there any disadvantage to the mileage?

From the alternative data base given to me I will randomly choose 3o cars and separate them from other cars taking in their age, mileage and make.

Hypothesises:

I predict that age, make and mileage are the three important factors that affect the second hand prices.

*AGE: -The older the car the lower its second hand price.

Middle

16

4

14.5

-2.5

6.25

86

1664

28

10

1

22

484

92

4693

17

5

10.5

2.5

6.25

96

3995

20

5

10.5

5.5

30.25

98

2748

24

6

8.5

10.5

110.25

= 209203

= 4284.20

Formula =  _ 6 d²

n(n²-1)

=  _ 6*4284.20

30 (900- 1)

=  _25704

26970

=  1- 0.953058956

= 0.046941045

= 0.05 (2 d.p)

You can compare two sets of ranking using Spearman’s coefficient of rank correlation.

You use the formulaρ=   _ 6

n (n²-1)

d is the difference between the two rankings of one item of data. n is the number of items of data.

ρ is Spearman’s coefficient of rank correlation.

The value of ρ will always be between -1 and +1.

-1                                                                       0                                                                           +1

ranking in                  weak negative               no                       Weak positive            same ranking

reverse order               correlation             correlation                   correlation           strong positive

strong  negative                                                                                                                 correlation

correlation

The spearman rank for the data is 0.05, which shows that there is almost no correlation between the age of the car and the second hand price of the car. This seems to contradict my hypothesis. However, I do feel that this is influenced by some of the cars brand new prices.

Display: Cumulative frequency

 Age of the cars Frequency Age of the cars (x) Cumulative frequency 1 2 <1 2 2 5 <2 7 3 4 <3 11 4 6 <4 17 5 2 <5 19 6 4 <6 23 7 2 <7 25 8 4 <8 29 9 0 <9 29 10 1 <10 30

In this cumulative frequency table the data shows that the model is 4. This supports my averages calculation. Therefore people prefer buying a second hand car at its least low age.  The cumulative diagram shows how the cumulative frequency changes as the data value increases. The cumulative frequency is shown on the vertical axis and the data is shown on the horizontal axis on continuous scale.  I have drawn the cumulative frequency curve on the next page on a graph paper.

I have used the cumulative frequency to find upper quartile, median, lower quartile and inter quartile to draw box and whisker diagram.

Display: Box and whisker diagram

To get an estimate of the median:

1. Divide the total cumulative frequency by 2.
2. Find this point on the cumulative frequency axis.
3. Draw a line across to the curve and down to the horizontal axis.
4. Read off the estimate of the median.

To get an estimate of the lower quartile:

1. Divide the total cumulative frequency by 4.
2. Find this point on the cumulative frequency axis.
3. Draw a line across to the curve and down to the horizontal axis.
4. Read off the lower quartile.

To get an estimate of the upper quartile:

1. Divide the total cumulative frequency by 4 and multiply by 3.
2. Find this point on the cumulative frequency axis.
3. Draw a line across to the curve and down to the horizontal axis.
4. Read off the lower quartile.

Inter quartile is upper quartile minus the lower quartile.

The box plot shows the median, lower quartile, upper quartile and the inter quartile, found out using the cumulative frequency curve for the age of the selected cars.

 Median 3.7 Lower quartile 2.1 Upper quartile 5.8 Inter quartile 5.8- 2.1 = 3.7

The median is nearly same as the median gained in the averages calculation. The inter quartile is the same value as the median. The box and whisker diagram also known as the box plot diagram is drawn at the back of the cumulative curve.

The box and whisker diagram has a positive skew. The median is not in the middle of the diagram. It is closer to the lower quartile.

Median- Lower Quartile < Upper Quartile- Median

M     -          LQ           <          UQ         -     M

Percentage depreciation:

 Car No: Make Price when new Second hand price Difference Age Percentage depreciation 2 Mercedes 16000 7999 8001 1 23.75043328 4 Vauxhall 8785 1595 7190 4 53.96160558 6 Renault 7875 1495 6380 4 63.26965467 10 Vauxhall 8748 1995 6753 4 40.79200592 14 Vauxhall 9105 2300 6805 2 43.87640449 19 Rover 12125 4295 7830 6 72.06683351 24 Fiat 11800 4700 7100 8 78.21969697 27 Toyota 8680 3200 5480 2 45.68884058 32 Rover 14505 8800 5705 8 82.22686879 37 Renault 13230 8250 4980 8 70.6401766 40 Fiat 13183 3495 9688 7 81.86653772 43 Ford 17780 7995 9785 7 77.19478738 47 Vauxhall 6590 1664 4926 6 78.8937409 50 Daewoo 15405 3995 11410 3 53.85826772 52 Ford 7310 1050 6260 3 64.57731959 54 Bentley 9995 2995 7000 8 77.76002248 57 Ford 16000 7999 8001 4 63.13364055 62 Peugot 8785 1595 7190 3 58.533309481 65 Ford 7875 1495 6380 3 37.64172336 66 Peugot 8748 1995 6753 1 17.80821918 70 Fiat 9105 2300 6805 2 53.79278446 72 Mercedes 12125 4295 7830 2 33.77483444 73 Porche 11800 4700 7100 6 40.9152902 77 Mercedes 8680 3200 5480 2 34.41250349 81 Ford 14505 8800 5705 4 55.03374578 84 Vauxhall 13230 8250 4980 4 36.53061224 86 Ford 13183 3495 9688 10 74.74962064 92 Volkswagen 17780 7995 9785 5 46.11940299 96 Ford 6590 1664 4926 5 74.06686141 98 Renault 15405 3995 11410 6 76.50277897

I have found out the percentage depreciation of the car by:

Percentage depreciation= Price when new- Second hand price

Price when new

This will help me to clarify the relationship between depreciation of price and age of the car.

Using the percentage depreciation I have calculated the four point moving averages for this data.

The results are followed…

Moving Averages

45.44443818

50.47506767

55.00137465

58.73888522

59.96283519

69.55045127

69.19378704                    *The graph is drawn on the next page on a graph paper.

70.10549723

77.98209262

77.14881065

72.95333343

68.6310289

68.77233767

64.83231259

66.00101936

59.2671203

44.27916948

41.94395545

35.75439036

40.72385315

41.03409348

41.72303793

50.18162054

43.97569235

57.86662432

67.859666

Moving averages are averages worked out for a given number of items of data as you work through the data.

A three- point moving average uses three items of data at a time.

A four- point moving averages uses four items of data and so on.

I have decided to do four point moving averages.

The moving averages show that the results are random. The trend line suggests that it has a negative trend. This shows that there would be a decrease in frequency with an increase in age.

Standard deviation:

The standard deviation, s, of a set of data is given by the formula:

The higher the standard deviation, the more spread out the data is.  The above formula gives the same results as the other formula but is much easier to work with, especially when the mean is not a whole number.

The other formula is:

I have decided to find out the standard deviation of the age of my 30 cars.

 Age (x) Frequency (f) fx x² f x² 1 2 2 1 2 2 5 10 4 20 3 4 12 9 36 4 6 24 16 96 5 2 10 25 50 6 4 24 36 144 7 2 14 49 98 8 4 32 64 256 9 0 0 81 0 10 1 10 100 100 = 138 = 799

Mean=  fx         = 138

x ,            30     =   4.6

The formula s=                                    becomes

Standard deviation=              2.339515619

= 2.34 (3 s.f)

The mean of the age of the second hand cars is 4.6 and the standard deviation is 2.34. This indicates that the age is bigger spread.

I have conducted spearman’s rank to find out whether if there is relationship between the percentage depreciation and the second hand price of the cars.

Spearman’s rank:

Car No:

Second- hand price

Rank 1

Percentage depreciation

Rank 2

Difference (d)

d²

2

10999

5

23.75043328

2

3

9

4

6595

11

53.96160558

14

-3

9

6

4999

13

63.26965467

18

-5

25

10

7499

9

40.79200592

7

2

4

14

4995

14.5

43.87640449

9

5.5

30.25

19

3795

21

72.06683351

21

0

0

24

1495

30

78.21969697

27

3

9

27

7495

10

45.68884058

10

0

0

32

1700

27

82.22686879

30

-3

9

37

1995

25.5

70.6401766

20

5.5

30.25

40

1500

29

81.86653772

29

0

0

43

1995

25.5

77.19478738

25

0.5

0.25

47

2900

23

78.8937409

28

-5

25

50

4395

18

53.85826772

13

5

25

52

4295

19

64.57731959

19

0

0

54

37995

1

77.76002248

26

-25

625

57

3200

22

63.13364055

17

5

25

62

5795

12

58.533309481

16

-4

16

65

8250

6

37.64172336

6

0

0

66

7500

8

17.80821918

1

7

49

70

4995

14.5

53.79278446

12

2.5

6.25

72

17500

3

33.77483444

3

0

0

73

19495

2

40.9152902

8

-6

36

77

11750

4

34.41250349

4

0

0

81

7995

7

55.03374578

15

-8

64

84

4976

16

36.53061224

5

11

121

86

1664

28

74.74962064

23

5

25

92

Conclusion

Like I mentioned in my introduction I do believe that my sample was large enough to represent the population fairly. The price of the car would decrease with increases in mileage and vice versa. The reputed and posh the make of the car more expensive it is. I have proven very clearly that the older the age less the value of its second hand price and lower the mileage higher the second hand car’s value.

If someone else carried out the same way investigation the chance of his or her findings matching my result is about 50 to 60 percent. This is due to the consideration of other factors, such the size of the sample, sampling method, the hypothesis, time provided and etc.

If I were to do the investigation again, the things I would prefer doing differently are:

1. The hypothesis: I would try and link the second hand price of the car with other factors.
2. The data: I might decide to collect the data myself.
3. I would also test the other sampling and link each of the findings.
4. Test each make separately.

I would also analyse my results with my sallow students to see whether id there is any match.

I think if someone else were to read my report it would be fairly easier to understand as I have shown all the calculations step by step and given brief description of all. I have repeated my hypothesis again and again and explained my graphs and what they show.

I don not think I have not concluded any irrelevant statistical calculations or irrelevant statistical diagrams or any inappropriate conclusions, Therefore, any one reading my course work would not find any misleading information.

This student written piece of work is one of many that can be found in our GCSE Gary's (and other) Car Sales section.

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