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  • Level: GCSE
  • Subject: Maths
  • Word count: 2135

The aim of this investigation is to find out what influences the price of used cars (second hand). Used cars usually cost less than brand new cars, but this can be affected by hypotheses' such as mileage, age, colour, how it has been used etc.

Extracts from this document...

Introduction

Cherry Robinson 10/7915/06/2007

Maths Coursework:

Strand one:

Aim:

The aim of this investigation is to find out what influences the price of used cars (second hand). Used cars usually cost less than brand new cars, but this can be affected by hypotheses’ such as mileage, age, colour, how it has been used etc. My investigation is to find out how these hypotheses affect a used cars’ price.

Hypotheses:

For my example I have chosen to use the hypotheses: 1: age, 2: mileage, 3: weather it was expensive when new and 4: make. I have chosen these because I think they are most likely to affect the price of used cars.

Sample:

Out of my database of 150 I need to randomly select at least 30 samples. Samples are used representatively. They represent the whole database. We use samples because it would take too long to investigate every piece of data on the database, so we only investigate the samples. This is acceptable because they are selected randomly.

The samples are very poor because there was a database of 150 and only 30 samples. The probability of getting a common car (i.e. Ford) is higher than less common cars (i.e. Mercedes)

Testing hypotheses one.

I’d expect to find negative correlation. But vintage cars might be worth more as they get older.

...read more.

Middle

56

28

61

29

70

30

33

Sample

% of original amount

1

59

2

18

3

13

4

67

5

18

6

76

7

14

8

29

9

35image00.png

10

46

11

26

12

24

13

56

14

50

15

76

16

19

17

78

18

37

19

40

20

18

21

23

22

18

23

30

24

68

25

50

26

55

image02.png

The graph shows that the higher the taken percentage of original value the newer the car is. It is negative correlation. I was right in my assumption that car prices drop by percentage per year. The graph is nearer perfect correlation than my age and price graph, but it’s still not perfect, this can be affected by other hypotheses such as milage.

Looking at the graph I can see there are a couple of cars that don’t fit into the negative correlation. We can use measures of dispersion to rule out these ‘odd’ cars. Using range and standard deviation we can create a cumulative frequency graph to show the majority range of used car prices. By finding this range we can dismiss extraneous values from the sample.

image03.png

This graph shows better negative correlation with just one extraneous value.

Conclusion

I have proved age affects the price of used cars. This graph shows the newer a car is the higher the percentage of original price. The price goes down by percentage every year.

P = -Sa + 80

M = -S

Y = Mx + C

P = Ma + C

Cherry Robinson 10/7915/06/2007

Maths coursework

Strand 2:

Testing hypotheses two: mileage

My second hypothesis is mileage. I need to prove mileage affects the price of used cars.

I need 30 samples again. I am going to use a stratified sampling method.

...read more.

Conclusion

2

48

24

78

-54

-108

-216

60 < m < 70

0

0

0

0

0

0

0

70 < m < 80

2

48

24

78

-54

-108

-216

delivery

2

48

24

78

-54

-108

-216

Median position = 15 + 1 / 2 = 8

40 < m < 50 this is the median interval. This is because the 8th number of frequency falls into this interval.

Range = 50 - 0 = 50

The range is in interval 50 < m < 60.

Mode = 20 < m < 30.

Mean = x = Efx / Ef = 12

The mean is interval 70 < m < 80

Lower quartile = n + 1 / 4 = 15 + 1 / 1 = 16, fourth number. Interval = 20 < m < 30

Upper quartile = 3(n + 1 / 4) = 48, twelfth number. Interval = 70 < m < 80

1:

Random

Mileage 000

Frequency

Angle

Fx Midpoint

mean

(x – x .)

(x – x .) squared

F (x – x .) squared

0 < m < 10

1

Freq / total x 360 = 24

12

156

-144

-288

-288

10 < m < 20

1

24

12

156

-144

-288

-288

20 < m < 30

1

24

12

156

-144

-288

-288

30 < m < 40

2

48

24

78

-54

-108

-216

40 < m < 50

5

120

60

31.2

28.8

784

3920

50 < m < 60

1

24

12

156

-144

-288

-288

60 < m < 70

1

24

12

156

-144

-288

-288

70 < m < 80

0

0

0

0

0

0

0

delivery

1

24

12

156

-144

-288

-288

Random:

Median position = 15 + 1 / 2 = 8

30 < m < 40 this is the median interval. This is because the 8th number of frequency falls into this interval.

Range = 50 - 0 = 50

The range is in interval 50 < m < 60.

Mode = 20 < m < 30.

Mean = x = Efx / Ef = 12

The mean is interval 60 <m < 70

Lower quartile = n + 1 / 4 = 15 + 1 / 1 = 16, fourth number. Interval = 30 < m < 40

Upper quartile = 3(n + 1 / 4) = 48, twelfth number. Interval = 60 < m < 70

1:

Random:

o square route Ef (x – x .) squared / Ef =  3206

Luxury:

o square route Ef (x – x .) squared / Ef =  1882

Random ois lowly dispersed.

Luxury is highly dispersed.

All this proves that Mileage affects the price of second hand cars.

...read more.

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