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As the height increases the weight increases and also the relationship between these the height and the weight will become stronger as they get older. I also think that there will be a difference in this between the boys and the girls.

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Introduction

Maths GCSE Handling Data Coursework

Mayfield High School – Year 10 & 11

     For my statistics coursework I am going to investigate the relationship between the height and the weight. I will use a wide range of mathematical techniques to present my data and findings in different ways. I will also focus to test my hypothesis which is: As the height increases the weight increases and also the relationship between these the height and the weight will become stronger as they get older. I also think that there will be a difference in this between the boys and the girls.

The table below shows the number of boys and girls there are in each year group:

Year Group

Number Of Boys

Number Of Girls

Total

10

106

94

200

11

84

86

170

Total

190

180

370

For my project I will take a random sample of 30 students. I will use the random sample button on a calculator to do this.

Year 10

No of boys: (106∕370) x 30 = 8.5946 ≈ 9

No of girls: (94∕370) x 30 = 7.6216 ≈ 8

Year 11

No of boys: (84/370) x 30 = 6.8108 ≈ 7

No of girls: (86/370) x 30 = 6.9730 ≈ 7

Total number of students = 31

...read more.

Middle

 is the mean of the data set and n is the number of values.

For height

Mean = 1.66233

Standard Deviation = 0.085286

(2 x 0.085286) = 0.255858

1.66233 - 0.170572= 1.491758

1.66233 + 0.170572 = 18.98045

Therefore, anything less than 1.491758 and more than 1.832902 is considered as an outlier.

In my sample, there are no outliers

For weight

Mean = 55.93333

Standard Deviation = 9.490225

(2 x 9.490225) = 18.98045

55.93333 – 18.98045 = 36.95288

55.93333 + 18.98045 = 74.91378

So anything less than 36.95288 and more than 74.91378 is an outlier.

Hence, there is no outlier in my sample.

Tally chart for the heightTally chart for the weight

Height is a continuous data, so you                      Weight is also a continuous               need to use class intervals.                                   data; the class interval I’ve used

I’ve used a class interval of 0.05 m.                     is 5kg

Height (cm)

Tally

Frequency

1.50≤H<1.55

I

1

1.55≤H<1.60

IIII I

6

1.60≤H<1.65

IIII II

7

1.65≤H<1.70

IIII

4

1.70≤H<1.75

IIII

5

1.75≤H<1.80

III

3

1.80≤H<1.85

IIII

4

Weight

Tally

Frequency

35≤W<40

II

2

40≤W<45

I

1

45≤W<50

IIII

4

50≤W<55

IIII I

6

55≤W<60

IIII I

6

60≤W<65

IIII I

6

65≤W<70

II

2

70≤W<75

III

3

...read more.

Conclusion

1.60

47

1.60

50

1.65

54

1.65

54

1.68

59

1.80

60

1.60

51

1.65

54

1.80

72

1.55

48

1.75

68

1.51

36

1.72

54

1.63

52

1.81

54

1.66

45

1.82

57

1.62

48

1.68

72

1.70

60

1.54

76

1.55

60

1.50

35

1.80

60

1.62

72

1.62

48

1.62

50

1.52

45

1.73

50

1.67

48

1.52

38

1.62

38

1.84

78

1.65

52

1.75

57

1.52

70

1.61

56

1.68

47

1.80

63

1.41

55

1.57

54

1.78

55

1.52

45

1.65

59

1.78

37

1.60

54

1.63

50

1.80

74

1.80

68

1.57

45

Check for an outlier

As I did for my first sample, I am going to do the same to check for the outlier.

The mean for the height of boys sample is:

(1.63+1.77+1.32+1.62+1.60+1.60+1.65+1.68+1.60+1.80+1.75+1.72+1.81+1.82+1.68+1.54+1.50+1.62+1.62+1.73+1.52+1.84+1.75+1.61+1.80+1.57+1.52+1.78+1.63+1.80)

30image01.png

Mean = 1.662667

Standard deviation = 0.11948

(2 x 0.11948) = 0.23896

1.662667 – 0.23896 = 1.429707

1.662667 + 0.23896 = 1.901627

Thus, anything less than 1.429707 and anything more than 1.901627 is an outlier.

In my sample there is only one outlier. I am going to leave this as it is because sometimes you might have someone shorter and he is not that short he is just approximately 0.10m shorter than the outlier range.

The mean for the weight of boys sample is:

(40+57+45+52+38+47+54+59+51+72+68+54+54+57+72+76+35+72+50+50+38+78+57+56+63+54+45+37+50+68)image01.png

30

Mean = 54.96667

Standard Deviation = 11.98126

(2 x 11.98126) = 23.96252

54.96667 – 23.96252 = 31.00415

54.96667 + 23.96252 = 78.92919

Anything less than 31.00415 and more than 78.92919 is an outlier.

There is no outlier

The mean for the

To compare my correlation I am going to use product momentum correlation coefficient. This is the accurate way to compare the correlation. It uses the mean of each set of data and looks at the distance away from the mean of each point.

The Formula is  image06.png

Where image03.pngand image07.pngare the means of the x and y values respectively

...read more.

This student written piece of work is one of many that can be found in our GCSE Height and Weight of Pupils and other Mayfield High School investigations section.

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