analysing corelations between the leaves from sun-exposed and shadowed side of the three

8/3/2019 ANALYSING CORELATIONS BETWEEN THE LEAVES FROM SUN-EXPOSED AND SHADOWED SIDE OF THE THREE

1/24

NATALIA KOCZKO 3IB

ANALYSING THE CORELATIONS BETWEEN THE LEAVES FROM

SUN-EXPOSED AND SHADOWED SIDE OF THE THREE

DATA COLLECTIONTable 1 . Dimensions of the leaves on the sun-exposed side of the tree

No. Width [cm] 0.1 Length [cm] 0.1

1 96 1042 85 1003 100 1114 95 1255 92 1056 91 110

7 89 978 115 120

9 92 11810 80 95

11 119 12312 91 115

13 94 10914 110 13515 105 10816 95 11217 83 9918 100 121

19 103 1082 90 100

021 84 11222 110 14523 90 11624 85 11025 96 11426 107 12027 110 12328 105 140

29 106 12830 105 117

31 90 12732 118 13533 60 9034 84 10635 98 13036 99 11037 102 132

38 92 11339 85 12040 117 135

41 75 9642 110 132

43 80 11044 98 12145 112 13546 90 11247 83 13048 105 114

49 85 97

50 107 122

Table 2. Dimensions of the leaves on the shadowed side of the tree

1


2/24

No. Width [cm] 0.1 Length [cm] 0.11 50 732 72 933 67 1004 70 1045 100 130

6 100 1237 42 56

8 52 110

9 60 8010 91 11611 70 9312 65 9213 68 9714 100 12415 43 60

16 92 13717 54 73

18 83 11619 61 83

20 58 78

21 31 4222 86 11623 45 6224 57 8825 75 102

26 62 9627 47 53

28 94 13029 82 115

30 91 12231 75 10332 74 93

33 55 8834 65 9435 57 7536 77 10537 54 64

38 73 8439 72 80

40 80 10041 81 8742 76 9143 77 10244 94 101

45 83 9046 67 8347 75 102

48 73 8749 53 65

50 86 90

2

DATA PROCESSING


3/24

H0: there is no significant difference between the width of the leaves on the sun-exposed and shadowed side of the tree

H1: The dimensions of the leaves on the sun-exposed side are greater than thoseof the shadowed side

Table 3 . Checking for normal distribution for width of the leaves on the sun-exposed side of the tree

Class Frequency

60 69 I

70 79 I

80 89 IIIIIIIIIII

90 99 IIIIIIIIIIIIIIIIII

100 109 IIIIIIIIIII

110 119 IIIII

3

Student t-testIs a statistical test that allows determination of marginal differences between theexamined sets of data. Test can be applied is the test statistic follow a normaldistribution. In this particular case it will be used to check if there is a significantdifference between the width or the length of the leaves on the sun-exposed andshadowed side.

The distribution is slightly skewed but can be still referred to as tonormal


4/24

26.9650

4813

samplesofnumber

xx

1

1 ===

Table 4. Calculating standarddeviation for the leaves on the sun-exposed side of the tree

No.Width [cm]

0.1 (X1)Width [cm]

0.1 (X1 )

1 96 9216

2 85 7225

3 100 10000

4 95 9025

5 92 8464

6 91 82817 89 7921

8 115 13225

9 92 8464

10 80 6400

11 119 14161

12 91 8281

13 94 8836

14 110 12100

15 105 11025

16 95 9025

17 83 688918 100 10000

19 103 10609

20 90 8100

21 84 7056

22 110 12100

23 90 8100

24 85 7225

25 96 9216

26 107 11449

27 110 12100

28 105 1102529 106 11236

30 105 11025

31 90 8100

32 118 13924

33 60 3600

34 84 7056

35 98 9604

36 99 9801

37 102 10404

38 92 8464

39 85 7225

40 117 13689

41 75 5625

42 110 12100

43 80 6400

44 98 9604

45 112 12544

46 90 8100

47 83 6889

48 105 11025

49 85 7225

50 107 11449

Sum: 4813 470607

4

( )[ ]1n

nxxS

22

=

Calculations

49

504813470607S

2

=

21.12S=

S = 12.21 constitutes less than 33% of

the mean (96.22) what means that itis a is a small standard deviation

70 % off the data (35 leaves) fallswithin one standard deviation and98% of the data (49 leaves) fallswithin two standard deviations

Calculating the mean


5/24

Table 5 . Sorting the widths of the leaves on the sun-exposed side of the

tree

No. Width [cm] 0.1 (X1) Sorted value

1 96 60

2 85 75

3 100 80

4 95 80

5 92 83

6 91 83

7 89 84

8 115 84

9 92 85

10 80 85

11 119 85

12 91 85

13 94 8914 110 90

15 105 90

16 95 90

17 83 90

18 100 91

19 103 91

20 90 92

21 84 92

22 110 92

23 90 94

24 85 9525 96 95

26 107 96

27 110 96

28 105 98

29 106 98

30 105 99

31 90 100

32 118 100

33 60 102

34 84 103

35 98 105

36 99 105

37 102 105

38 92 105

39 85 106

40 117 107

41 75 107

42 110 110

43 80 110

44 98 110

45 112 110

46 90 112

47 83 115

48 105 117

49 85 118

5


6/24

50 107 119

Table 6. Checking for normal distribution for width of the leaves on theshadowed side of the tree

Class Frequency

30 39 I

40 49 IIII

50 59 IIIIIIIII

60 69 IIIIIIII

70 79 IIIIIIIIIIIII

80 89 IIIIIII

90 99 IIIII

100 109 III

6

There is a good normal distribution


7/24

3.7050

3515

samplesofnumber

x

x

2

2 ===

Table 7 . Calculating standard deviation for the leaves on the shadowed

side of the tree

Table 8 . Sorting the widths of the leaves on the shadowed side of the tree

No.Width [cm] (X2) 0.1

X2 0.1

1 50 2500

2 72 51843 67 4489

4 70 4900

5 100 100006 100 100007 42 17648 52 27049 60 3600

10 91 828111 70 4900

12 65 422513 68 4624

14 100 1000015 43 184916 92 8464

17 54 291618 83 688919 61 372120 58 336421 31 961

22 86 739623 45 2025

24 57 324925 75 562526 62 384427 47 220928 94 8836

29 82 672430 91 828131 75 5625

32 74 547633 55 3025

34 65 422535 57 324936 77 592937 54 291638 73 532939 72 518440 80 640041 81 6561

42 76 577643 77 5929

44 94 883645 83 6889

46 67 448947 75 562548 73 532949 53 280950 86 7396

Sum:

3515 260521

No. Width [cm] (X2) 0.1 Sorted value1 50 31

2 72 423 67 43

4 70 455 100 47

6 100 50

7 42 528 52 53

9 60 5410 91 54

11 70 5512 65 57

13 68 5714 100 58

15 43 6016 92 61

17 54 6218 83 65

19 61 6520 58 67

21 31 6722 86 68

23 45 70

24 57 7025 75 72

26 62 7227 47 73

28 94 7329 82 74

30 91 75

31 75 7532 74 7533 55 76

34 65 77

35 57 77

36 77 8037 54 81

38 73 82

39 72 83

40 80 8341 81 86

42 76 86

43 77 91

44 94 91

45 83 92

46 67 94

47 75 94

48 73 100

49 53 100

50 86 100

7

Calculations

( )[ ]

1n

nxxS

22

=

49

503515260521S

2

=

55.16S=


S = 16.55 constitutes less than 33% othe mean (70.03) what means that iis a is a small standard deviation

68 % off the data (34 leaves) falls

within one standard deviation and98% of the data (49 leaves) fallswithin two standard deviations


8/24

26.9650

4813

samplesofnumber

xx

1

1 ===

3.7050

3515

samplesofnumber

xx

2

2 ===

14.149150

)50/23164969(470607

1n

)n/)x((xs

1

1

2

1

2

12

1=

=

=

81.273150

)50/12355225(260521

1n

)n/)x((xs

2

2

2

2

2

22

2=

=

=

Table 9 . Calculating mean and varianceMean formula and calculations:

Variance formula and calculations:

Variance for X1 =

Variance for X2 =

No.

Width [cm]0.1

Sun-exposed

side(X1)

Width [cm]0.1

Darkened

side(X2)

1 96 502 85 72

3 100 674 95 705 92 1006 91 1007 89 428 115 529 92 6010 80 91

11 119 7012 91 65

13 94 6814 110 10015 105 4316 95 9217 83 5418 100 8319 103 6120 90 58

21 84 3122 110 86

23 90 45

24 85 5725 96 7526 107 6227 110 4728 105 9429 106 8230 105 91

31 90 7532 118 74

33 60 5534 84 65

35 98 57

36 99 7737 102 5438 92 7339 85 7240 117 80

41 75 8142 110 76

43 80 7744 98 94

45 112 8346 90 6747 83 75

48 105 7349 85 5350 107 86

mean 96.26 70.3variabl

e149.14 273.81

8


9/24

Graph 1. Widths of the leaves on the sun-exposed and shadowed side

Calculating value oft:

The formula for t:

Calculating degree of freedom

T value of 8. 95 for 98 degrees of freedom is less than 1%, therefore thedifference between the two sets is highly significant. This implies that widtha ofthe leaves are not random and that the null hypothesis can be rejected. Widths ofthe leaves on the sun-exposed side are grater than those of the leaves placed onthe shadowed side.

9

95.890.2

96.25

48.598.2

96.25

50

81.273

50

14.149

30.7026.96

n

s

n

s

xxt

2

2

2

1

2

1

21==

+

=

+

=

+

=

982)nn(df 21 =+=

0

20

40

60

80

100

120

140

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

shadowed side

sun-exposed side

Sample number

Size[cm]


10/24


11/24

14.11650

5807

samplesofnumber

xx

1

1 ===

Table 11 . Calculating standard deviation for the leaves on the shadowed sideof the tree

Table 12. Sorting the lengths of the

leaves on the sun-exposed side of the tree

No.Length

[cm] (X1) 0.1

X1 0.1

1 104 108162 100 100003 111 12321

4 125 156255 105 11025

6 110 121007 97 9409

8 120 144009 118 1392410 95 902511 123 1512912 115 1322513 109 1188114 135 1822515 108 11664

16 112 1254417 99 9801

18 121 1464119 108 11664

20 100 1000021 112 1254422 145 2102523 116 1345624 110 1210025 114 12996

26 120 1440027 123 15129

28 140 1960029 128 1638430 117 1368931 127 16129

32 135 1822533 90 810034 106 1123635 130 16900

36 110 1210037 132 17424

38 113 1276939 120 14400

40 135 1822541 96 921642 132 1742443 110 12100

44 121 1464145 135 1822546 112 1254447 130 16900

48 114 1299649 97 9409

50 122 14884Sum: 5807 682589

11

Calculations

( )[ ]1n

nxxS

22

=

49

505807682589S

2

=

91.12S=


S = 12.91 constitutes less than 33% ofthe mean (116.14) what means that itis a is a small standard deviation

62 % off the data (34 leaves) fallswithin one standard deviation and

98% of the data (49 leaves) fallswithin two standard deviations


12/24

Table 13 . Checking for normal distribution for length of the leaves on the sun-exposed side of the tree

No. Length [cm] (X1) 0.1 Sorted value

1 104 90

2 100 95

3 111 96

4 125 97

5 105 97

6 110 997 97 100

8 120 100

9 118 104

10 95 105

11 123 106

12 115 108

13 109 108

14 135 109

15 108 110

16 112 110

17 99 110

18 121 11019 108 111

20 100 112

21 112 112

22 145 112

23 116 113

24 110 114

25 114 114

26 120 115

27 123 116

28 140 117

29 128 118

30 117 120

31 127 120

32 135 120

33 90 121

34 106 121

35 130 122

36 110 123

37 132 123

38 113 125

39 120 127

40 135 128

41 96 13042 132 130

43 110 132

44 121 132

45 135 135

46 112 135

47 130 135

48 114 135

49 97 140

50 122 145

12


13/24

Table 14. Calculating standard deviation forthe leaves on the shadowed side of the tree

Class Frequency

40 49 I

50 59 II60 69 IIII

70 79 IIII80 89 IIIIIIIII

90 99 IIIIIIIIII

100 109 IIIIIIIII110 119 IIIII120 129 II

130 139 INo. Length

[cm] (X1) 0.1

X1 0.1

1 73 53292 93 8649

3 100 100004 104 108165 130 169006 123 151297 56 31368 110 121009 80 6400

10 116 13456

11 93 864912 92 8464

13 97 940914 124 1537615 60 360016 137 1876917 73 53298 116 13456

19 83 6889

20 78 608421 42 176422 116 13456

23 62 384424 88 774425 102 1040426 96 921627 53 280928 130 1690029 115 1322530 122 14884

31 103 10609

32 93 864933 88 774434 94 8836

35 75 562536 105 1102537 64 409638 84 705639 80 640040 100 10000

41 87 756942 91 8281

43 102 10404

44 101 1020145 90 810046 83 688947 102 1040448 87 756949 65 422550 90 8100Sum:

4648453968

13

Calculations

( )[ ]1n

nxxS

22

=

49

504648453968S

2

=

13.21S=


There is a good normal distribution


14/24

96.9250

4648

samplesofnumber

xx

2

2 ===

Table 15. Sorting the lengths of the leaves on the shadowed side of the tree

No. Length [cm] (X1) 0.1 Sorted value

1 73 42

2 93 53

3 100 56

4 104 60

5 130 62

6 123 64

7 56 65

8 110 73

9 80 73

10 116 75

11 93 78

12 92 80

13 97 80

14 124 83

15 60 83

16 137 84

17 73 87

18 116 87

19 83 8820 78 88

21 42 90

22 116 90

23 62 91

24 88 92

25 102 93

26 96 93

27 53 93

28 130 94

29 115 96

30 122 97

31 103 100

32 93 100

33 88 101

14

S = 21.13 constitutes less than 33% ofthe mean (92.96) what means that it is ais a small standard deviation

62 % off the data (34 leaves) fallswithin one standard deviation and

96% of the data (48 leaves) falls within two standard deviations


15/24

14.11650

5807

samplesofnumber

xx

1

1 ===

96.9250

4648

samplesofnumber

xx

2

2 ===

61.166150

)50/33721249(682589

1n

)n/)x((xs

1

1

2

1

2

12

1 =

=

=

73.446150

)50/21603904(453968

1n

)n/)x((xs

2

2

2

2

2

22

2=

=

=

34 94 102

35 75 102

36 105 102

37 64 103

38 84 104

39 80 105

40 100 110

41 87 115

42 91 11643 102 116

44 101 116

45 90 122

46 83 123

47 102 124

48 87 130

49 65 130

50 90 137

Table 16. Calculating mean andvariance

Mean formula and calculations:

Variance formula and calculations:

Variance for X1 =

Variance for X2 =

No.

Length [cm]0.1Sun-exposed

side(X1)

Length [cm] 0.1Darkened

side(X2)

1 104 73

2 100 933 111 100

4 125 1045 105 1306 110 1237 97 568 120 1109 118 8010 95 11611 123 93

12 115 9213 109 97

14 135 12415 108 6016 112 13717 99 7318 121 11619 108 8320 100 78

21 112 4222 145 11623 116 62

24 110 8825 114 10226 120 9627 123 5328 140 13029 128 11530 117 12231 127 103

32 135 93

33 90 8834 106 9435 130 75

36 110 10537 132 6438 113 8439 120 8040 135 10041 96 87

42 132 9143 110 102

44 121 101

45 135 9046 112 8347 130 10248 114 8749 97 6550 122 90

mean 116.14 92.96variabl

e166.61 446.73

15


16/24

Graph 2 . Widths of the leaves on the sun-exposed and shadowed side

Calculating value oft:

The formula for t:

Calculating degree of freedom

16

62.650.3

18.23

93.833.3

18.23

50

73.446

50

61.166

96.9214.116

n

s

n

s

xxt

2

2

2

1

2

1

21==

+

=

+

=

+

=

982)nn(df 21 =+=

0

20

40

60

80

100

120

140

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

shadowed side

sun-exposed side

Sample number

Size

[cm]


17/24

T value of 8. 95 for 98 degrees of freedom is less than 1%, therefore thedifference between the two sets is highly significant. This implies that lengths ofthe leaves are not random and that the null hypothesis can be rejected. Lengthsof the leaves on the sun-exposed side are grater than those of the leaves whichare on the shadowed side.

Table 17 . Ranking the dimensions of the leaves from sun-exposed side

17

The Spearman Rank CorrelationIs a statistical test that assesses the correlation between two sets of variables. The measure ofcorrelation lies between +1 and 1, where +1 stands for a perfect positive correlation while -1determinates perfect negative correlation. In this case the test will be used to determinatewhether there is a correlation between the length and the width of the leaves on each side of thetree (the sun-exposed and the shadowed).


18/24

Table 18 . Calculating the differences between ranks

No.Length[cm]

0.1Rank R1

Width[cm] 0.1

Rank R2 D=R1 R2 D

1 104 9 96 26.5 -17,5 306,25

2 100 7.5 85 9.25 -17,5 306,253 111 19 100 31.5 -12,5 156,254 125 38 95 24.5 13,5 182,255 105 10 92 20.33 -10,33 106,71

No.

Length[cm] (X1)

0.1

Sortedvalue

Rank

Width[cm] (X1)

0.1

Sortedvalue

Rank

1 104 90 1 96 60 12 100 95 2 85 75 2

3 111 96 3 100 80 3.54 125 97 4.5 95 80 3.5

5 105 97 4.5 92 83 5.5

6 110 99 6 91 83 5.57 97 100 7.5 89 84 7.5

8 120 100 7.5 115 84 7.59 118 104 9 92 85 9.2510 95 105 10 80 85 9.25

11 123 106 11 119 85 9.2512 115 108 12.5 91 85 9.25

13 109 108 12.5 94 89 1314 135 109 14 110 90 14.2515 108 110 15.25 105 90 14.25

16 112 110 15.25 95 90 14.2517 99 110 15.25 83 90 14.2518 121 110 15.25 100 91 18.5

19 108 111 19 103 91 18.520 100 112 20.33 90 92 20.33

21 112 112 20.33 84 92 20.3322 145 112 20.33 110 92 20.3323 116 113 23 90 94 23

24 110 114 24.5 85 95 24.525 114 114 24.5 96 95 24.526 120 115 26 107 96 26.5

27 123 116 27 110 96 26.528 140 117 28 105 98 28.5

29 128 118 29 106 98 28.530 117 120 30.33 105 99 3031 127 120 30.33 90 100 31.5

32 135 120 30.33 118 100 31.5

33 90 121 33.5 60 102 3334 106 121 33.5 84 103 3435 130 122 35 98 105 35.2536 110 123 36.5 99 105 35.25

37 132 123 36.5 102 105 35.2538 113 125 38 92 105 35.2539 120 127 39 85 106 39

40 135 128 40 117 107 40.541 96 130 41.5 75 107 40.5

42 132 130 41.5 110 110 42.2543 110 132 43.5 80 110 42.2544 121 132 43.5 98 110 42.25

45 135 135 45.25 112 110 42.25

46 112 135 45.25 90 112 4647 130 135 45.25 83 115 47

48 114 135 45.25 105 117 4849 97 140 49 85 118 49

50 122 145 50 107 119 50

18


19/24

6 110 15.25 91 18.5 -3,25 10,567 97 4.5 89 13 -8,5 72,258 120 30.33 115 47 -16,67 277,899 118 29 92 20.33 8,67 75,17

10 95 2 80 3.5 -1,5 2,2511 123 36.5 119 50 -13,5 182,25

12 115 26 91 18.5 7,5 56,2513 109 14 94 23 -9 81

14 135 45.25 110 42.25 3 9

15 108 12.5 105 35.25 -22,75 517,5616 112 20.33 95 24.5 -4,17 17,3917 99 6 83 5.5 0,5 0,2518 121 33.5 100 31.5 2 4,0019 108 12.5 103 34 -21,5 462,252 100 7.5 90 14.25 -6,75 45,56

21 112 20.33 84 7.5 12,83 164,61

22 145 50 110 42.25 7,75 60,0623 116 27 90 14.25 12,75 162,56

24 110 15.25 85 9.25 6 3625 114 24.5 96 26.5 -2 4

26 120 30.33 107 40.5 -10,17 103,43

27 123 36.5 110 42.25 -5,75 33,0628 140 49 105 35.25 13,75 189,0629 128 40 106 39 1 130 117 28 105 35.25 -7,25 52,5631 127 39 90 14.25 24,75 612,56

32 135 45.25 118 49 -3,75 14,0633 90 1 60 1 0 0,00

34 106 11 84 7.5 3,5 12,2535 130 41.5 98 28.5 13 169,00

36 110 15.25 99 30 -14,75 217,5637 132 43.5 102 33 10,5 110,2538 113 23 92 20.33 2,67 7,13

39 120 30.33 85 9.25 21,08 444,3740 135 45.25 117 48 -2,75 7,5641 96 3 75 2 1 142 132 43.5 110 42.25 1,25 1,5643 110 15.25 80 3.5 11,75 138,06

44 121 33.5 98 28.5 5 2545 135 45.25 112 46 -,75 0,56

46 112 20.33 90 14.25 6,08 36,9747 130 41.5 83 5.5 36 129648 114 24.5 105 35.25 -10,75 115,5649 97 4.5 85 9.25 -4,75 22,5650 122 35 107 40.5 -5,5 30,25

Sum: 6939,94

H0: there is no correlation between the length and the width of the leaves on the sun-exposed side of the tree

H1: There is a positive correlation between the length and width of the leaves on thesun-exposed side of the tree

Calculating the Spearman Rank correlation (rs)

Formula:

19

33.0124950

64.41639

249950

6939,946

)1n(n

D61r

2

2

s ==

=

=


20/24

The critical value for 50 sample pairs is 0.28. the calculated value of spearmanrank correlation is 0.33, what indicates that there is positive correlation betweenexamined sets of data. Therefore the null hypothesis can be rejected and H1supported as the greater the length of the leaf the greater its width.

Graph 3. The length and the width of the leaves on the sun-exposed side of thetree

Graph 4. Correlation between the length and the width of the leaves on thesun-exposed side of the tree

20

1 4 710

13

16

19

22

25

28

31

34

37

40

43

46

49

0

20

40

60

80

100

120

140

160 width

lenght

Sample number

Size[cm]

50

60

70

80

90

100

110

120

130

80 90 100 110 120 130 140 150

Widthoftheleaf

[cm]

Length of the leaf [cm]


21/24

Table 19 . Ranking the dimensions of the leaves from the shadowed side

Table 20. Calculating the differences between ranks

No. Length[cm]

Rank R1 Width[cm]

Rank R2 D=R1 R2 D

No.

Length[cm] (X1)

0.1

Sortedvalue

Rank

Width[cm] (X1)

0.1

Sortedvalue

Rank

1 73 42 1 50 31 12 93 53 2 72 42 23 100 56 3 67 43 3.5

4 104 60 4 70 45 3.55 130 62 5 100 47 5.5

6 123 64 6 100 50 5.57 56 65 7 42 52 7.58 110 73 8.5 52 53 7.59 80 73 8.5 60 54 9.510 116 75 10 91 54 9.511 93 78 11 70 55 1112 92 80 12.5 65 57 12.513 97 80 12.5 68 57 12.5

14 124 83 14.5 100 58 1415 60 83 14.5 43 60 15

16 137 84 16 92 61 1617 73 87 17.5 54 62 17

18 116 87 17.5 83 65 18.519 83 88 19.5 61 65 18.520 78 88 19.5 58 67 20.521 42 90 21.5 31 67 20.522 116 90 21.5 86 68 2223 62 91 23 45 70 23.5

24 88 92 24.5 57 70 23.525 102 93 25.33 75 72 25.5

26 96 93 25.33 62 72 25.527 53 93 25.33 47 73 27.5

28 130 94 28 94 73 27.529 115 96 29 82 74 2930 122 97 30 91 75 30.3331 103 100 31.5 75 75 30.3332 93 100 31.5 74 75 30.3333 88 101 33 55 76 3334 94 102 34.33 65 77 34.535 75 102 34.33 57 77 34.5

36 105 102 34.33 77 80 3637 64 103 37 54 81 37

38 84 104 38 73 82 3839 80 105 39 72 83 39.5

40 100 110 40 80 83 39.541 87 115 41.5 81 86 41.542 91 116 42.33 76 86 41.543 102 116 42.33 77 91 43.544 101 116 42.33 94 91 43.545 90 122 45 83 92 45

46 83 123 46 67 94 46.547 102 124 47 75 94 46.5

48 87 130 48.5 73 100 47.3349 65 130 48.5 53 100 47.3350 90 137 50 86 100 47.33

21


22/24

0.1 0.11 73 8,5 96 26,5 -18 3242 93 25,33 85 9.25 16,08 258,56643 100 31,5 100 31,5 0 04 104 38 95 24.5 13,5 182,255 130 48,5 92 20,33 28,17 793,5489

6 123 46 91 18,5 27,5 756,257 56 3 89 13 -10 100

8 110 40 115 47 -7 49

9 80 12,5 92 20,33 -7,83 61,308910 116 42,33 80 3,5 38,83 1507,76911 93 25,33 119 50 -24,67 608,608912 92 24 91 18,5 5,5 30,2513 97 30 94 23 7 4914 124 47 110 42,25 4,75 22,562515 60 4 105 35,25 -31,25 976,5625

16 137 50 95 24,5 25,5 650,2517 73 8,5 83 5,5 3 9

18 116 42,33 100 31,5 10,83 117,288919 83 14,5 103 34 -19,5 380,25

2 78 11 90 14,25 -3,25 10,5625

21 42 1 84 7,5 -6,5 42,2522 116 42,33 110 42,25 0,08 0,006423 62 5 90 14,25 -9,25 85,562524 88 19,5 85 9,25 10,25 105,062525 102 34,33 96 26,5 7,83 61,3089

26 96 29 107 40,5 -11,5 132,2527 53 2 110 42,25 -40,25 1620,063

28 130 48,5 105 35,25 13,25 175,562529 115 41 106 39 2 4

30 122 45 105 35,25 9,75 95,062531 103 37 90 14,25 22,75 517,562532 93 25,33 118 49 -23,67 560,2689

33 88 19,5 60 1 18,5 342,2534 94 28 84 7,5 20,5 420,2535 75 10 98 28,5 -18,5 342,2536 105 39 99 30 9 8137 64 6 102 33 -27 729

38 84 16 92 20,33 -4,33 18,748939 80 12,5 85 9,25 3,25 10,5625

40 100 31,5 117 48 -16,5 272,2541 87 17,5 75 2 15,5 240,2542 91 23 110 42,25 -19,25 370,562543 102 34,33 80 3,5 30,83 950,488944 101 33 98 28,5 4,5 20,25

45 90 21,5 112 46 -24,5 600,2546 83 14,5 90 14,25 0,25 0,062547 102 34,33 83 5,5 28,83 831,1689

48 87 15,5 105 35,25 -19,75 390,062549 65 7 85 9,25 -2,25 5,0625

50 90 21,5 107 40,5 -19 361Sum: 16271,46

H0: there is no correlation between the length and the width of the leaves on theshadowed side of the tree

H1: There is a positive correlation between the length and width of the leaves on theshadowed side of the tree

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23/24


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results of Spearman Rank Correlation statistical tests it can be observed thatthere is a strong positive correlation (the greater the length of the leaf, thegreater its width) between the leaves on both the sun-exposed and theshadowed side of the tree. However the gradient of correlation appears to bemuch grater on the shadowed side as the value of rs there is 0.78 which is graterthan 0.33 on the sun-exposed side.

analysing corelations between the leaves from sun-exposed and shadowed side of the three

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