An ecologist studied two species in an area and his observation are shown in the given table. Draw a scatter graph to show the data in the table.

The scatter graph shown above represent the correlation of two species i.e the dependence of one species on the other. The ecologists use Spearman's correlation coefficient to find whether one variable increases with the increase in the other (positive correlation) or decreases with the decrease in the second variable (negative correlation).

An ecologist studied two species in an area and his observation are shown in the given table. Calculate Spearman's correlation coefficient for the data in this table.

The equation used to find Spearman's correlation coefficient is where n=number of pairs of items in the sample. D is the difference between each pair of ranked measurements. ∑ is ‘the sum of’. Quadrat C.vulgaris (A) Rank A V.myrtillus (B) Rank B D (RankA - Rank B) D2 1 30 3 15 4 -1 1 2 37 1 23 2 -1 1 3 15 6.5 6 7 -0.5 0.25 4 15 6.5 10 6.5 0 0 5 20 5 11 5 0 0 6 9 8 10 6.5 1.5 2.25 7 3 10 3 9 1 1 8 5 9 1 10 -1 1 9 10 7 5 8 -1 1 10 25 4 17 3 1 1 11 35 2 30 1 1 1 ∑D 2 =9.5 Thus, r = 1 - 6 × 9 . 5 1331 - 11 = 1 - 0 . 431 = 0 . 57 Thus, we can say that the value of Spearman's correlation coefficient is 0.57. Since, the value is near to +1, the ecologist can analyse that there is a positive correlation between the given species.

An ecologist studied two species in an area and his observation are shown in the given table. Calculate Spearman's correlation coefficient for the data in this table. Why the the ecologist was able to reject the null hypothesis.

In the given example, to find out the correlation between two variables, we make a null hypothesis that there is no correlation between the abundance of species. The equation used to find Spearman's correlation coefficient is where n=number of pairs of items in the sample. D is the difference between each pair of ranked measurements. ∑ is ‘the sum of’. Quadrat C.vulgaris (A) Rank A V.myrtillus (B) Rank B D (RankA - Rank B) D 2 1 30 3 15 4 -1 1 2 37 1 23 2 -1 1 3 15 6.5 6 7 -0.5 0.25 4 15 6.5 10 6.5 0 0 5 20 5 11 5 0 0 6 9 8 10 6.5 1.5 2.25 7 3 10 3 9 1 1 8 5 9 1 10 -1 1 9 10 7 5 8 -1 1 10 25 4 17 3 1 1 11 35 2 30 1 1 1 ∑D 2 =9.5 r = 1 - 6 × 9 . 5 1331 - 11 = 0.57 Thus, we can say that the value of Spearman's correlation coefficient is 0.57. Since, the value is near to +1, the ecologist can analyse that there is a positive correlation between the given species. Thus, we can reject the null hypothesis that there is no correlation between the two variables and can make an alternative hypothesis that the abundance of both the species is positively related.

Number of tree Circumference of tree/metres Mean width of crack/mm 1 1.77 50 2 1.65 28 3 1.81 60 4 0.89 24 5 1.97 95 6 2.15 51 7 0.18 2 8 0.46 15 9 2.11 69 10 2.00 64 11 2.42 74 12 1.89 69 Draw a scatter graph of the data given in the table.

The scatter graph shown below represents the correlation of two variables i.e the dependence of one variable on the other. The ecologists can use Pearson's correlation coefficient to find whether one variable increases with the increase in the other (positive correlation) or decreases with the decrease in the second variable (negative correlation).

HARD

AS and A Level

IMPORTANT

Earn 100

Explain the advantage of using Simpson’s index of diversity rather than a species list as a way to represent the biodiversity of an area.

Important Questions on Classification Biodiversity and Conversation

HARD

AS and A Level

IMPORTANT

Biology for Cambridge International AS & A Level coursebook 2nd Edition Digital Access>Chapter 18 - Classification Biodiversity and Conversation>Question>Q 12

The Simpson’s index of diversity for the vegetation in an area of open grassland was 0.8 for a similar sized area of vegetation beneath some conifer trees it was 0.2. What do you conclude from these results.

MEDIUM

AS and A Level

IMPORTANT

Biology for Cambridge International AS & A Level coursebook 2nd Edition Digital Access>Chapter 18 - Classification Biodiversity and Conversation>Question>Q 12

In a survey of trees in a dry tropical forest. Some students identified five tree species (A to E). They counted the numbers of trees in an area 100 m * 100 m. Their results are as follows:

Species	No. of individuals
A	56
B	48
C	12
D	6
E	3

Looking at table write a set of instructions for making a kite diagram

HARD

AS and A Level

IMPORTANT

Biology for Cambridge International AS & A Level coursebook 2nd Edition Digital Access>Chapter 18 - Classification Biodiversity and Conversation>Question>Q 13

An ecologist studied two species in an area and his observation are shown in the given table.

Question Image

Draw a scatter graph to show the data in the table.

HARD

AS and A Level

IMPORTANT

Biology for Cambridge International AS & A Level coursebook 2nd Edition Digital Access>Chapter 18 - Classification Biodiversity and Conversation>Question>Q 13

An ecologist studied two species in an area and his observation are shown in the given table.

Question Image

Calculate Spearman's correlation coefficient for the data in this table.

HARD

AS and A Level

IMPORTANT

Biology for Cambridge International AS & A Level coursebook 2nd Edition Digital Access>Chapter 18 - Classification Biodiversity and Conversation>Question>Q 13

An ecologist studied two species in an area and his observation are shown in the given table.

Question Image

Calculate Spearman's correlation coefficient for the data in this table. Why the the ecologist was able to reject the null hypothesis.

HARD

AS and A Level

IMPORTANT

Biology for Cambridge International AS & A Level coursebook 2nd Edition Digital Access>Chapter 18 - Classification Biodiversity and Conversation>Question>Q 13

An ecologist studied two species in an area and his observation are shown in the given table.

Question Image

Calculate Spearman's correlation coefficient for the data in this table. State the conclusion that the ecologist can make from this investigation.

HARD

AS and A Level

IMPORTANT

Biology for Cambridge International AS & A Level coursebook 2nd Edition Digital Access>Chapter 18 - Classification Biodiversity and Conversation>Question>Q 13

An ecologist studied two species in an area and his observation are shown in the given table.

Question Image

The ecologist think that this relationship is due to habitat preference that both species prefer dry soil. Describe an investigation that he could carry out to test this hypothesis.

HARD

AS and A Level

IMPORTANT

Biology for Cambridge International AS & A Level coursebook 2nd Edition Digital Access>Chapter 18 - Classification Biodiversity and Conversation>Question>Q 14

Number of tree	Circumference of tree/metres	Mean width of crack/mm
1	1.77	50
2	1.65	28
3	1.81	60
4	0.89	24
5	1.97	95
6	2.15	51
7	0.18	2
8	0.46	15
9	2.11	69
10	2.00	64
11	2.42	74
12	1.89	69

Draw a scatter graph of the data given in the table.

Explain the advantage of using Simpson’s index of diversity rather than a species list as a way to represent the biodiversity of an area.

Important Questions on Classification Biodiversity and Conversation

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