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Spearmen Rank

Spearmen Rank

 

Used for:

  • A significant association between 2 sets of measurements from the same sample

5-30 pairs of measurementsMethod

Structure

  1. Create a table with the headings
Variable 1 Rank 1 Variable 2 Rank 2 Difference (d) d2

If two or more variables are the same, add the rank positions the number would have if they were the next ones in the sequence.

For example, if you got 2 ‘4’s in a row

Variable 1 Rank 1
5 2
4 1.5 (1*)
4 1.5 (2*)

For their rank, add 1+2 (*their standard rank) = 3

Then divide 2 by how many there are, which is also 2

So 3 ÷ 2 = 1.5

  1. if they are however the next numbers in the sequence, then divide by the number of positions used
  2. calculate the difference (d) between the two ranks
  3. square the difference of each
  4. Add up all the values of dto get the ∑
  5. Multiply the ∑ by 6
  6. To find n, it is the amount of data (the highest rank)
  7. Cube the value of n
  8. Subtract the value of nby the original value of n

The value will be between -1 (perfect negative correlation) and + 1 (perfect positive correlation). The closer the value is to 0, the weaker the correlation

e.g. “Some students carried out a belt transect across some sand dunes. They counted the number of banded snails they found as they moved away from the shoreline

Distance from Shore (m) Number of banded snails
0 10
5 2
10 8
15 7
20 5
25 8
30 2

results

Variable 1 Rank 1 Variable 2 Rank 2 Difference (d) d2
0 1 10 7 -6 36
5 2 2 1.5 0.5 0.25
10 3 8 5.5 -2.5 6.25
15 4 7 4 0 0
20 5 5 3 2 4
25 6 8 5.5 0.5 0.25
30 7 2 1.5 5.5 30.25
Total ∑ 77

∑ = 77
= -0.375

Structure
*we use 0.05 as that is a 5% factor of error

As n = 7 (as there are 7 variables) and we use the 0.05 grid, the result is:

Critical value: 0.6786

The correlation is not significant as there is a 5% chance the results are due to chance, therefore we agree with the null hypothesis

 

 

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Chi Squared Test

Chi Squared Test

Chi Squared Test, tests for variance which is the difference between observed and expected numbers

Structure
O = Observed Number

E= Expected number

The numbers of degree of freedom = Number of Categories – 1

If the value is greater than the table value then null hypothesis is rejected

e.g. A gardener wanted to compare the number of worms found in his compost heap with the number in the middle of the field

results:

Compost Heap

Lawn

150

50

Null Hypothesis: There is no significant difference between the distribution of the worms in the compost heap compared to the lawn

Compost Heap

Lawn

Total

O

150

50

200

E

100

100

200

(O-E)

50

-50

0

(O-E)2

2500

-2500

0

(O-E)2

E

25

25

50

*according to the null hypothesis, there are equal quantities of the worms in both the compost and the law, therefore our “expected” (E) is the total of the Observed results divided by the 2 groups (150 + 50 = 200 200 / 2 = 100

*Remember that it is a 5% error margin, therefore use

0.05

table
*Remember that it is a 5% error margin, therefore use 0.05

Degrees of Freedom: 2 – 1 = 1

(2 Variables: Compost Heap and Lawn)

Critical Value = 3.84

The critical value is 3.84 which is lower than the , so the value is rejected therefore the null hypothesis is also rejected. There is less than a 5% probability that the difference is due to chance, so the difference is significant.

T-test

T-test

  1. Used for:

  2. To find the difference between 2 means
  3. The data is normally distributed
  4. Structure

    1 <= mean of the first sample

    = mean of the second sample

    S= Standard Deviation of first sample

    S= Standard Deviation of second sample

    n= Number of measurements in first sample

    n= Number of measurements in second sample

    degrees of freedom = (n1 + n2) – 2

  5. Compare the calculated T-value, with the number of degrees of freedom to the critical T-value from the T-distribution table at a chosen confidence level
  6. Reject the null hypothesis when: Calculated T value ≥ Critical T-value

Standard Error

Standard Error

  • Standard error is a measurement which determines the accuracy of results according to 95% confidence.

Structure
SD= standard Deviation

n = sample size

  • It is calculated by dividing the Standard Deviation of the mean by the square root of the sample size
  • The 95% confidence interval allows the results to have:
  • The true mean value of the results from which the sample was taken lies between the upper and lower confidence limit
  • If the intervals of two calculated means do not overlap, then the results are 95% confident that these means are different
Standard Deviation

Standard Deviation

Structure

 

  1. Find the mean (In this example it is 1+2+3+4+5/5 = 3)
  2. Find the total amount of data (In this example there are 5 as there are 5 given modes of transport, therefore N = 5)
  3. To complete the top half of the fraction, , add an extra column to the table. For each piece of data subtract the given value from the mean ( in this example) and square the result

i.e. car = 2 people, so (2 – mean(3)) 2 = 4

 

  1. Add up the total of all the results from step 3 (The sum of) This result can now replace the top half of the fraction ( in this example, it’s 4+1+0+1+4=10)
  2. To find the bottom half of the fraction, like the equation says to do so subtract 1 from the total amount of data in the dataset, 5-1 = 4
  3. The fraction should look like this now:
  4. Pop all the calculation into the calculator

e.g.

How do you get to school? Total
Bus 1 (1-3) = 4
Car 2 (2-3) = 1
Bike 3 (3-3) = 0
Walk 4 (4-3)= 1
Helicopter 5 (5-3) = 4
Total 15 10

n = 5-1 = 4

S = 1.58

 

Introduction to Biological Statistics

Introduction to Biological Statistics

 

The statistical tests are not particularly challenging so don’t get too worried about the equations looking scary.

There are 3 main tests that you do need to know:

flowchart

(Also note that Spearmen Rank and Chi Squared are primarily found in A2.)

Null Hypothesis

  • A null hypothesis dictates that there is no differences between the samples being studied.
  • An experiment must be made which tries to disproves this.
  • The ending results either disproves the null hypothesis or fails to disprove the null hypothesis
  • It can never prove a hypothesis to be true

Significance Levels

  • The probability that the observed data would occur by chance in a given single null hypothesis
  • This is 5% significance level
  • If the results produced are likely to be due by chance, then the results are ‘non-significant’ and the null hypothesis is cannot be rejected.
  • If the results are not produced by chance, the results are ‘significant’ then the null hypothesis can be rejected

Critical Values

  • The Critical Value is the value corresponding to the given significance level p=0.05. This given value is the boundary between the samples resulting in a test statistic that leads to rejecting the null hypothesis
  • If the absolute value of your test statistic is greater than the critical value, you can declare statistical significance and reject the null hypothesis.

 

Diversity Index

Diversity Index

Measuring Diversity

Assessment to test for the monitoring environmental change, damage or the success of conservation effects

Extreme Environments

Species Diversity is low in extreme environments which are dominated by abiotic factors (non-living factors such as temperature, pH etc) where there is a population that may fluctuate drastically

Species Diversity

Species diversity is the number of different species and the number of individuals of each species within any one community

It is more useful to measure the species diversity compared to the species richness due to it focusing more on the amount of induvial in a species rather than just how many species there are.

Diversity and the Ecosystem

Ecosystems that undergo abiotic factors may cause a low-level organism in the food change to be unable to produce food for the next trophic level.

In a diverse ecosystem, the higher trophic level can change to another food course which is available, however in a low diversity ecosystem, another food source may not be available

Index of Biodiversity

 

Structure

 

N = total number of all organisms

n = Population size of the particular species

e.g.

Species

Number of individuals seen

Standard Deviation

n x (n-1)

Magpie

11

11 x 10 = 110

Black Headed Gull

4

4 x 3 = 12

Crow

4

4 x 3 = 12

Blackbird

1

1 x 0 = 0

Starling

37

37 x 36 = 1332

Sparrow

7

7 x 6 = 42

Total

64

1508

d = 1508

64 (64-1)

d = 0.374

The closer the value is to 0, that means there is infinite diversity, whereas closer to 1 means there is no diversity.