Quadrats

Square sample areas marked out with a frame.

Repeatedly place a quadrat

at random positions in a

habitat

Record number of organisms present each

time

- Shape of known area- Randomly placed in each area- Small quadrats = many times, larger quadrats = fewer

times- Random number tables used to avoid bias- Count number of individuals of species inside quadrat- Many samples must be taken to make it

representative- Population density = number of individual/area

Quadrat Rules

Quadrats

Random numbers

generated to create

coordinates where the quadrat is

placed in an area (reduces

bias)

There are two main kinds of data that we can gather for the daisies in a field:– Qualitative: ‘there are lots of daisies in the field’– Quantitative: ‘there are 5087 daisies in the field’

Counting daisies

How many daisies in the field?You have 15 seconds…

How did you estimate the number of daisies?– Did you try to count them all?– Or did you use another method?

We need a quantitative estimate for the number of daisies – it doesn’t have to be perfect but it should be as close as possible to the real number.

Write your first estimate down, then try again, seeing if this will help…

How many did you count?

How did you use the grid to estimate the number of daisies? Did it help?– There are 78 daisies

If you were asked to count the number of daisies in the school field, it would be impractical to count each one.

– How could you use the grid method to get an accurate, reproducible estimate?

Use the following steps:– Select at least three quadrats and count how many daisies are in each (eg 4,

8, 3)– Then find the mean number per quadrat (4 + 8 + 3 = 15. 15/3 = 5

daisies per quadrat)– Multiply the mean by the number of quadrats that would fit into the field to

get your estimated total number of daisies. (5 x 20 = 100 daisies estimated in the field)

Is your estimate the same?

There were 103 daisies in the field.

How close were you?

How many daisies were there?

They only work for immobile/slow moving populations.

The more data you collect, the more reproducible your result…the more samples the better!

Quadrats should be placed randomly to avoid bias.

Quadrats: Top Tips

Chi-squared test

Chi Squared Test (stats test)

Test for an association between

the species

- If species always are in the same

quadrat (positive)

- If species are never in the same

quadrate (negative)

Hypotheses

H0 = null hypothesis = two species are distributed independently (there is no association)

H1 = two species associated (either positively or negatively)

My results

Species Frequency

Heather only 9

Moss Only 7

Both species 57

Neither species 27

Total samples 100

Heather absent

Heather present

Total

Moss absent 27 9 36

Moss present

7 57 64

Total 34 66 100

Start with a contingency table

Expected values

Expected = row total x column total grand

total Heather absent

Heather present

Total

Moss absent (34x36)/100 =

(66x36)/100 =

36

Moss present (34x64)x100 =

(66x64)/100=

64

Total 34 66 100

Expected values

Expected = row total x column total grand

total Heather absent

Heather present

Total

Moss absent (34x36)/100 = 12

(66x36)/100 = 24

36

Moss present (34x64)x100 = 22

(66x64)/100= 42

64

Total 34 66 100

Degrees of Freedom

(m – 1) x (n – 1)

m = number of rowsn = number of columns

(measure of how many values can vary)

Degrees of Freedom

(m – 1) x (n – 1)

m = number of rows = 2n = number of columns = 2

(2 – 1) x (2 – 1)

= 1 x 1

Degrees of freedom for this test = 1

Critical value

Find critical value from a table of chi-squared values

Significance level of 5% (0.05)

Chi-squared Tests (x2)

• Degrees of freedom = (m– 1)(n-1)• If x2 is > critical value, H0 is rejected

Critical region

Find critical region from a table of chi-squared values

Significance level of 5% (0.05)

For this test = 3.841

Chi Squared Test

χ2 = (O – E)2

E

(O – E)2

E(O – E)2

E(O – E)2

E(O – E)2

E

Chi Squared Test

χ2 = (O – E)2

E

(27 – 12)2

12(7 – 22)2

22(9 – 24)2

24(57 – 42)2

42

Chi Squared value

18.75 + 10.23 + 9.38 + 5.36

Chi squared = 43.72

Which hypothesis

Chi squared = 43.72Critical region = 3.841

Compare calculated chi squared value to critical region

Higher than critical region = reject null hypothesis (there is an association)

Equal to or lower than critical region = keep null hypothesis (there is no association)

Since the x2 value of 43.72 is greater than critical value of 3.841, the null hypothesis is rejected.

Therefore, we can be 95% sure that there is a relationship between the heather and moss.

How do you think the abundance of bluebells changes depending on how deep into this woodland you go?

Finding a trend

By placing one quadrat each metre along a straight line you can find the % cover for different distances. This is called ‘sampling along a transect’.

Because we have quantitative results we can specifically say how the trend develops – it starts at 18% for 1m, increases rapidly to 61% for 6m but levels out at 68% for 8m.

If we’d only had qualitative results we’d only be able to say ‘there are more bluebells the further in you go’ – not very useful!

Quant vs. Qual

• Create a quadrat of 5cm x 5cm using A4 paper• Do at least 50 throws• Use the previous example and the instructions

to complete the rest of the worksheet

Complete your own!