You would not go and count every single flower

How would you know if the flowers are associated

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

Quadrat Rules 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

Quadrats Random numbers generated to create coordinates where the quadrat is placed in an area (reduces bias)

Counting daisies 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’

How many daisies in the field? You have 15 seconds… 78

How many did you count? 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…

78

Is your estimate the same? 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)

103

How many daisies were there? There were 103 daisies in the field. How close were you?

Quadrats: Top Tips 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.

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)

H1 = two species associated (either positively or negatively) 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

Start with a contingency table Heather absent Heather present Total Moss absent 27 9 36 Moss present 7 57 64 34 66 100

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 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 34 66 100

Degrees of Freedom (m – 1) x (n – 1) m = number of rows n = number of columns (measure of how many values can vary)

Degrees of Freedom (m – 1) x (n – 1) m = number of rows = 2 n = 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) http://www.real-statistics.com/chi-square-and-f-distributions/independence-testing/ 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 (O – E)2 E (O – E)2 E (O – E)2 E (O – E)2 E χ2 = (O – E)2 E  (O – E)2 E (O – E)2 E (O – E)2 E (O – E)2 E

 Chi Squared Test (O – E)2 E (27 – 12)2 12 (7 – 22)2 22 (9 – 24)2 24 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 = 43.72 Critical 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 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.

Finding a trend How do you think the abundance of bluebells changes depending on how deep into this woodland you go? 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’.

Quant vs. Qual 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!

Complete your own! 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