From last lesson…

Ecological techniques Describe the ecological techniques used to assess abundance and distribution of organisms in a natural habitat, including types of quadrat, transects, ACFOR scales, percentage cover and individual counts. Select appropriate ecological techniques according to the ecosystem and organisms to be studied Be able to use statistical tests to analyse data, including t-test, correlation coefficient and Spearman’s rank.

Ecological Techniques Abundance It is the numbers of an organism relative to the numbers of other organisms in the same habitat Distribution Arrangement of organisms in the environment

Why might studying distribution and abundance be important?

Quadrats Individual counts Percentage cover ACFOR Scale Random sampling Line transect Belt transect These will be explained outside and you can add to your notes when we return

ACFOR scale Sometimes we will assign numbers to each letter so that data can then be manipulated and statistical tests can be performed A - abundant C - common F - frequent O - occasional R - rare

Belt transect

Line transect

William Gosset (aka ‘Student’) (1876-1937) Worked in quality control at the Guinness brewery and could not publish under his own name. Former student of Karl Pearson 10

The t-test What can this test tell you? If there is a statistically significant difference between two means, when: The sample size is less than 25. The data is normally distributed 11

Our mini investigation To see if there is a significant difference in the number of plant species at Site A and Site B

What is the null hypothesis (H0)? H0 = there is no statistically significant difference between the number of different plant species at Site A compared with site B HA = there is a significant difference between the number of different plant species at Site A compared with Site B. If the value for t exceeds the critical value (P = 0.05), then you can reject the null hypothesis. What is the null hypothesis (H0)? 13

(x – x)2 1 2 3 4 5 6 7 8 9 10 Mean  Quadrat number Number of different species at Site A (x – x)2 Number of different species at Site B 1 2 3 4 5 6 7 8 9 10 Mean 

(x – x)2 1 2 3 4 5 6 7 8 9 10 Mean  Quadrat number Number of different species at Site A (x – x)2 Number of different species at Site B 1 2 3 4 5 6 7 8 9 10 Mean 

t-test t =  x1 – x2  (s12/n1) + (s22/n2) (x – x)2 n – 1 SD = (x – x)2 n – 1  x1 = mean of first sample x2 = mean of second sample s1 = standard deviation of first sample s2 = standard deviation of second sample n1 = number of measurements in first sample n2 = number of measurements in second sample 16

Worked example Does the pH of soil affects seed germination of a specific plant species? Group 1: eight pots with soil at pH 5.5 Group 2: eight pots with soil at pH 7.0 50 seeds planted in each pot and the number that germinated in each pot was recorded.

What is the null hypothesis (H0)? H0 = there is no statistically significant difference between the germination success of seeds in two soils of different pH HA = there is a significant difference between the germination of seeds in two soils of different pH If the value for t exceeds the critical value (P = 0.05), then you can reject the null hypothesis. What is the null hypothesis (H0)? 18

Construct the following table… Pot Group 1 (pH5.5) (x – x)2 Group 2 (pH7.0) 1 38 1.21 39 20.25 2 41 3.61 45 2.25 3 43 15.21 6.25 4 0.01 46 5 37 4.41 48 6 7 8 36 9.61 44 0.25 Mean 39.1 4.86 43.5 10.25  38.88 82.0 19

Construct the following table… Pot Group 1 (pH5.5) (x – x)2 Group 2 (pH7.0) 1 38 1.27 39 20.25 2 41 3.52 45 2.25 3 43 15.02 6.25 4 0.02 46 5 37 4.52 48 6 7 8 36 9.77 44 0.25 Mean 39.1 43.5  38.88 82.0 20

Calculate standard deviation for both groups SD = (x – x)2 n – 1  38.88 = 2.36 =  8 – 1 Group 2: SD = (x – x)2 n – 1  82.0 =  = 3.42 8 – 1

Using your means and SDs, calculate value for t x1 – x2  (s12/n1) + (s22/n2) t = 39.1 – 43.5 -4.4 t = =  0.696 + 1.462  (2.362/8) + (3.422/8) t = 2.99

Compare our calculated value of r with the relevant critical value in the stats table of critical values Our value of t = 2.99 Degrees of freedom = n1 + n2 – 2 = 14 D.F. Critical Value (P = 0.05) 14 2.15 15 2.13 16 2.12 17 2.11 18 2.10 Our value for t exceeds the critical value, so we can reject the null hypothesis. We can conclude that there is a significant difference between the two means, so pH does affect the germination rate for this plant. 23

Plenary Evaluate the use of a frame quadrat Describe in your own words random sampling, belt transect and line transect State the purpose of the t-test? In the worked example, analyse the data to explain the difference between the groups of results.