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Ecology is a Science – Queen of Sciences

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Pattern Observation

Rigorously Describe

Model

Explanation or theory (maybe >1)

Hypothesis

Prediction deduced from model

Generate null hypothesis – H0: Falsification test

*

Test

- Experiment
- IF H0 rejected – model supported
- IF H0 accepted – model wrong

*

Statistics

Ecology is a Science – Queen of Sciences

Follows Scientific Method

Hypothetico-deductive approach (Popper) based on principle of falsification: theories are disproved because proof is logically impossible. A theory is disproved if there exists a logically possible explanation that is inconsistent with it

Can only really test hypotheses by experimentation

Give off light when attacked by copepods to attract fish (to eat the copepods)

Model

Explanation or theory (maybe >1)

Hypothesis

Prediction deduced from model

Generate null hypothesis – H0: Falsification test

H0: Bioluminescence has no effect on predation of copepods by fish (or decreases predation)

H1: Bioluminescence increases predation of copepods by fish

- Experiment
- IF H0 rejected – model supported
- IF H0 accepted – model wrong

Test

Notiluca give off light when disturbed

Pattern Observation

Rigorously Describe

Statistics – summary, analysis and interpretation of data eat the copepods)

Data pl (datum, s) are observations, numerical facts

RAW MATERIAL OF SCIENCE

Types of Data

Nominal data – gender, colour, species, genus, class, town, country, model etc

Continuous data – concentration, depth, height, weight, temperature, rate etc

Discrete data – numbers per unit space, numbers per entity etc

Often referred to as VARIABLES because they vary

The type of data collected influences their analysis

Variability – key feature of the natural world eat the copepods)

Genotypic/Phenotypic variation – differences between individuals of the same species (blood-type, colour, height etc)

Variability in time/space – changes in numbers per unit space, time

Uniform

Random

Clumped

Space/Time

Measurement variability – experimental error (bias)

Variability = Uncertainty eat the copepods)

Variability means that it is impossible to describe data exactly –Accuracy, Precision

20 cm + eat the copepods)

6 mm +

300 μm +

Accuracy – how close a measure is to the real value

20.63 cm

20.631506542 cm

Accept a level of measurement error: be upfront

20.755 eat the copepods)

20.632

20.623

21.102

20.710

19.986

22.356

20.493

20.578

Precision – how close repeat measures are to each other

Describing data and variability eat the copepods)

Population – the entire collection of measurements, e.g. mass of 19 yr old elephants, the blood pressure of women between 16-18 yrs of age, number of earthworms on UWC rugby field, height of UWC BSc II students, oxygen content of water

If population small, then possible to obtain all measurements in the population. However, if population very large, then impractical or impossible to measure all - must take Samples

One sample from a large population is meaningless –need to take replicate samples and obtain an average sample measure, which is then assumed to be representative of the population

When taking samples it is vital that they are Random and Independent

Earthworms eat the copepods)

D

C

B

A

100 m

100 m

How many earthworms in the field of

25 0000 m2?

500 m

A – 1 (25)

B – 17 (375)

C – 10 (250)

D – 4 (100)

Mean = 8 (200)

500 m

N = 106 eat the copepods)

Σ = 182.4

Mean = 1.72

Mean = Σx

N

Population mean = μ; sample mean = x

We use x as a proxy for μ

Describing data and variability

How high is a UWC BSc II student?

What is the NO3 concentration in the Black River?

Measures of Central Tendency

Arithmetic mean or Average

=COUNT(DATA:RANGE) eat the copepods)

=SUM(DATA:RANGE)

=TOTAL / N

MSExcel also allows you to calculate the mean from a data series…

=AVERAGE(DATA:RANGE)

Enter data (x) into MSExcel spreadsheet

Calculate N

Calculate Total

Calculate Mean

Mode – eat the copepods)the most commonly represented value

How? Construct a frequency table from the data: whichever “class” of data occurs at the highest frequency is the MODE

Classes should be calculated in EVEN intervals from smallest to largest value of x

MSExcel allows you to calculate a frequency table

It also allows you to calculate MODE: = MODE(DATA:RANGE)

If there are an odd number of data points this is easy eat the copepods)

If there are an even number of data points you will need to interpolate

The middle data point lies half way between that associated with observation no 5 (1.75) and observation no 6 (1.8) = 1.775

Can be calculated as either (1.75 + 1.8) / 2

OR as ((1.8 – 1.75) / 2) + 1.75

MSExcel also allows you to calculate the median from a data series…

=MEDIAN(DATA:RANGE)

Median – the middle value in a ranked data set

Step 1 – Order the data from low to high

Step 2 – Determine the middle data point

Range: eat the copepods)

Essentially the lowest and highest value in the data set

Measures of Dispersion – how data are distributed around the mean

N.B. Subject to measurement errors, typographic mistakes and freaks

In MSExcel: =MIN(DATA:RANGE), =MAX(DATA:RANGE)

¼ of the way through this ranked data set of 9 values = observation number 2.25 (=9 x 0.25)

Calculate the data point that would be associated with observation number 2.25 by interpolation between observation numbers 2 (1.45) and 3 (1.6) i.e. = ((1.6 – 1.45) * 0.25) + 1.45

= ((0.15) * 0.25) + 1.45

= 0.0375 + 1.45

= 1.4875 (Lower Quartile)

Inter-Quartile Range:

In a ranked data set, those values corresponding to ¼ (lower or 25% quartile) and ¾ (upper or 75% quartile) of the observations: 50% of the observations lie between these two values

DITTO for 75% Quartile………………..

To give us an interquartile range: 1.4875 – 1.8875

If you have to use a range, use the inter-quartile range as it ignores outliers

MEDIAN observation number 2.25 (=9 x 0.25)

LOWER

UPPER

Can also calculate Cumulative Frequency

THEN

Draw a Graph of Cumulative Frequency (Y) against Ordered Data (X) on an X-Y PLOT

THEN

Calculate Lower and Upper Quartiles from Figure

Always = Zero observation number 2.25 (=9 x 0.25)

Mean Deviation

Σ

Convert negatives to positives to give overall deviation from the mean; SUM, Divide by N to give average deviation of any data point from the mean – MEAN DEVIATION

N

mean

Always = Zero observation number 2.25 (=9 x 0.25)

The values of x = 4, sample variance (2.25) and sample standard deviation (1.5) ALL refer to the sample of 16 measures

Sum of Squares (Sample)

Mean Sum of Squares (sample)

(Variance)

Σ

Σ

Σ

√

N

N

N

(Variance) =

mean

mean

mean

Standard Deviation (sample)

s = 1.5

16

2.25

Length (mm) of Drosophila melanogaster Instar III larvae

Variance and Standard Deviation

There is another way to remove the negatives – and that is to square the (x – mean) values

Square units?

X observation number 2.25 (=9 x 0.25)

3

If you had collected only the first four of the measures (in pink), then the total would be 18. In order for you to get a mean of 5 from five measures, the last value would HAVE TO BE seven (7).

In other words the last number is not independent of the others, and when we deal with the population we have to use independent data.

Consequently when we calculate σ2 we divide the sum of squares by (n-1) and NOT n (as previously): σ is still calculated as

In this table of five measures, the total is 25 and x is 5

4

X

5

3

6

4

7

5

6

Total

25

7

N

5

Mean

5

Total

25

N

5

Mean

5

√

σ2

Are they the best estimators of these properties for the population?

In the case of the mean (x), there is no reason to suppose that the mean of all observations in the sample will not provide the best estimator of the population mean (μ)

However, we cannot use sample variance and sample standard deviation as estimators of σ2 and σ respectively!

WHY? Because not all the measures are completely independent of each other.

(n-1) = Degrees of Freedom = v

Sample derived estimates of population variance and population standard deviation are referred to as s2 and s respectively

Mean observation number 2.25 (=9 x 0.25)

N

s2

s

The smaller the standard deviation, the closer the data are to the mean

The bigger the standard deviation, the greater the spread of data around the mean – the greater the variability

So…… observation number 2.25 (=9 x 0.25)

Mean – measure of central tendency of sample data

Variance and Standard Deviation – index of dispersion of data around the sample and/or population mean

Two other commonly reported measures of central tendency:

Standard Error – index of dispersion of sample means around population mean

95% confidence intervals – describes limits around your sample mean within which you are 95% confident that the REAL value of the population mean lies

To calculate the last two measures, it is necessary to digress a little……

A die has six sides: observation number 2.25 (=9 x 0.25)

The probability of throwing = 1/6 = 0.167

The probability of NOT throwing a = 1 – 0.167 = 0.833

Variability = Uncertainty

View uncertainty in terms of probability

- What is the probability of a particular event occurring?
- What is the probability of a particular observation being made?

NB – the sum of probabilities = 1.0

A coin has two sides: 1 heads and 1 tails: 1 + 1 = 2

The probability of throwing heads = ½ = 0.5

P(heads) = 0.5

P(tails) = 1 – P(heads) = 1 – 0.5 = 0.5

16 observation number 2.25 (=9 x 0.25)

14

12

10

8

Frequency

6

4

2

0

2

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2.1

2.2

Height (m)

What is probability the picking a student of 1.65 m high from the class?

Depends on how the data are distributed

If the total number of students in the class is 106, and 10 of them are 1.65 m high, then the chance of picking (at random) a student measuring 1.65 m is 10 in 106: P(1.65) = 0.094

If the total number of students in the class is 106, and 96 (106-10) of them are NOT 1.65 m high, then the chance of picking (at random) a student NOT measuring 1.65 m is 96 in 106: P(NOT 1.65) = 0.906

P(NOT 1.65) = 1 – P(1.65) = 1 – 0.094 = 0.906

When data are displayed as a frequency distribution, the area under any part of the curve reflects the number of observations involved.

In this case, 10 observations are of 1.65 m (in red) 96 (in blue) are not

16

14

12

10

16

8

Frequency (%)

14

6

12

4

10

2

8

Frequency

0

6

2

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2.1

2.2

4

Frequency distributions do not only have to be displayed in terms of numbers, they can also be displayed as proportions or percentages.

Height (m)

2

0

2

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2.1

2.2

Height (m)

Same rules – the area under any part of the curve reflects the proportion of observations involved...or PROBABILITIES

In this case, 0.094 (9.4%) are of 1.65 m (in red) and 0.906 (90.6%) are not (in blue)

The total area under the curve = the total number of observations

The total area under the curve = 1.0

16 area under any part of the curve reflects the number of observations involved.

14

12

10

8

Frequency (%)

6

4

2

0

2

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2.1

2.2

Height (m)

Most of the data are clustered around the mean, which means that there is a fairly good chance (high probability) of your picking at random from the class a student with a height close to the mean

On the other hand, there is a relatively small chance that you will pick a student (by random) that is either very tall or very short: i.e. those whose measures are located in the tails of the distribution

500 area under any part of the curve reflects the number of observations involved.

12

N = 4064

10

Σ = 100%

400

8

300

6

Frequency (%)

Frequency

200

4

2

100

0

0

0

2

4

6

8

10

12

14

16

18

20

22

24

0

2

4

6

8

10

12

14

16

18

20

22

24

No Worms per quadrat

No Worms per quadrat

The shape of the curve depends on the variance or standard deviation: the spread of values about the mean

Mean = 10

Most data that scientists collect is what we call normally distributed – but NOT all.

s2 = 4

s2 = 8

s2 = 12

s2 = 16

12 area under any part of the curve reflects the number of observations involved.

10

Σ = 100%

8

6

Frequency (%)

4

2

0

0

2

4

6

8

10

12

14

16

18

20

22

24

No Worms per quadrat

- For data that are normally distributed:
- The mean, median and mode are the same
- The frequency distribution is completely symmetrical either side of the mean
- The area under the curve is proportional to number of observations
- The normal curve has fixed mathematical properties, irrespective of
- The scale on which it is drawn
- The magnitude or units of its mean
- The magnitude or units of its Standard Deviation
- …….and these render it susceptible to statistical analysis

Z = (x – area under any part of the curve reflects the number of observations involved.μ)

σ

Calculating proportions of a Normal Distribution

To calculate the probability of a particular value x being drawn from a normally distributed population of data, you need to know the mean AND the standard deviation of the data

Equation 1

μ = population mean, σ = population standard deviation

What Z describes is the difference between the mean and any value x, expressed as a proportion of the standard deviation, i.e. how many standard deviations away from the mean is the value x

Obviously, the smaller the value of Z, the closer the value of x is to the mean

Because Z is based on data that are normally distributed, it too is normally distributed (the Z distribution).

With a knowledge of Z, we can go to statistical tables drawn up based on the normal distribution and calculate the associated probability

Frequency area under any part of the curve reflects the number of observations involved.

Z = 1.33

Z = (x – μ)

0.45

0.55

0.65

0.75

0.85

0.95

1.05

1.15

1.25

1.35

1.45

1.55

1.65

1.75

1.85

1.95

2.05

2.15

2.25

2.35

2.45

2.55

Height (m)

σ

Z = (0.4)

Z = (1.95 – 1.55)

0.3

0.3

Frequency

0.00

0.33

0.67

1.00

1.33

1.67

2.00

2.33

2.67

3.00

3.33

-3.67

-3.33

-3.00

-2.67

-2.33

-2.00

-1.67

-1.33

-1.00

-0.67

-0.33

Z

e.g. if μ = 1.55 m, σ = 0.3 m, what is the probability of a student measuring more than 1.95 m being drawn at random from the population?

?

A student measuring 1.95 m is 1.33 times the standard deviation away from the mean, and this corresponds to a value of 0.0918 from the Z Tables

0.0918

n area under any part of the curve reflects the number of observations involved.

N

16

15

14

13

12

11

10

9

8

7

6

5

4

σ2

σ2

σ2

=

= population variance of the mean

x

x

n

The Distribution of Means

If random samples of size n are drawn from a normal population, the means of those samples will form a normal distribution

The variance of the distribution of means will decrease as n increases

Equation 2

Just as is a normal deviate referring to the normal distribution of Xi values

σ

=

√

Z = (x – μ)

n

σ

So is a normal deviate referring to the normal distribution of means

σ2

√

=

Z = (x – μ)

n

4

4

N = 9, X = 50.0 mm, μ = 47.0 mm, σ = 12.0 mm

Z = (50.0 – 47.0) = 3 = 0.75

= 4

√

= 12.0

= 12

9

σ

σ

σ

σ

3

= standard error of the mean

x

x

x

x

So…

What is the probability of obtaining a random sample of nine measurements with a mean greater than 50.0 mm, from a population having a mean of 47 mm and a standard deviation of 12.0 mm?

What is the probability of obtaining a random sample of nine measurements with a mean greater than 50.0 mm, from a population having a mean of 47 mm and a standard deviation of 12.0 mm?

4

4

N = 9, X = 50.0 mm, μ = 47.0 mm, σ = 12.0 mm

Z = (50.0 – 47.0) = 3 = 0.75

= 4

√

= 12.0

= 12

9

σ

3

x

Looking up 0.75 on the Z Tables gives – 0.2266

The observant amongst you will have noted that in the last couple of equations for Z we have used the population parameters: μ, σ and

Trouble is we don’t usually have access to population data and must make do with sample estimators x, s and

IF n is large: we use Z distribution to calculate normal deviates

=

Z = (x – μ)

t = (x – μ)

V = 100

V = 10

V = 5

V = 1

s

s

σ

s

σ

σ

-9

-8

-7

-6

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

7

8

9

x

x

x

x

x

x

t

Equation 3

IF n is small, then must use t distribution:

Shape of the t distribution varies with v (Degrees of Freedom: n-1): the bigger the n, the less spread the distribution

t distribution given rise to many statistical tests!

α couple of equations for Z we have used the population parameters: (1)

α (1)

0.1

0.1

-4

-3

- 2

-1

0

1

2

3

4

-4

-3

- 2

-1

0

1

2

3

4

t

1.372

-1.372

t

α (2)

0.05

0.05

- 2

-4

-3

-1

0

1

2

3

4

t

-1.812

1.812

One-Tailed

Two-Tailed

- Because it is based on the normal distribution, the t distribution has all the attributes of the normal distribution:
- Completely symmetrical
- Area under any part of the curve reflects proportion of t values involved
- etc….

For a particular area of the curve we can calculate the associated t values, using t-tables at the end of most text books on statistics

For example: if our sample size is 11 (v = 10), what is the value of t beyond which 10% (0.1) of the curve is enclosed? – Two possible answers

α couple of equations for Z we have used the population parameters: (1)

α (2)

0.1

0.05

0.05

- 2

-4

-3

-1

0

1

2

3

4

-4

-3

- 2

-1

0

1

2

3

4

t

-1.372

t

-1.812

1.812

One-Tailed

Two-Tailed

How do you get the t values from the t-tables?

s couple of equations for Z we have used the population parameters:

=

√

n

H0: μ = 22

H1: μ ≠ 22

t = (x – μ)

How? – use t-test

=

=

= 2.629

2.23

(24.23 – 22)

0.848

0.848

x = 24.23 s = 4.24 μ = 22.00 n = 25

4.24

4.24

=

=

√

5

25

= 0.848

s

s

x

x

The mean nitrate concentration of water in all the upstream tributaries of a large river prior to intensive agriculture is 22 mg.l-1.

Afterwards the mean nitrate concentration in 25 of these tributaries is 24.23 mg.l-1 and s = 4.24 mg.l-1

We can now use the t distribution to demonstrate the term statistical significance – which is something that you will get confronted with regularly when reading EIA reports…

This is an observation, and we want to determine if the intensification of agricultural practices has resulted in any change to the nitrate concentration of the freshwater resources.

Step 1: establish the hypotheses

Step 2: Need to determine the probability that a random sample (size 25) will generate a mean of 24.23 mg.l-1 from a population with a mean of 22 mg.l-1?

α couple of equations for Z we have used the population parameters: (1)

α (2)

H0: μ = 22

H1: μ ≠ 22

0.05

0.025

0.025

t

t

One-Tailed

Two-Tailed

Go to the hypotheses

Step 3: Determine, from the t-tables, the (critical) value of t, beyond which we consider such a random sample mean as being unlikely

Generally we consider an event as being unlikely if it occurs in the extreme 5% of the normal distribution

So we need to determine the (critical) value of t, beyond which 5% of the curve is enclosed – for v = 24

But do we use α (1) or α (2)?

The critical value of t, couple of equations for Z we have used the population parameters: α (2) 0.05, v = 24, is 2.064

0.025

0.025

-4

-3

- 2

-1

0

1

2

3

4

t

-2.064

2.064

2.629

Our value of t is 2.629, which lies beyond the critical value of t

That means it is very unlikely that a random sample (size 25) would generate a mean of 24.23 mg.l-1 from a population with a mean of 22 mg.l-1

So unlikely, in fact, that we don’t believe it can happen by chance

Reject H0 and accept H1

What we can then say, is that the before and after nitrate levels in the water are (statistically) significantly different from each other (p < 0.05)

We are not making any judgment about whether there is more nitrate in the water after than before, only that the concentrations are different – though some things are self evident!

You will frequently come across the terms p<0.05, p<0.01: these mean that the probability of a particular event occurring by chance alone are less than 5% and 1% respectively, which is unlikely

On the other hand if results are reported as p>0.05, it means that the probability of a particular event occurring by chance alone is greater than 5%, which is possible.

The t-Distribution allows us to calculate the 95% (or 99%) confidence intervals around an estimate of the population mean

In other words, what are limits around our estimate of the population mean, WITHIN which we 95% (or 99%) confident that the REAL value of the population mean lies

α (2)

0.025

0.025

t

Two-Tailed

To do this, we need a set of t-tables, and V (N-1)

s

s

x

x

t = (x – μ)

*

Difference between population and sample mean

IF confidence intervals around an estimate of the population mean

= 42.3 mm

N

= 26 (V = 25)

= 2.15

* t

= 2.15 *2.06 = 4.429

ά

2

The expression is then written as:

42.3 mm 4.43 mm

±

To do this, we need a set of t-tables, and V (N-1)

s

s

s

x

x

x

x

Then the 95% CI around the mean will be

16 confidence intervals around an estimate of the population mean

14

12

You can covert continuous data to discrete data, by assigning data to data classes

10

8

Frequency

6

4

2

0

2

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2.1

2.2

Height (m)

Testing Patterns in Discrete (count) Data: the Chi-Square Test

Examples of count data:

Number of petals per flower

Number of segments per insect leg

Number of worms per quadrat

Number of white cars on campus

etc

Often want to determine if the population from which you have obtained count data conform to a certain prediction

DATA DO NOT HAVE TO BE NORMALLY DISTRIBUTED

For Example: A geneticist has raised 134 progeny from a cross that is hypothesized to result in a 3:1 ratio of yellow-flowered to green-flowered plants.

She counts 113 yellow-flowering and 21 green-flowering plants amongst the progeny

Theoretically she should have obtained 100.5 yellow-flowering plants and 33.5 green-flowering plants (100.5 = 134 x 0.75, 33.5 = 134 x 0.25: 0.75 =3/ (3+1), 0.25 =1/ (3+1))

Does the OBSERVED ratio (113:21) differ (SIGNIFICANTLY) from the Expected (100.5:33.5) ratio?

Establish Hypotheses

H0: Population sampled has yellow:green flowering plants in ratio 3:1

H1: Population sampled does not have yellow:green flowering plants in ratio 3:1

χ have obtained count data conform to a certain prediction

χ

χ

2

2

2

Does the OBSERVED ratio (113:21) differ (SIGNIFICANTLY) from the Expected (100.5:33.5) ratio?

[ ]

[ ]

(113 – 100.5)2

(21 – 33.5)2

=

1.55 + 4.66 = 6.22

+

=

100.5

33.5

[ ]

2

Σ

(O – E)

=

Equation 4

E

Where O = Observed, E = Expected

Obviously, the bigger the difference between O and E, the greater the

When there is no difference, the value will be ZERO: hence Goodness of Fit

NB: MUST ALWAYS USE FREQUENCIES: not PERCENTAGES OR PROPORTIONS

χ have obtained count data conform to a certain prediction

χ

χ

2

2

2

Our value of greater than that corresponding to 0.025 (2.5%) but less

than that corresponding to 0.01 (1%) – from the Tables.

Do we accept or reject the Null Hypothesis?

[ ]

[ ]

(113 – 100.5)2

(21 – 33.5)2

=

1.55 + 4.66 = 6.22

+

=

100.5

33.5

Degrees of Freedom (v) = K – 1, where K = Number of categories (in this case two: yellow-flowering or green-flowering) = 2 – 1 = 1

A plant geneticist has done some crossing between plants and come up with the following numbers of different seeds

Has the geneticist sampled from a population having a ratio of 9:3:3:1 ?

What are the hypotheses being tested?

How many degrees of freedom are there?

H0: Population sampled has YS:YW:GS:GW seeds in the ratio 9:3:3:1

H1: Population sampled does not have YS:YW:GS:GW seeds in the ratio 9:3:3:1

K – 1 = 4 – 1 = 3

χ come up with the following numbers of different seeds

χ

χ

2

2

2

= 8.97

What is the critical value of

Our value of is greater than the critical value: Reject the Null Hypothesis that sample drawn from a population showing 9:3:3:1 ratio of YS:YW:GS:GW

χ come up with the following numbers of different seeds

2

The greatest contributor to the value is GW

v = K – 1 = 3 – 1 = 2

You can go further – and look to find whereabouts the pattern fails to conform to predictions

Do the other observations conform to the ratio 9:3:3 (YS:YW:GS)?

Establish Hypotheses

YES

v come up with the following numbers of different seeds = K - 1 = 2 – 1 = 1

Having confirmed that the observations fit the model (9:3:3), we can now combine them and test if the ratio of GW to the others is 1:15

Establish Hypotheses

Reject Null Hypothesis and draw the conclusion that there is a problem with GW

IF K = 2 (i.e. v = 1), then you must correct for continuity by subtracting 0.5 from each (O-E) before squaring and dividing by E. When you do this, ALL the (O-E) values must be positive – so convert them

Other Issues

IF Expected Counts are LESS than ONE, then you must combine the categories

NB: By combining data you reduce value of K and also v

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