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

Model Explanation or theory (maybe >1)

Hypothesis Prediction deduced from modelGenerate null hypothesis – H0: Falsification test

Test Experiment•IF H0 rejected – model supported•IF H0 accepted – model wrong

Pattern Observation Rigorously Describe

**

*

StatisticsCan only really test hypotheses by experimentation

Notiluca give off light when disturbed

Pattern Observation

Rigorously Describe

Model Explanation or theory (maybe >1)

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

Hypothesis Prediction deduced from modelGenerate 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

TestExperiment•IF H0 rejected – model supported•IF H0 accepted – model wrong

Statistics – summary, analysis and interpretation of data

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

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

RAW MATERIAL OF SCIENCE

Often referred to as VARIABLES because they vary

Types of Data

The type of data collected influences their analysis

Variability – key feature of the natural world

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

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

Accuracy – how close a measure is to the real value

20 cm +

20.63 cm

6 mm +

300 μm +

20.631506542 cm

Accept a level of measurement error: be upfront

Precision – how close repeat measures are to each other

20.632

19.986

21.102

20.493 20.578

20.710

22.356

20.623

20.755

Describing data and variability

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

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

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

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

What is the weight of a UWC, Second year student?

Number Mass 1 (kg) Mass 2 (kg) Height 1 (m) Height 2 (m) GenderYear of study

Age

Student No

Estimate Scale 1 Estimate Measure 1 M or F 2, 3 etc Months

What do you notice from the data?

Differences – variability! – natural and machine

Population – the entire collection of measurements

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

When might your sample be the population?

IF: what is the weight of a BDC222 student in 2012?

Describing data and variability

How high is a UWC BSc II student?What is the NO3 concentration in the Black River?

N = 20Σ = 69

Mean = 3.45Mean = Σx

N

Sample measure - Central Tendency

Arithmetic mean or Average

Population mean = μ; sample mean = x

We use x as a proxy for μ

Units?

Units?

25424175214321546326

Enter data (x) into MSExcel spreadsheet

Calculate N

Calculate Total

Calculate Mean

=COUNT(DATA:RANGE)

=SUM(DATA:RANGE)

=TOTAL / N

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

=AVERAGE(DATA:RANGE)

N = 20Σ = 69

Mean = 3.45Mean = Σx

N

Units?

What we actually doing here?

In the numerator we are probably doing this…..

2 + 5 + 4 + 2 + 4 + 1 + 7 + 5…etc etc

But there is another of looking at it…..

(3*1)+(5*2)+(2*3)+(4*4)+(3*5)+(2*6)+(1*7)

In other words, we are summing the products of x and F(x) in our data set

We have now generated a Frequency Table

25424175214321546326

11122222334444555667

Minimum

Maximum

Sum x.F(x) = 69

Sum F(x) = 20 – what is this?

N = 20Σ = 69

Mean = 3.45

Mean = Σx

N

Units?}Even and

equal CLASS intervals

X F X.F1 3 32 5 103 2 64 4 165 3 156 2 127 1 7

SUMS 20 69

2 83 92 14 45 53 76 34 48 56 73 32 5

1) Calculate the mean femur length

2) Construct a frequency table of femur length

Length (cm) of femur from domestic rabbits shot on Robben Island during March 2010

Femur Length (cm) Frequency

1

2

3

3) Recalculate the mean femur length

MSExcel allows you to calculate a frequency table

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1.2

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2.2

Height (m)

Fre

qu

en

cy

Height (m) Frequency1.2 1

1.25 11.3 1

1.35 21.4 2

1.45 31.5 5

1.55 61.6 8

1.65 101.7 12

1.75 151.8 11

1.85 91.9 7

1.95 52 3

2.05 22.1 1

2.15 12.2 1

Mode – 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

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

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1 2 3 4 5 6 7 8 9 10

Data Class

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0123456789

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1 2 3 4 5 6 7 8 9 10

Data Class

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Data Class

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UNIMODAL BIMODAL TRIMODAL

Datum No 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27Datum 1.2 1.25 1.3 1.35 1.35 1.4 1.4 1.45 1.45 1.45 1.5 1.5 1.5 1.5 1.5 1.55 1.55 1.55 1.55 1.55 1.55 1.6 1.6 1.6 1.6 1.6 1.6

Datum No 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54Datum 1.6 1.6 1.65 1.65 1.65 1.7 1.7 1.65 1.65 1.65 1.65 1.65 1.7 1.7 1.7 1.7 1.7 1.7 1.7 1.7 1.7 1.7 1.7 1.7 1.75 1.75 1.75

Datum No 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81Datum 1.75 1.75 1.75 1.75 1.75 1.8 1.8 1.75 1.75 1.75 1.75 1.75 1.8 1.8 1.8 1.8 1.8 1.8 1.8 1.8 1.8 1.8 1.8 1.85 1.85 1.85 1.85

Datum No 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106Datum 1.85 1.85 1.85 1.85 1.85 1.9 1.9 1.9 1.9 1.9 1.9 1.9 1.95 1.95 1.95 1.95 1.95 2 2 2 2.05 2.05 2.1 2.15 2.2

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

If there are an odd number of data points this is easy

Data (x) Ordered Data Order No1.85 1.4 11.6 1.45 2

1.95 1.6 31.65 1.65 41.45 1.8 51.9 1.85 61.4 1.9 71.8 1.9 81.9 1.95 9

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

interpolateData (x) Ordered Data Order No

1.85 1.4 11.6 1.45 2

1.95 1.6 31.65 1.65 41.45 1.75 51.9 1.8 61.4 1.85 71.8 1.9 81.9 1.9 9

1.75 1.95 10

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)

24.625.925.926.625.825.425.224.724.125

24.824.526.425.325

25.425.825.325.923.6

Unordered data Ordered data

In MSExcel: Highlight data range to be sorted (including variable header row), select data, sort….

1.Tick “data has headers” box…2.Select variable to sort by…3.Sort from smallest to largest…

THINGS TO REMEMBER…

….COUNTER…..

Rank Order X1 23.62 24.13 24.54 24.65 24.76 24.87 258 259 25.210 25.311 25.312 25.413 25.414 25.815 25.816 25.917 25.918 25.919 26.420 26.6

Measures of Dispersion – how data are distributed around the mean

1.85 1.65 1.55 1.91.6 1.95 1.7 1.7

1.95 1.75 1.8 1.71.65 1.55 1.65 1.751.45 1.85 1.85 1.81.9 1.75 1.7 2.051.4 2 1.35 21.8 1.65 1.5 1.81.9 2.1 1.8 1.5

1.75 1.2 1.5 2.151.3 1.7 1.6 1.55

1.85 1.45 1.8 1.851.5 1.75 1.75 1.251.8 1.95 1.75 21.9 1.7 1.8 1.9

1.75 1.85 1.8 1.751.7 1.9 1.45 1.65

1.35 1.65 1.7 1.61.75 1.5 1.55 1.551.6 1.8 1.75 1.85

2.05 1.6 1.85 1.71.65 1.7 1.4 1.751.95 1.9 1.65 1.61.75 1.65 1.7 1.851.8 1.75 1.95 1.65

1.55 2.2 1.751.7 1.6 1.6

Range:

Essentially the lowest and highest value in the data set

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

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

Mean DeviationData (x) x - mean !(x - mean)!

3 -1 14 0 05 1 16 2 27 3 32 -2 23 -1 14 0 05 1 16 2 23 -1 12 -2 23 -1 14 0 05 1 12 -2 264 0 2016 154 1.33

ΣN

mean

Always = Zero

Data (x) x - mean !(x - mean)!3 -1 14 0 05 1 16 2 27 3 32 -2 23 -1 14 0 05 1 16 2 23 -1 12 -2 23 -1 14 0 05 1 12 -2 264 0 2016 154 1.33

Data (x) x - mean !(x - mean)!3 -1 14 0 05 1 16 2 27 3 32 -2 23 -1 14 0 05 1 16 2 23 -1 12 -2 23 -1 14 0 05 1 12 -2 264 0 2016 154 1.33

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

2 83 92 14 45 53 76 34 48 56 73 32 5

Calculate the Mean Deviation

Length (cm) of femur from domestic rabbits shot on Robben Island during March 2010

Variance and Standard Deviation

Data (x) x - mean (x - mean)2

3 -1 14 0 05 1 16 2 47 3 92 -2 43 -1 14 0 05 1 16 2 43 -1 12 -2 43 -1 14 0 05 1 12 -2 464 0 3616 154 2.4

ΣN

mean

Length (mm) of Drosophila melanogaster Instar III larvae

Data (x) x - mean (x - mean)2

3 -1 14 0 05 1 16 2 47 3 92 -2 43 -1 14 0 05 1 16 2 43 -1 12 -2 43 -1 14 0 05 1 12 -2 464 0 3616 154 2.4

ΣN

mean

Always = Zero

(Variance) = √ Standard Deviation (sample)

s = 1.5

Square units?

Sum of Squares (Sample)

Mean Sum of Squares (sample)(Variance)

Data (x) x - mean (x - mean)2

3 -1 14 0 05 1 16 2 47 3 92 -2 43 -1 14 0 05 1 16 2 43 -1 12 -2 43 -1 14 0 05 1 12 -2 464 0 3616 154 2.4

ΣN

mean16

2.25

The values of x = 4, sample variance (2.25) and sample standard deviation

(1.5) ALL refer to the sample of 16 measures

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

– mean) values

(n-1) = Degrees of Freedom = v

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.

X

5Mean

5N

25Total

7

6

5

4

3

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

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

X

5Mean

5N

25Total

7

6

5

4

3

√σ2

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

IF the sample mean is 13.5, N = 8 AND

X1 = 5X2 = 18X3 = 22X4 = 38X5 = 2X6 = 10X7 = 19

WHAT IS THE VALUE OF X8?

2 83 92 14 45 53 76 34 48 56 73 32 5

Calculate the Variance and Standard Deviation….

Length (cm) of femur from domestic rabbits shot on Robben Island during March 2010

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

A B C5 2 06 4 127 6 115 8 16 10 07 2 125 4 116 6 17 8 126 10 06 6 610 10 10

0.67 8.89 35.110.82 2.98 5.93

Mean

s2

N

s

Calculating Variance and Standard Deviation using MSExcel

=VAR(DATA:RANGE)

=STDEV(DATA:RANGE)

2 83 92 14 45 53 76 34 48 56 73 32 5

Length (cm) of femur from domestic rabbits shot on Robben Island during March 2010

Sum x.F(x) = 109

Sum F(x) = 24 – what is this?

Calculating Variance and Standard Deviation from a frequency table

X F XF X-Mean (X-Mean)2 F.(X-Mean)2

1 1 1 -3.54 12.54 12.542 3 6 -2.54 6.46 19.383 5 15 -1.54 2.38 11.884 4 16 -0.54 0.29 1.175 4 20 0.46 0.21 0.846 2 12 1.46 2.13 4.257 2 14 2.46 6.04 12.098 2 16 3.46 11.96 23.929 1 9 4.46 19.88 19.88

N 24 109 105.96Mean 4.54

Variance 4.61STDEV 2.15

REMEMBER……

N = 24Σ = 109

Mean = 4.54

Mean = Σx

N

Units?

Sum F.(X-mean)2 = 105.96

Sum F = 24

Var = Sum F.(X-mean)2 / (Sum F -1) = 4.61

X F XF X-Mean (X-Mean)2 F.(X-Mean)2

1 1 1 -3.54 12.54 12.542 3 6 -2.54 6.46 19.383 5 15 -1.54 2.38 11.884 4 16 -0.54 0.29 1.175 4 20 0.46 0.21 0.846 2 12 1.46 2.13 4.257 2 14 2.46 6.04 12.098 2 16 3.46 11.96 23.929 1 9 4.46 19.88 19.88

N 24 109 105.96Mean 4.54

Variance 4.61STDEV 2.15

So……

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……

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?

Variability = Uncertainty

A coin has two sides: 1 heads and 1 tails: 1 + 1 = 2The probability of throwing heads = ½ = 0.5P(heads) = 0.5P(tails) = 1 – P(heads) = 1 – 0.5 = 0.5

A die has six sides:

The probability of throwing = 1/6 = 0.167

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

NB – the sum of probabilities = 1.0

What is the probability of picking a student of 1.65 m high from the class?Depends on how the data are distributed

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Height (m)

Fre

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Height (m) Frequency1.2 1

1.25 11.3 1

1.35 21.4 2

1.45 31.5 5

1.55 61.6 8

1.65 101.7 12

1.75 151.8 11

1.85 91.9 7

1.95 52 3

2.05 22.1 1

2.15 12.2 1

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

If you know the probability of picking a student of 1.65 m high from the class (0.094), and you know how many students there are in TOTAL (106) in the class, you can calculate the number of students that are 1.65 m high.

106 * 0.094 = 10

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

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Height (m)

Fre

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of 1.65 m (in red) 96 (in blue) are not

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

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(%

)

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

0

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Height (m)

Fre

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(%

)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

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

No Worms per quadrat

Fre

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100

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300

400

500

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24

N = 4064

No Worms per quadrat

Fre

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(%

)

02468

1012

0 2 4 6 8 10 12 14 16 18 20 22 24

Σ = 100%

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

0

0.05

0.1

0.15

0.2

0.25

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

X

Fre

qu

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Mean = 10

s2 = 4s2 = 8s2 = 12s2 = 16

For data that are normally distributed:The mean, median and mode are the sameThe frequency distribution is completely symmetrical either side of the meanThe 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

No Worms per quadrat

Fre

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(%

)

02468

1012

0 2 4 6 8 10 12 14 16 18 20 22 24

Σ = 100%

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

Z = (x – μ)

σ

μ = population mean, σ = population standard deviation

Equation 1

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

Calculating proportions of a Normal Distribution

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?

Z = (x – μ)

σ

Z = (1.95 – 1.55)

0.3Z = (0.4)

0.3

Z = 1.33

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

0.450.550.650.750.850.951.051.151.251.351.451.551.651.751.851.952.052.152.252.352.452.55

Height (m)

Fre

qu

ency

-3.67-3.33-3.00-2.67-2.33-2.00-1.67-1.33-1.00-0.67-0.330.000.330.671.001.331.672.002.332.673.003.33

Z

Fre

qu

ency

?

0.0918

Z 0 1 2 3 4 5

0.0 0.5000 0.4960 0.4920 0.4880 0.4840 0.48010.1 0.4602 0.4562 0.4522 0.4483 0.4443 0.44040.2 0.4207 0.4168 0.4129 0.4090 0.4052 0.40130.3 0.3821 0.3783 0.3745 0.3707 0.3669 0.36320.4 0.3466 0.3409 0.3372 0.3336 0.3300 0.3264

0.5 0.3085 0.3050 0.3015 0.2981 0.2946 0.29120.6 0.2743 0.2709 0.2676 0.2643 0.2611 0.25780.7 0.2420 0.2389 0.2358 0.2327 0.2297 0.22660.8 0.2119 0.2090 0.2061 0.2033 0.2005 0.19770.9 0.1841 0.1814 0.1788 0.1762 0.1736 0.1711

1.0 0.1587 0.1562 0.1539 0.1515 0.1492 0.14691.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.12511.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.10561.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.08851.4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735

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

σ2

nσ2

x = Equation 2

σ2

x = population variance of the mean

1 2 3 4 5 6 7 8 9 10 11 12 133 3.5 4.0 4.5 5.0 4.5 4.3 4.3 4.3 4.5 4.4 4.2 4.14 4.5 5.0 5.5 4.8 4.5 4.4 4.5 4.7 4.5 4.3 4.2 4.25 5.5 6.0 5.0 4.6 4.5 4.6 4.8 4.6 4.3 4.2 4.2 4.26 6.5 5.0 4.5 4.4 4.5 4.7 4.5 4.2 4.1 4.1 4.2 4.07 4.5 4.0 4.0 4.2 4.5 4.3 4.0 3.9 3.9 4.0 3.82 2.5 3.0 3.5 4.0 3.8 3.6 3.5 3.6 3.7 3.53 3.5 4.0 4.5 4.2 3.8 3.7 3.8 3.9 3.74 4.5 5.0 4.5 4.0 3.8 3.9 4.0 3.85 5.5 4.7 4.0 3.8 3.8 4.0 3.86 4.5 3.7 3.5 3.6 3.8 3.63 2.5 2.7 3.0 3.4 3.22 2.5 3.0 3.5 3.23 3.5 4.0 3.54 4.5 3.75 3.52

Mean 4.0 4.1 4.1 4.1 4.1 4.1 4.1 4.1 4.1 4.1 4.1 4.1 4.1Variance 2.40 1.40 0.87 0.51 0.30 0.20 0.17 0.17 0.16 0.12 0.08 0.02 0.01

n

N 16 15 14 13 12 11 10 9 8 7 6 5 4

Get students to do this for themselves

σx = standard error of the meanσ

xσ

n

=

√ So… = √

σ2

n

Z = (x – μ)

σ

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

Z = (x – μ)

σx

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

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?

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

σx

= 12.0

√ 9= 4= 12

3

Z = (50.0 – 47.0) = 3 = 0.75

4 4

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?

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

σx

= 12.0

√ 9= 4= 12

3

Z = (50.0 – 47.0) = 3 = 0.75

4 4

Looking up 0.75 on the Z Tables gives – 0.2266

Z 0 1 2 3 4 5

0.0 0.5000 0.4960 0.4920 0.4880 0.4840 0.48010.1 0.4602 0.4562 0.4522 0.4483 0.4443 0.44040.2 0.4207 0.4168 0.4129 0.4090 0.4052 0.40130.3 0.3821 0.3783 0.3745 0.3707 0.3669 0.36320.4 0.3466 0.3409 0.3372 0.3336 0.3300 0.3264

0.5 0.3085 0.3050 0.3015 0.2981 0.2946 0.29120.6 0.2743 0.2709 0.2676 0.2643 0.2611 0.25780.7 0.2420 0.2389 0.2358 0.2327 0.2297 0.22660.8 0.2119 0.2090 0.2061 0.2033 0.2005 0.19770.9 0.1841 0.1814 0.1788 0.1762 0.1736 0.1711

1.0 0.1587 0.1562 0.1539 0.1515 0.1492 0.14691.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.12511.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.10561.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.08851.4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735

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

σx

sx

σx

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

IF n is small, then must use t distribution: t = (x – μ)

sx

Equation 3

Z = (x – μ)

σx

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

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

t

V = 100V = 10V = 5V = 1

t distribution given rise to many statistical tests!

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

-4 -3 - 2 -1 0 1 2 3 4

t

α (1)

0.1

1.372

-4 -3 - 2 -1 0 1 2 3 4

t

α (1)

0.1

-1.372

One-Tailed

0.05 0.05

-4 -3 - 2 -1 0 1 2 3 4

t1.812-1.812

α (2)

Two-Tailed

-4 -3 - 2 -1 0 1 2 3 4

t

α (1)

0.1

-1.372

One-Tailed

0.05 0.05

-4 -3 - 2 -1 0 1 2 3 4

t1.812-1.812

α (2)

Two-Tailed

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

α (2) 0.5 0.2 0.1 0.05 0.02

α (1) 0.25 0.1 0.05 0.025 0.011 1.000 3.078 6.314 12.7062 0.816 1.886 2.920 4.3033 0.765 1.638 2.353 3.1824 0.741 1.533 2.132 2.7765 0.727 1.476 2.015 2.571

6 0.718 1.440 1.943 2.4477 0.711 1.415 1.895 2.3658 0.706 1.397 1.860 2.3069 0.703 1.383 1.833 1.26210 0.700 1.372 1.812 2.228

11 0.697 1.363 1.796 2.20112 0.695 1.356 1.782 2.17913 0.694 1.350 1.771 2.16014 0.692 1.345 1.761 2.14515 0.691 1.341 1.753 2.131

16 0.690 1.337 1.746 2.12017 0.689 1.333 1.740 2.11018 0.688 1.330 1.734 2.10119 0.688 1.328 1.729 2.09320 0.687 1.325 1.725 2.086

21 0.686 1.323 1.721 2.08022 0.686 1.321 1.717 2.07423 0.685 1.319 1.714 2.06924 0.685 1.318 1.711 2.06425 0.684 1.316 1.708 2.060

v

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…

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

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 H0: μ = 22 H1: μ ≠ 22

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?

How? – use t-test t = (x – μ)

sxx = 24.23

s = 4.24 μ = 22.00 n = 25

sx

s

n

=

√ 4.24

25

=

√ 4.24

5= = 0.848

(24.23 – 22)

0.848

= 2.23

0.848

= = 2.629

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

t

α (1)

0.05

One-Tailed

0.025 0.025

t

α (2)

Two-Tailed

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)?

Go to the hypotheses H0: μ = 22 H1: μ ≠ 22

The critical value of t, α (2) 0.05, v = 24, is 2.064

α (2) 0.5 0.2 0.1 0.05 0.02

α (1) 0.25 0.1 0.05 0.025 0.011 1.000 3.078 6.314 12.7062 0.816 1.886 2.920 4.3033 0.765 1.638 2.353 3.1824 0.741 1.533 2.132 2.7765 0.727 1.476 2.015 2.571

6 0.718 1.440 1.943 2.4477 0.711 1.415 1.895 2.3658 0.706 1.397 1.860 2.3069 0.703 1.383 1.833 1.26210 0.700 1.372 1.812 2.228

11 0.697 1.363 1.796 2.20112 0.695 1.356 1.782 2.17913 0.694 1.350 1.771 2.16014 0.692 1.345 1.761 2.14515 0.691 1.341 1.753 2.131

16 0.690 1.337 1.746 2.12017 0.689 1.333 1.740 2.11018 0.688 1.330 1.734 2.10119 0.688 1.328 1.729 2.09320 0.687 1.325 1.725 2.086

21 0.686 1.323 1.721 2.08022 0.686 1.321 1.717 2.07423 0.685 1.319 1.714 2.06924 0.685 1.318 1.711 2.06425 0.684 1.316 1.708 2.060

v

-4 -3 - 2 -1 0 1 2 3 4

t 2.064-2.064

0.025 0.025

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

2.629

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

0.025 0.025

t

α (2)

Two-Tailed

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

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

t = (x – μ)sx *

Difference between population and sample mean

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

IF

N

sx

x = 42.3 mm

= 26 (V = 25)

= 2.15

α (2) 0.5 0.2 0.1 0.05 0.02

α (1) 0.25 0.1 0.05 0.025 0.011 1.000 3.078 6.314 12.7062 0.816 1.886 2.920 4.3033 0.765 1.638 2.353 3.1824 0.741 1.533 2.132 2.7765 0.727 1.476 2.015 2.571

6 0.718 1.440 1.943 2.4477 0.711 1.415 1.895 2.3658 0.706 1.397 1.860 2.3069 0.703 1.383 1.833 1.26210 0.700 1.372 1.812 2.228

11 0.697 1.363 1.796 2.20112 0.695 1.356 1.782 2.17913 0.694 1.350 1.771 2.16014 0.692 1.345 1.761 2.14515 0.691 1.341 1.753 2.131

16 0.690 1.337 1.746 2.12017 0.689 1.333 1.740 2.11018 0.688 1.330 1.734 2.10119 0.688 1.328 1.729 2.09320 0.687 1.325 1.725 2.086

21 0.686 1.323 1.721 2.08022 0.686 1.321 1.717 2.07423 0.685 1.319 1.714 2.06924 0.685 1.318 1.711 2.06425 0.684 1.316 1.708 2.060

v

Then the 95% CI around the mean will be

sx

* tά 2

= 2.15 *2.06 = 4.429

The expression is then written as:

42.3 mm 4.43 mm±