maths for biology july 6 th 2015 christian bokhove carys hughes hilary otter rebecca d’silva nicky...

Maths for Biology

July 6th 2015

Christian BokhoveCarys HughesHilary Otter

Rebecca D’SilvaNicky Miller

Why this day

• Changes in the A level Biology curriculum• More maths• Maths used interdisciplinary• Principles

– Instruction but also hands-on tasks– Collaborative, doing it together, ask questions– Want to customise the course to where you

need support

Web page with resources

http://is.gd/maths4bio



Introductions

• Christian Bokhove– Lecturer in Mathematics Education

• Carys Hughes• Hilary Otter• Rebecca D’Silva• Nicky Miller

Objectives of the day

• Update and strengthen knowledge on some topics for the new Biology A level curriculum;– Exponential growth and decay, logarithms– Statistical tests

• Hear and discuss ideas for teaching them;• Want to hear your opinions for

improvements;• You leave with:

– Ideas, knowledge and some resources

Schedule for the day

When What9:00 – 9:30 Welcome, introductions9:30 – 10:15 Exponential growth and decay10:15 – 12:30 Logarithms and log paper12:30 – 13:30 Lunch break13:30 – 14:00 Statistical tests14:00 – Mini workshops statistics

MATHS CONFIDENCE SURVEYhttp://is.gd/m4bsurvey

EXPONENTIAL GROWTH AND DECAY, LOGARITHMS

Slide exam curriculum

From (i) A.0 - arithmetic and numerical computation

A.0.5 Use calculators to find and use power, exponential and logarithmic functions Candidates may be tested on their ability to: estimate the number of bacteria grown over a certain length of time

From (ii) A.2 – algebra A.2.5 Use logarithms in relation to quantities that range over several orders of magnitude Candidates may be tested on their ability to: use a logarithmic scale in the context of microbiology, e.g. growth rate of a microorganism such as yeast

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Standard form and powers

ACTIVITYLink to task

H

VERY LARGE AND VERY SMALL• Negative and positive standard

form

VERY LARGE AND VERY SMALL

EXPONENTIAL GROWTH AND DECAY

(Before we can look at logarithms we need to deal with exponential growth and decay)

19

ACTIVITY

Exponentials

Take an A4 sheet of paper.

How many times can you fold it in half?

MYTHBUSTERS

21

How many layers do you produce? HANDOUT

Number of folds

(x)

Number of layers

(y)

Mult Power Height

(cm)0 1 0.011 2 22 2*23 2*2*24 2*2*2*2 24

5

Table of results

H

22

Plot your values on graph paper:

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

- 100

100

200

300

x

y

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

- 100

100

200

300

x

y

23

Exponential growth

Imagine you contracted a virus (such as SARS) where you infected the first five people that you met, and they each infected the first five people that they met and so on…. There are 186,701 people living in Southampton. How many interactions would it take until everyone was infected?

24

How many infections?

Number of interactions (x)

Number of infected people (y)

0 11 52 25345

Table of results

25

General form:

y=bt

where b is the base and x is the power (or exponent)

26

The exponential graph

- 2 2 4 6 8

- 200

200

400

x

y

ACTIVITY: Use Desmos or Geogebra online to graph

27

How many layers do you produce?

Number of folds

(x)

Number of layers

(y)

Height

(cm)0 1 0.011 2 22 2*23 2*2*24 2*2*2*2 24

5

Table of results

y=2x

28

Common features of y=bx

all curves pass through (0,1)exponential growth (and decay) takes place very rapidlyb > 0b 0b 1b > 1 has a positive gradient (PLOT THIS!)0 < b < 1 has a negative gradient (PLOT THIS!)

https://www.desmos.com/calculator/auubsaje fh

https://www.desmos.com/calculator/auubsaje

https://www.desmos.com/calculator/auubsaje

29

HANDOUT

Exponential growth and decay worksheet

Exponential growth and decay practice sheet(and answers)

We are not doing all of these during the session.

30

Logarithms

Logarithm is another name for a power

532log

232

2

5

So let’s say you know there are 32 layers in the folding task. How many times has someone

made a fold?

You could say taking ‘logarithms’ is the opposite of exponential growth or decay.

Exponential form

Logarithmic form

Log examples: positive numbers

32

ACTIVITY

In pairs decide whether or not each statement (selection)

33

Further practice

There is further practice of conversion between logs and powers on:

www.bbc.co.uk/education/asguru

and theLogarithms practice sheet

http://www.bbc.co.uk/education/asguru

34

Why do we even need log paper?The exponential graph

- 2 2 4 6 8

- 200

200

400

x

y

ACTIVITY: Now with log paper. Demonstrate with Geogebra

Log paper

• Log paper and powers of 2 5 10• Step by step example• Use of calculator• Most tools are rather poor at log paper!

TASK LOG PAPER – SARS - PLOT

Number of interactions (x)

Number of people with the disease (y)

0 11 52 253 1254 6255 3,1256 15,6257 78,1258 390,625

PEDAGOGY

How would you teach these topic?

From sample exam question

Discuss exam question

Lunch break

Statistical tests

These slides partly rely on the excellent resources from SteveJ64 from the TES

website

Slide exam curriculumFrom (i) A.1 - handling data A.1.9 Select and use a statistical test Candidates may be tested on their ability to select and use: • the Chi squared test to test the significance of the

difference between observed and expected results • the Students t-test • the correlation coefficient A.1.11 Identify uncertainties in measurements and use simple techniques to determine uncertainty when data are combined Candidates may be tested on their ability to: • calculate percentage error where there are

uncertainties in measurement

Normal distribution

45

Suppose we have a crate of apples which are to be sorted by weight into small, medium and large. If we wanted 25% to be in the large category, we would need to know the lowest weight a “large” apple could be.

To solve a problem like this we can use a statistical model.

A model often used for continuous quantities such as weight, volume, length and time is the Normal Distribution.

The Normal Distribution is an example of a probability model.

46

The Normal Distribution curve

2To fit the curve we use the mean, m, and variance, , of the data. These are the parameters of the model.

),(~ 2NXIf X is the random variable “ the weight of apples”, we write

Characteristics of the Normal Distribution

The Normal distribution model is a symmetric bell-shaped curve. We fit it as closely as possible to the data.

Reminder: s is the standard deviation.

47

e.g.

)1,5(~ NX)1,4(~ NX

The axis of symmetry of the Normal distribution passes through the mean.

48

e.g.

)50,4(~ 2NX

A smaller variance “squashes” the distribution closer to the mean.

)1,4(~ NX

49

The percentages of the Normal Distribution lying within the given number of standard deviations either side of the mean are approximately:

SUMMARY

1 s.d. : 68%

2 s.d. : 95% 3 s.d. : 99·8%

68%

22

95%

33

99·8%

Statistical tests

• Type of data collected– Measurements– Frequencies

• What are you looking for?– Associations– Differences

MINI WORKSHOPS

• Break up in smaller groups to study a particular stats topic

• Slides and materials are available• Sample exam questions can also help

guide the work• At end we come back together and share

knowledge and experiences

Chi-squared (χ2) test

KARL PEARSON(1857-1936)

British mathematician, ‘father’ of modern statistics and a pioneer of eugenics!

(Pearson’s)

Chi-squared (χ2) test

• This test compares measurements relating to the frequency of individuals in defined categories e.g. the numbers of white and purple flowers in a population of pea plants.

• Chi-squared is used to test if the observed frequency fits the frequency you expected or predicted.

How do we calculate the expected frequency?

• You might expect the observed frequency of your data to match a specific ratio. e.g. a 3:1 ratio of phenotypes in a genetic cross.

• Or you may predict a homogenous distribution of individuals in an environment. e.g. numbers of daisies counted in quadrats on a field.

Note: In some cases you might expect the observed frequencies to match the expected, in others you might hope for a difference between them.

Example 1: GENETICS

Comparing the observed frequency of different types of maize grains with the expected ratio calculated using a Punnett square.

The photo shows four different phenotypes for maize grain, as follows:

Purple & Smooth (A), Purple & Shrunken (B), Yellow & Smooth (C) and Yellow & Shrunken (D)

Gametes PS Ps pS ps

PS PPSS PPSs PpSS PpSs

Ps PPSs PPss PpSs Ppss

pS PpSS PpSs ppSS ppSs

ps PpSs Ppss ppSs ppss

The Punnett square below shows the expected ratio of phenotypes from crosses of four genotypes of maize.

A : B : C : D = 9 : 3 : 3 : 1

H0 = there is no statistically significant difference between the observed frequency of maize grains and the expected frequency (the 9:3:3:1 ratio)

HA = there is a significant difference between the observed frequency of maize grains and the expected frequency

If the value for χ2 exceeds the critical value (P = 0.05), then you can reject the null hypothesis.

What is the null hypothesis (H0)?

• How critical value• P-value• For the tests do not do all step-by-step recipes• Use the datasets from the separate

presentations to ask ‘what test would be appropriate here’.

• Central: chi squared, mini workshops– T-test– Spearman rank– SE and confidence intervals

• Mini workshop interpretation and reporting.

Calculating χ2

χ2 = (O – E)2

E

O = the observed resultsE = the expected (or predicted) results

Phenotype O E(9:3:3:1)

O-E (O-E)2 (O-E)2

E

A 271 244 27 729 2.99

B 73 81 -8 64 0.88

C 63 81 -18 324 4.00

D 26 27 -1 1 0.04

433 433 χ2= 7.91

Compare your calculated value of χ2 with the critical value in your stats table

Our value of χ2 = 7.91Degrees of freedom = no. of categories - 1 = 3

D.F. Critical Value (P = 0.05)

1 3.842 5.993 7.824 9.495 11.07

Our value for χ2 exceeds the critical value, so we can reject the null hypothesis.

There is a significant difference between our expected and observed ratios. i.e. they are a poor fit.

Example 2: ECOLOGY

• One section of a river was trawled and four species of fish counted and frequencies recorded.

• The expected frequency is equal numbers of the four fish species to be present in the sample.

H0 = there is no statistically significant difference between the observed frequency of fish species and the expected frequency.

HA = there is a significant difference between the observed frequency of fish and the expected frequency



Calculating χ2

χ2 = (O – E)2

E

O = the observed resultsE = the expected (or predicted) results

Species O E O-E (O-E)2 (O-E)2

E

Rudd 15 10 5 25 2.5

Roach 15 10 5 25 2.5

Dace 4 10 -6 36 3.6

Bream 6 10 -4 16 1.6

40 40 χ2= 10.2

Compare your calculated value of χ2 with the critical value in your table of critical values.

Our value of χ2 = 10.2Degrees of freedom = no. of categories - 1 = 3


1 3.842 5.993 7.824 9.495 11.07


There is a significant difference between our expected and observed frequencies of fish species.

Example 3: CONTINGENCY TABLES

You can use contingency tables to calculate expected frequencies when the relationship between two quantities is being investigated.

In this example we will look at the incidence of colour blindness in both males and females.

H0 = there is no statistically significant difference between the observed frequency of colour blindness in males and females.

HA = there is a significant difference between the between the observed frequency of colour blindness in males and females



Observed frequencies Males Females

Colour blind 56 14

Not colour blind 754 536

e.g.The expected frequency for colour blind males =

(56 + 14) x (56 + 754)

1360= 42

Expected Cell Frequency = (Row Total x Column Total)

n

Observed: Males Females

•Colour blind 56 14•Not colour blind 754 536

Expected: Males Females

•Colour blind 42 28

•Not colour blind 768 522

Males Females

•Colour blind 4.7 14•Not colour blind 754 536

χ2 =… (O – E)2

E = 4.7 + 14 + 754 + 536 = 12.33

(O – E)2 / E

Compare your calculated value of χ2 with the critical value in your table of critical values

Our value of χ2 = 12.33Deg of Freedom = (2 rows - 1) x (2 cols – 1) = 1


1 3.842 5.993 7.824 9.495 11.07


There is a significant difference between our expected and observed frequencies.

The fraction of males with colour blindness is greater than that in females. The difference cannot be attributed to chance alone.

Spearman rank

Charles Spearman(1863-1945)British Psychologist and pioneer of IQ theory

What can this test tell you?

Whether there is a statisticallysignificant correlation between two measurements from the same sample when you have 5-30 pairs of data.

If the correlation is negative or positive

Spearman’s Rank Correlation Coefficient

What is the correlation coefficient r ?

Does being good at maths make you better at biology?

Student Maths exam score

Biology exam score

Anand 57 83

Bernard 45 37

Charlotte 72 41

Demi 78 85

Eustace 53 56

Ferdinand 63 85

Gemma 86 77

Hector 98 87

Ivor 59 70

Jasmine 71 59

Is there a statistically significant correlation between these two sets of results?

Spearman’s Rank Correlation Coefficient: rs

Where:

N = the number of individuals in the sample

D = difference in the rank of the two measurements made on an individual

rs will be a number between – 1 and +1

This number can be compared with those in a table of critical values

rs = 1 – [ 6 x∑D2

N3 – N) ]

H0 = there is no statistically significant correlation between Maths scores and Biology scores

HA = there is a statistically significant correlation between Maths scores and Biology scores

A negative value for rs implies a negative correlation

A positive value for rs implies a positive correlation

If the value for rs exceeds the critical value, then you can reject the null hypothesis

Step 1: Rank each set of data


Maths rank

Biology exam score

Biology rank

Alex 57 83

Bernard 45 37

Charlotte 72 41

Demi 78 85

Eustace 53 56

Ferdinand 63 85

Gemma 86 77

Hector 98 87

Ivor 59 70

Jasmine 71 59

3

1

7

8

2

5

9

10

4

6

(lowest to highest)

Where two or more scores are tied each is assigned an average rank

7

1

2

8

3

9

6

10

5

4

8.5

8.5


Maths rank

Biology exam score

Biology rank D D2

Alex 57 3 83 7

Bernard 45 1 37 1

Charlotte 72 7 41 2

Demi 78 8 85 8.5

Eustace 53 2 56 3

Ferdinand 63 5 85 8.5

Gemma 86 9 77 6

Hector 98 10 87 10

Ivor 59 4 70 5

Jasmine 71 6 59 4

∑D2 =

Step 2: Work out the differences in ranks (maths – biology)

- 4

0

5

- 0.5

- 1

- 3.5

3

0

- 1

2

Step 3: Work out the square of the differences

Step 4: Work out the sum of the square of the differences

16

0

25

0.25

1

12.25

9

0

1

4

68.5

Step 5: Work out the correlation coefficient, rs

N = 10

∑D2 = 68.5

rs = 1 - 6 x 68.5)103 – 10)

= 1 - 6 x 68.5)1000 - 10

= 1 - 411990

= 1 – 0.415 = 0.585

rs = 1 – [ 6 x∑D2

N3 – N) ]Where:

Step 6: Compare your calculated value of rs with the relevant critical value in your stats table

For N = 10 and P = 0.05, the critical value of rs is 0.65

Our value of rs is 0.585

Because this is below the critical value, we must accept H0

There is no statistically significant correlation between Maths scores and Biology scores

rs critical values (P=0.05)

No. of pairs Critical value

8 0.74

9 0.68

10 0.65

12 0.59

14 0.54

Mad Geoff’s Chaotic Firework Factory

Do peoples stress levels increase the closer they live to Mad Geoff’s Chaotic Firework

Factory?

Cortisol is a stress hormone

The more stressed an individual is, the higher their blood cortisol levels will be

Acacia Ave.

Resident Address Blood cortisol level (μg/ml)

Karl (Caretaker) 2 (Factory) 13.4

Lillie 8 22.6

Melanie 10 23.4

Nigel 12 18.6

Olga 12 17.4

Peter 14 16.8

Quentin 16 15.2

Rajesh 16 10.2

Susan 18 9.8

Toni 18 12.6

Uri 18 8.8

Vanessa 20 7.5

H0 = there is no significantly significant correlation between proximity to the fireworks factory and blood cortisol levels

Resident Address Address rank

Blood cortisol

level (μg/ml)

Cortisolrank D D2

Karl 2 13.4

Lillie 8 22.6

Melanie 10 23.4

Nigel 12 18.6

Olga 12 17.4

Peter 14 16.8

Quentin 16 15.2

Rajesh 16 10.2

Susan 18 9.8

Toni 18 12.6

Uri 18 8.8

Vanessa 20 7.5

∑D2

1

2

3

4.5

4.5

6

7.5

7.5

10

10

10

12

6

11

12

10

9

8

7

4

3

5

2

1

-5

-9

-9

-5.5

-4.5

-2

0.5

3.5

7

5

8

11

25

81

81

30.25

20.25

4

0.25

12.25

49

25

64

121

513

rs = 1 - 6∑D2

N3 – N)

N = 12

∑D2 = 513

rs = 1 - 6(513)123 - 12

= 1 - 6(513)1728 - 12

= 1 - 30781716

= 1 – 1.794 = - 0.794

Where:

When comparing rs to the critical value ignore the sign on rs

For N = 12 and P = 0.05, the critical value of rs is 0.587

Our value of rs is 0.794

Because this is above the critical value, we can reject H0

There is a statistically significant correlation between proximity to the fireworks factory and blood cortisol levels

rs is negative so there is a negative correlation...

Therefore: the further one lives from the firework factory, the lower one’s blood cortisol levels.

Pearson

KARL PEARSON(1857-1936)British mathematician, ‘father’ of statistics and a pioneer of eugenics!


If there is a statisticallysignificant correlation between two measured variables, X and Y, and….

If that correlation is negative or positive

Pearson’s Correlation Coefficient

Note: The data must show normal distribution

What is the correlation coefficient r ?

Is there a significant correlation between an animal’s nose-to-tail length and its body mass?

Animal Mass (arbitrary units)

Length (arbitrary units)

1 1 2

2 4 5

3 3 8

4 4 12

5 8 14

6 9 19

7 8 22

If yes, then is the correlation positive (does a long tail mean a larger mass)?

Pearson’s Correlation Coefficient

Where:n = the number of values of X and Y

r will always be a number between –1 and +1

This number can be compared with those in a table of critical values using: n – 2 degrees of freedom.

∑XY – [(∑X)(∑Y)]/n

{∑X2-[(∑X)2/n]} {∑Y2-[(∑Y)2/n]}r =

H0 = there is no statistically significant correlation between length and body mass

HA = there is a statistically significant correlation between length and body mass

A negative value for r implies a negative correlation

A positive value for r implies a positive correlation

If the value for r exceeds the critical value, then you can reject the null hypothesis.


Construct the following results table

Animal(n = 7)

Mass (X)

Length (Y)

X2 Y2 XY

1 1 2

2 4 5

3 3 8

4 4 12

5 8 14

6 9 19

7 8 22

Total

Mean

Construct the following table:

Animal Mass

(X)Length

(Y)X2 Y2 XY

1 1 2 1

2 4 5 16

3 3 8 9

4 4 12 16

5 8 14 25

6 9 19 81

7 8 22 64

Total X = 37

Mean 5.29

Calculate values for X:

Animal Mass

(X)Length

(Y)X2 Y2 XY

1 1 2 1

2 4 5 16

3 3 8 9

4 4 12 16

5 8 14 25

6 9 19 81

7 8 22 64

Total X = 37 X2 = 251

Mean 5.29

Calculate values for Y:

Animal Mass

(X)Length

(Y)X2 Y2 XY

1 1 2 1 4

2 4 5 16 25

3 3 8 9 64

4 4 12 16 244

5 8 14 25 196

6 9 19 81 361

7 8 22 64 484

Total X = 37 Y = 82 X2 = 251 Y2 = 1278

Mean 5.29 11.71

Calculate values for XY:

Animal Mass

(X)Length

(Y)X2 Y2 XY

1 1 2 1 4 2

2 4 5 16 25 20

3 3 8 9 64 24

4 4 12 16 244 48

5 8 14 25 196 112

6 9 19 81 361 152

7 8 22 64 484 176

Total X = 37 Y = 82 X2 = 251 Y2 = 1278 XY = 553

Mean 5.29 11.71

Use values obtained to populate the equation:

X = 37 Y = 82 X2 = 251 Y2 = 1278 XY = 553

∑XY – [(∑X)(∑Y)]/n

{∑X2-[(∑X)2/n]} {∑Y2-[(∑Y)2/n]}r =

553 – (37 x 82)/7

{251-[372/7]} {1278-[822/7]}r =

n = 7

=

r = 0.901

119.57

132.60

Compare your calculated value of r with the relevant critical value in your stats table

Our value of r = + 0.901Degrees of freedom = n - 2 = 5


3 0.88

4 0.81

5 0.75

6 0.71

7 0.67

Our value for r exceeds the critical value, so we can reject the null hypothesis.

The + sign shows that any correlation is positive.

We can conclude that there is a significant positive correlation between the length of an animal and its body mass

i.e. a long tail is associated with a large body mass!

Two sample t-test

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


If there is a statistically significant difference between two means, when:

The sample size is less than 25.The data is normally distributed

The t-test

t-test

x1 = mean of first sample x2 = mean of second samples1 = standard deviation of first samples2 = standard deviation of second samplen1 = number of measurements in first samplen2 = number of measurements in second sample

x1 – x2

(s12/n1) + (s2

2/n2)

t =

SD = (x – x)2

n – 1

Worked exampleDoes 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.

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.


Pot Group 1 (pH5.5)

(x – x)2 Group 2 (pH7.0)

(x – x)2

1 38 1.27 39 20.25

2 41 3.52 45 2.25

3 43 15.02 41 6.25

4 39 0.02 46 6.25

5 37 4.52 48 20.25

6 38 1.27 39 20.25

7 41 3.52 46 6.25

8 36 9.77 44 0.25

Mean 39.1 1.27 43.5 20.25

38.88 82.0

Construct the following table…

Johnson

Click to reveal data

Calculate standard deviation for both groups

SD = (x – x)2

n – 1

SD = (x – x)2

n – 1

Group 1:

Group 2:

= 38.888 – 1

=82.0

8 – 1

= 2.36

= 3.42

Using your means and SDs, calculate value for t

x1 – x2

(s12/n1) + (s2

2/n2)

t =

39.1 – 43.5

(2.362/8) + (3.422/8)

t =

t = 2.99

-4.4

0.696 + 1.462=

Compare our calculated value of r with the relevant critical value in the stats table of critical values

Our value of t = 2.99Degrees of freedom = n1 + n2 – 2 = 14


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.

Standard error and confidence limits


If there is a statistically significant difference between two means, when:

The sample size is at least 30.The data are normally distributed.

NB: You can use this test to assess up to 5 means on the same graph.

Standard Error with 95% Confidence Limits

Worked example

A student investigated the variation in the length of mussel shells on two different locations on a rocky shore.

The student measured the shell length of 30 mussels at each location.

H0 = there is no statistically significant difference between the means of the two samples of mussels

HA = there is a significant difference between the means of the two samples of mussels

If the 95% confidence limit around the means do not overlap, then you can reject the null hypothesis.


Step 1: Calculate SD for both groups

SD = (x – x)2

n – 1

SD = (x – x)2

n – 1

Group 1:

Group 2:

= 361630 – 1

=1416

30 – 1

= 11.2

= 7.0

Step 2: Calculate the SE for both groups

SD

n

SE =

SD

n

SE =

Careful with rounding off roundig offsGroup 1:

Group 2:

7.0

30 =

11.2

30

== 2.04

= 1.28

Step 3: Calculate the 95% confidence limitsMean ± 2 x SE

Group 1:Upper limit = 61 + (2 x 2.04) = Lower limit = 61 – (2 x 2.04) =

Group 2:Upper limit = 33 + (2 x 1.28) = Lower limit = 33 – (2 x 1.28) =

Note: in recent exam specs maybe 1.96 instead of 2 (more

precise)

0.00

10.00

20.00

30.00

40.00

50.00

60.00

70.00

0.5 1 1.5 2

With no overlap, we can conclude that there is a significant difference between the two means. Shell lengths differ significantly between the two locations.

Step 3: Plot means and confidence limits

Mean ± 2 x SE

EXAM QUESTIONS

• Should guide you quite a lot:

Conclusion

• Hopefully some skills, knowledge and confidence added

• Please fill in the evaluation form and http://is.gd/m4bsurvey2

• In a month’s time we would like to send you some follow-up questions.– Impact form– Maths confidence

• Thank you for your attention.

maths for biology july 6 th 2015 christian bokhove carys hughes hilary otter rebecca d’silva nicky...

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