Ma & THs sTaTs · M. Harris, G. Taylor & J. Taylor second ediTion Ma & THs sTaTs Ma for ife and THe...
Transcript of Ma & THs sTaTs · M. Harris, G. Taylor & J. Taylor second ediTion Ma & THs sTaTs Ma for ife and THe...
M. H a r r i s , G . Tay l or & J . Tay l or
se cond ed i t ion
MaTHs & sTaTs f o r T H e l i f e a n d M e d i c a l s c i e n c e s
MaTHs & sTaTs
Catch Up Maths & Stats covers the core maths skills you will need on any life or medical sciences course including: • working with fractions and powers • using and understanding graphs • preparation of a dilution series • analysing enzyme kinetics • how to calculate standard deviation • choosing the right statistical test
Over 200 examples are provided to show the relevance and application of maths and stats to your course. This new edition also now provides further coverage of differentiation and integration alongside nearly 70 more questions and answers to help you test your understanding.
Catch Up Maths & Stats will bring you up to speed in the shortest possible time, even if you didn’t study maths at advanced level.
r e v ie w s of T He f ir s T ed i T ion:“This is an excellent book, explains the fundamentals but also explains more
complicated concepts such as standard deviations in a way that is easy to understand and useable.”
“Great book, I just started university for a biology degree and hadn’t taken maths at A level - immediately felt behind. This book starts with simple
concepts and moves on to the really difficult stuff but explains in an easy way. Also has questions to check learning. I would recommend to anybody
who feels their maths skills aren’t up to scratch.”
“This is an invaluable book, very easy to use and understand even with no prior knowledge of maths. A real must-have!”
Harris, Taylor & TaylorP HIL IP BR A DL E Y & J A NE C A LV ER T
SE COND ED I T ION
BIOLOGY F O R T H E M E D I C A L S C I E N C E S
MI T CH F R Y A ND EL IZ A BE T H PA GE
SE COND ED I T ION
CHEMISTRY F O R T H E L I F E A N D M E D I C A L S C I E N C E S
MaTHs & sTaTs
9 781904 842903
ISBN 978-1-904842-90-3
w w w. s c i o n p u b l i s h i n g . c o m
second edition
Contents
Preface vii
About the authors viii
Acknowledgements viii
1 How to use this book 1
Mathematics section
2 Handling numbers 3
3 Working with fractions 9
4 Percentages 13
5 Powers 17
6 Circles and spheres 21
7 Approximation and errors 23
8 Introduction to graphs 27
9 The gradient of a graph 33
10 Algebra 41
11 Polynomials 47
12 Algebraic equations 51
13 Quadratic equations 53
14 Simultaneous equations 57
15 Sequences and series of numbers 61
16 Working with powers 65
17 Logarithms 69
18 Exponential growth and decay 73
19 Differential calculus 77
20 Further differential calculus 89
21 Integral calculus 95
22 Further integral calculus 101
23 Using graphs 107
24 Recognising patterns in graphs 115
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vi Catch Up Maths & Stats
Applications of mathematics section
25 SI units 121
26 Moles 125
27 pH 131
28 Buffers 135
29 Kinetics 137
Statistics section
30 The language of statistics 141
31 Describing data: measuring averages 145
32 Standard deviation 151
33 Checking for a normal distribution 157
34 Degrees of freedom 159
35 How to use statistics to make comparisons 161
36 The standard error of the mean 163
37 Confidence intervals 165
38 Probability 169
39 Significance and P values 171
40 Tests of significance 175
41 t tests 179
42 Analysis of variance 185
43 The chi-squared test 189
44 Correlation 193
45 Regression 199
46 Bayesian statistics 205
Answers to “test yourself” questions 207
Appendix 1: Flow chart for choosing statistical tests 225
Appendix 2: Critical values for the t distribution 226
Appendix 3: Critical values for the chi-squared distribution 227
Index 229
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151
32 Standard deviation
Standard deviation (SD) is used for data which are normally distributed (see Section 31.1). It provides an indicator of how much the data are spread around their mean.
32.1 Standard deviation
The “deviation” is the difference between any individual reading and the mean of all the readings.
The standarddeviation is a kind of average of the individual deviations.
So, standard deviation indicates how much a set of values is spread around the mean of those values.
A range of one standard deviation above and below the mean (abbreviated to ±1 SD) includes 68.2% of the values.
±2 SD includes 95.4% of the data.
±3 SD includes 99.7%.
ExamplE
Let’s say that a group of patients has a normal distribution for weight. The mean weight of the patients is 80 kg. For this group, the SD is calculated to be 5 kg.
• 1 SD below the average is 80 - 5 = 75 kg.
• 1 SD above the average is 80 + 5 = 85 kg.
±1 SD will include 68.2% of the subjects, so 68.2% of patients will weigh between 75 and 85 kg.
95.4% will weigh between 70 and 90 kg (±2 SD).
99.7% of patients will weigh between 65 and 95 kg (±3 SD).
See how this relates to this graph of the data.
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14
12
10
8
6
4
2
060 65 70 75 80 85 90 95 100
±1 SD (68.2%)
±2 SD (95.4%)
±3 SD (99.7%)
Weight (kg)
Num
ber
of p
atie
nts
Graphshowingnormaldistributionofweightsofpatients,mean80kg,SD5kg
32.2 Calculating the standard deviation of a whole populationCalculating the SD of a whole population involves five steps.
1) First we calculate the mean, m, of the population, N, by adding all the values and dividing the result by the number of values, as in Section 31.1.
2) Then we need to work out the difference (deviation) between each individual value and the mean, by subtracting the mean from each value, symbolised by:
x - m
3) Next we calculate the square of each deviation, multiplying each value by itself. Remember that any negative deviations become positive when squared.
The process so far is:
(x - m)2
4) Then we calculate the sum of those squared deviations, known as the sumofsquares.
 (x - m)2
5) Then we need the mean of the sum of squares, so we divide the sum by the number of observations. This gives a value called the populationvariance, s2:
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Chapter 32 Standard deviation 153
s2 = Â (x - m)2
N6) The SD is the square root of the variance.
The whole process can be symbolised by:
s = ! (x - m)2
N
s, pronounced “sigma”, is the symbol for standard deviation.
ExamplE
All 10 salmon, Salmo salar, in a lake weigh 1.6, 1.7, 1.8, 1.8, 2.3, 2.4, 2.6, 2.8, 3.1 and 3.3 kg.
NB: normally, whole populations will number far more than 10. We are using a small number so that it is easier to follow the calculations.
1) The sum of the values is 23.4 kg, giving a mean of 2.34 kg.
m = 2.34 kg
2) The difference (deviation) between each individual value and the mean is as follows:
1.6 - 2.34 = -0.741.7 - 2.34 = -0.641.8 - 2.34 = -0.541.8 - 2.34 = -0.542.3 - 2.34 = -0.042.4 - 2.34 = 0.062.6 - 2.34 = 0.262.8 - 2.34 = 0.463.1 - 2.34 = 0.763.3 - 2.34 = 0.96
3) Calculating the square of each deviation gives:
-0.742 = 0.5476-0.642 = 0.4096-0.542 = 0.2916-0.542 = 0.2916-0.042 = 0.0016
0.062 = 0.00360.262 = 0.06760.462 = 0.21160.762 = 0.57760.962 = 0.9216
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4) The sum of these squares is 3.324.
5) The mean of the squares is the sum of squares divided by the number of salmon in the population, giving the variance, s2:
s2 = Â (x - m)2
N =
3.32410
= 0.3324
6) The square root of the variance gives the SD:
s = w0.3324 = 0.577 kg to three significant figures.
32.3 Calculating standard deviation rangesHaving calculated the SD, we can work out the ranges that include 68.2% (± 1 SD), 95.4% (± 2 SD) and 99.7% (± 3 SD) of a population.
ExamplE
The mean fasting triglyceride level of 183 patients is found to be 2.2 mmol l-1.
The SD is calculated to be 0.3 mmol l-1.
1 SD below the average is 2.2 - 0.3 = 1.9 mmol l-1.
1 SD above the average is 2.2 + 0.3 = 2.5 mmol l-1.
±1 SD will include 68.2% of the data, so we expect 68.2% of the population to have a triglyceride level between 1.9 and 2.5 mmol l-1.
95.4% will be between 1.6 and 2.8 mmol l-1 (±2 SD).
99.7% of triglyceride levels will be between 1.3 and 3.1 mmol l-1 (±3 SD).
See how this relates to the following graph of the data (99.7% line not shown for clarity).
1 1.5 2 2.5 3 3.50
25
20
15
10
5
Triglyceride (mmol l–1)
±1 SD (68.2%)
±2 SD (95.4%)
Fre
quen
cy
Graphshowingnormaldistributionoffastingtriglyceridelevelswithmean2.2mmoll-1andSD0.3mmoll-1
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32.4 Comparing different standard deviations
If we have two populations with the same mean but different standard deviations, then the population with the larger SD has a wider spread than the population with the smaller SD.
ExamplE
If another population of patients has the same mean fasting triglyceride level of 2.2 mmol l-1 but an SD of only 0.2 mmol l-1, ±1 SD will include 68.2% of the subjects, so 68.2% of the patients will have a fasting triglyceride level of between 2.0 and 2.4 mmol l-1.
1 1.5 2 2.5 3 3.50
5
10
15
20
25
Triglyceride (mmol l–1)
Fre
quen
cy
Graphshowinganormaldistributionoffastingtriglyceridelevelswithmean2.2mmoll-1andSD0.2mmoll-1
Compare this graph with the graph of the previous example which is drawn to the same scale. A smaller SD gives a taller, narrower normal distribution.
32.5 Calculating the standard deviation of a sample
When calculating the SD of a sample, the variance (denoted as V for a sample but s2 for a population) is calculated by dividing the sum of the squared deviations by one fewer than the number in the sample, i.e. n - 1. The value n - 1 is used rather than n as this gives a better estimate of the SD.
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ExamplE
Let’s say that the 10 salmon in the example in Section 32.2 were a sample of salmon in a lake. Compare each step with the equivalent calculation above for a population of 10.
1) The sum of the values is still 23.4 kg, giving the same mean of 2.34 kg.
However, this is now symbolised by xw = 2.34. The use of xw (rather than m) denotes that this a sample mean, not the mean of a whole population.
2) The difference (deviation) between each individual value and the mean is also unchanged.
3) Similarly, the calculation for the square of each deviation is unchanged.
4) The sum of squares is also unchanged at 3.324.
5) However, we now divide the sum of squares by (number of salmon - 1) to get the sample variance, V:
V = Â (x - xw)2
n - 1 =
3.3249
= 0.3693
6) The square root of the variance gives the SD:
SD = w0.3693 = 0.607 kg to three significant figures.
Test yourselfThe answers are given on page 218.
Question 32.1The haemoglobin levels in the blood of five volunteers are 11.7, 11.9, 12.2, 12.7 and 13.0 gl-1.What is the standard deviation?
Question 32.2For the sample of aphids on 25 lettuce plants in Question 31.3, calculate the standard deviation for the number of aphids on each plant.
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