Calculations & Statistics - OCR A Level Biology

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42 Terms

1
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You want to know if there is an association between different measurement from the same sample, use...

Spearman's Rank Correlation

2
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You have counted the number of individuals in two or more categories, use...

Chi squared test

3
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You want to look at the spread of data about the mean, use...

Standard deviation

4
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You want to compare two sets of data, the data sets are collected from the same individuals, use...

paired t-test

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You want to compare two sets of data, the data sets are no collected from the same individual, use...

unpaired t-test

6
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When can you use chi squared?

sample size over 20, only discontinuous data

7
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Null hypothesis for chi squared

no difference between observed and expected results

8
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If x2 is greater than critical value...

difference is not due to chance and the null hypothesis is rejected

9
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How do you choose the number of degrees of freedom in chi squared?

n-1

(n= number of categories)

10
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chi squared

knowt flashcard image
11
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What is the significance level?

5%

what percentage probability we are happy to accept for there to be a mistake in our conclusion

12
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What is the null hypothesis of unpaired t-test?

The means of both groups are equal.

13
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How do you choose the number of degrees of freedom in the unpaired t-test?

(nA + nB) - 2

(ie number of categories in group A but number of categories in group B -2)

<p>(nA + nB) - 2</p><p>(ie number of categories in group A but number of categories in group B -2)</p>
14
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unpaired t-test

x= mean s= standard deviation n= number of categories

15
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If t is bigger than the critical value in an unpaired t-test...

reject the null hypothesis, the two means are not equal

16
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paired t-test

t =(d√n)/ sd

d= mean difference n= number of categories sd= standard deviation

17
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How do you decide the number of degrees of freedom for the paired t-test?

n-1

(number of categories - 1)

18
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If t is much bigger than the critical value in the paired t-test...

the difference is not due to chance

19
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Spearman's rank- how to rank data

1 onwards, 1 being largest

2 groups, 1 labelled and ranked as x, the other as y

20
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Spearman's rank- after ranking the data...

find the difference between ranks (d), then find d^2 and the total of d^2

21
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Spearman's rank

d^2= square of difference between data ranks n= number of paired data items

<p>d^2= square of difference between data ranks n= number of paired data items</p>
22
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If rs is bigger than the critical value...

then we have evidence to reject the null hypothesis (H0) and accept H1- with 95% confidence there is a positive correlation between x and y

23
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standard deviation

quantitative measure of the spread of data about the mean

<p>quantitative measure of the spread of data about the mean</p>
24
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interpreting standard deviation

• The middle 68% of the data lie within one standard deviation either side of the mean • 95% of the data lie within 2 standard deviations either side of the mean.

25
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Simpson's index of diversity

n= number of a particular species N= total number of organisms

<p>n= number of a particular species N= total number of organisms</p>
26
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The higher the value of D...

the higher the diversity

27
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exponential graph

knowt flashcard image
28
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exponential log graph

knowt flashcard image
29
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percentage change=

(new-orginal/original) x100

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percentage yield=

(actual/ theoretical)x100

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percentage error (uncertainty)=

(2x absolute uncertainty/ quantity measured) x 100

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magnification=

size of image/ size of real object

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Rf=

distance moved by solute/ distance moved by solvent

34
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rate =

change in quantity/ time taken

35
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ratio=

SA/ V

36
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proportion of polymorphic gene loci (genetic biodiversity)=

number of polymorphic gene loci/ total number of loci

37
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cardiac output =

heart rate × stroke volume

38
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Respiratory quotient

CO2 produced/ O2 consumed

39
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Microorganism population growth N=

N0 × 2^n

40
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Efficiency of biomass transfers=

(biomass transferred/ biomass intake) x 100

41
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Temperature coefficient (Q10)=

R2/ R1

42
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Hardy Weinberg principle

(p +q)^2= 1 (100%) so p^2 + 2pq + q^2 = 1 (100%)

p= dominant allele p^2= homozygous dom

q= recessive allele q^2= homozygous recessive 2pq= heterozygous