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Unit 7 populations and inheritance a level biology £7.16
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Unit 7 populations and inheritance a level biology

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Notes made by a college (UK) student on Unit 7 A level Biology which includes the subject of Populations and inheritance. Includes in depth notes with some diagrams

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  • June 27, 2024
  • 16
  • 2023/2024
  • Lecture notes
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Student T test
e Biomedicalscientists carried out a study to look for a significant difference
between the mean cardiac output of patients with and without angina.
There were 6 patients studied in the group with angina, and there were 8
patients studied in the group without angina.
Write a null hypothesisfor this investigation

042




Why wouldusing a student T test be an appropriatestatistical test for this
investigation?




Using the results of the investigation,scientists carried out a T test and
obtained a value of 2.45.
Using the table below, write a conclusionto determinewhetherto accept or reject
the null hypothesis

A tablo showing the critical values of j for difforont degrees of freedom.

Dogroos Critical Degrees Critical
of val ue of value
freedom froodom
4 2.78
5 2.57 15 2.13
6 2.48 16 2.12
7 2.37 1B 2.10
8 2.31 20 2.09
9 2.26 22 2.07
10 2.23 24 2.06
11 2.20 26 2.06
12 2.18 28 2.05
13 2,16 30 2.04
14 2.15 2.02
The numbcr of dogrocs of fruodom (n, '12) — 2
(6+6)

Vauxc
so
Uvac»kOne duffer enc e 03 clue
veoeceeck

, Spearman's Rank Correlation test

Ecologists carried out an investigation to look for a significant correlation
betweenthe length of grass in different fields and the numberof grasshoppers
present
Ecologists collected data from 8 different fields
Write a null hypothesisfor this investigation
noe Seqnc
c%VCkEs cluJPee_r QÄcks ounce
Of asskoppers
VAUXVYÄbe_r- presen
e

Why wouldusing a Spearman's Rank test be an appropriate statistical test for this
investigation?

Com Z va-rvoÖo\-CS


Using the results of the investigation,scientists carried out a Spearman'sRank
test and obtained a value of 0.97.
Using the table below, write a conclusionto determine whether to accept or
reject
the null hypothesis

A table showing the critical values of rs for different numbers of paired values.



Number of pairs
Critical value
of measurements
5 I .00
6 0.89
7 0.79
8 0.74

9 0.68
10 0.65
12 0.59
14 0.54
16 0.51

0.48


voLwe 08 es

eo cho.mce

, Chi Squared test clcv€c,e

blood group determined.
A group of 25patients were randomlyselected to have their
12 patients were found to have
Of the 24, 7 patients were found to have blood group A,
AB.
blood group B, and 6 were found to have blood group

Write a null hypothesis for this:
es seovüRccuae beecoeeh
e OJccenes A ,
AB
v o wp


2
Observed Expected



Blood group A 0.20
Blood group AB
Blood group B
Sum =
Average-6.3


Write a conclusionto this investigation:
A table showing the critical values of for differentdegrees of freedom.

Degrees of Critical value
freedom
1 3.84
2 5.99
2
— 791
4 9
11.07
12.59
7 14.07
15.51
9 16.92
10 18 31


c.ateoori0b—1
The number degrees of freedom c numb0f of



oe one
60
eeel
CYÄ0-.n c e

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