1. Are our results reliable enough to
support a conclusion?
Geoff Browne
Anglo-European School
Essex, UK
2. Imagine we chose two children at random from two class
rooms…
D8 C1
… and compare their height …
3. D8 C1
… we find that
one pupil is
taller than the
other
WHY?
4. REASON 1: There is a significant difference between the
two groups, so pupils in C1 are taller than
pupils in D8
D8 C1
YEAR 7 YEAR 11
5. REASON 2: By chance, we picked a short pupil from D8
and a tall one from C1
D8 C1
TITCH HAGRID
(Year 9) (Year 9)
6. How do we decide which reason is
most likely?
MEASURE MORE STUDENTS!!!
7. If there is a significant difference between the two groups…
D8 … the average or C1
mean height of the
two groups should
be very…
… DIFFERENT
8. If there is no significant difference between the two groups…
D8 … the average or C1
mean height of the
two groups should
be very…
… SIMILAR
9. Remember:
Living things normally show
a lot of variation, so…
10. It is VERY unlikely that the mean height of our two samples
will be exactly the same
C1 Sample D8 Sample
Average height = 162 cm Average height = 168 cm
Is the difference in average height of the samples
large enough to be significant?
11. 16
We can analyse the 14 C1 Sample
spread of the heights of 12
Frequency
the students in the 10
8
samples by drawing
6
histograms 4
2
Here, the ranges of the 140- 150- 160- 170- 180-
149 159 169 179 189
two samples have a Height (cm)
small overlap, so… 16
14
12
D8 Sample
… the difference between Frequency 10
8
the means of the two 6
samples IS probably 4
significant. 2
140- 150- 160- 170- 180-
149 159 169 179 189
Height (cm)
12. 16
Here, the ranges of 14 C1 Sample
the two samples have 12
Frequency
a large overlap, so… 10
8
6
4
… the difference 2
between the two 140- 150- 160- 170- 180-
samples may NOT be 149 159 169 179 189
Height (cm)
significant. 16
14
12
D8 Sample
The difference in Frequency
10
means is possibly due 8
to random sampling 6
4
error
2
140- 150- 160- 170- 180-
149 159 169 179 189
Height (cm)
13. To decide if there is a significant difference between two
samples we must compare the mean height for each
sample…
… and the spread of heights in each sample.
Statisticians calculate the standard deviation of a sample
as a measure of the spread of a sample
You can calculate standard deviation using the formula:
Where:
(Σx)2 Sx is the standard deviation of sample
Sx = Σx2 - Σ stands for ‘sum of’
n
x stands for the individual measurements in
n-1 the sample
n is the number of individuals in the sample
14. It is much easier to use the statistics functions on a scientific
calculator!
e.g. for data 25, 34, 13
Set calculator on statistics mode
MODE 2 (CASIO fx-85MS)
Clear statistics memory
SHIFT CLR 1 (Scl) =
Enter data
2 5 DT (M+ Button) 3 4 DT 1 3 DT
15. Calculate the mean
AC SHIFT S-VAR (2 Button) 1 ( x ) = 24
Calculate the standard deviation
AC SHIFT S-VAR 3 (xσn-1) = 10.5357
16. Student’s t-test
The Student’s t-test compares the averages and standard
deviations of two samples to see if there is a significant
difference between them.
We start by calculating a number, t
t can be calculated using the equation:
( x1 – x2 ) Where:
t= x1 is the mean of sample 1
(s1)2 (s2)2 s1 is the standard deviation of sample 1
+ n1 is the number of individuals in sample 1
n1 n2
x2 is the mean of sample 2
s2 is the standard deviation of sample 2
n2 is the number of individuals in sample 2
17. Worked Example: Random samples were taken of pupils in
C1 and D8
Their recorded heights are shown below…
Students in C1 Students in D8
Student 145 149 152 153 154 148 153 157 161 162
Height
154 158 160 166 166 162 163 167 172 172
(cm)
166 167 175 177 182 175 177 183 185 187
Step 1: Work out the mean height for each sample
C1: x1 = 161.60 D8: x2 = 168.27
Step 2: Work out the difference in means
x2 – x1 = 168.27 – 161.60 = 6.67
18. Step 3: Work out the standard deviation for each sample
C1: s1 = 10.86 D8: s2 = 11.74
Step 4: Calculate s2/n for each sample
C1: (s1)2
= 10.862 ‚ 15 = 7.86
n1
D8: (s2)2
= 11.742 ‚ 15 = 9.19
n2
20. Step 7: Work out the number of degrees of freedom
d.f. = n1 + n2 – 2 = 15 + 15 – 2 = 28
Step 8: Find the critical value of t for the relevant number of
degrees of freedom
Use the 95% (p=0.05) confidence limit
Critical value = 2.048
Our calculated value of t is below the critical value for 28d.f.,
therefore, there is no significant difference between the
height of students in samples from C1 and D8
21. Do not worry if you do not understand
how or why the test works
Follow the
instructions
CAREFULLY
You will NOT need to remember how to do this for your exam
22. Uploaded (with permission from Geoff Browne)
by Stephen Taylor
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