Transcript Mr. Howell`s Stats Presentation

Topic 1: Statistical Analysis
• Assessment Statements
– 1.1.1 State that error bars are a graphical representation of the
variability of data
– 1.1.2 Calculate the mean and standard deviation of a set of
values.
– 1.1.3 State that the term standard deviation is used to
summarize the spread of values around the mean, and that 68%
of the values fall within one standard deviation of the mean.
– 1.1.4 Explain how the standard deviation is useful for comparing
the means and the spread of data between 2 or more samples
– 1.1.5 Deduce the significance of the difference between two
sets of data using calculated values for t and the appropriate
tables.
– 1.1.6 Explain that the existence of a correlation does not
establish that there is a causal relationship between two
variables.
Why use statistics?
• In science, we are always making
observations about the world around us.
• Many times these observations result in
the collection of measurable, quantitative
data.
• Using this data, we can ask questions
based upon some of these observations
and try to answer those questions.
Essential Statistics Terms
• Mean: the average of the data points
• Range: the difference between the largest and
smallest observed values in a data set (the
measure of the spread of the data)
• Standard Deviation: a measure of how the
individual observations of a data set are
dispersed or spread out around the mean
• Error bars: Graphical representation of the
variability of data
Standard Deviation
• Standard deviation is used to summarize the
spread of values around the mean and to
compare the means of data between two or
more samples.
– In a normal distribution, about 68% of all
values lie within ±1 standard deviation of the
mean. 95% will lie within ±2 standard
deviations of the mean.
– This normal distribution will form a bell curve
– The shape of the curve tells us how close all
of the data points are to the mean.
Calculating Standard Deviation
• We can use the formula to solve for standard
deviation…
Σ(xi - (Σx)2
Sx =
n
n-1
Example data: Ward pg.5
For calculator help, go to http://www.heinemann.co.uk/hotlinks ,enter code
4242p and click weblink 1.4a or 1.4b
HOWEVER the easiest way to calculate…find an app online!!!!
Are our results reliable enough to
support a conclusion?
Imagine we chose two children at random from two class
rooms…
D8
C1
… and compare their height …
D8
… we find that
one pupil is
taller than the
other
C1
WHY?
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
REASON 2: By chance, we picked a short pupil from D8
and a tall one from C1
D8
C1
Sammy
HAGRID
(Year 9)
(Year 9)
How do we decide which reason is
most likely?
MEASURE MORE STUDENTS!!!
If there is a significant difference between the two groups…
D8
… the average or
mean height of the
two groups should
be very…
… DIFFERENT
C1
If there is no significant difference between the two groups…
D8
… the average or
mean height of the
two groups should
be very…
… SIMILAR
C1
Remember:
Living things normally show
a lot of variation, so…
It is VERY unlikely that the mean height of our two samples
will be exactly the same
C1 Sample
Average height = 162 cm
D8 Sample
Average height = 168 cm
Is the difference in average height of the samples
large enough to be significant?
16
C1 Sample
14
Frequency
We can analyse the
spread of the heights of
the students in the
samples by drawing
histograms
12
10
8
6
4
2
Here, the ranges of the
two samples have a
small overlap, so…
16
14
Frequency
… the difference between
the means of the two
samples IS probably
significant.
140149
150- 160- 170159
169
179
Height (cm)
180189
D8 Sample
12
10
8
6
4
2
140149
150- 160- 170159
169
179
Height (cm)
180189
C1 Sample
14
Frequency
Here, the ranges of
the two samples have
a large overlap, so…
16
12
10
8
6
4
… the difference
between the two
samples may NOT be
significant.
2
140149
16
The difference in
means is possibly due
to random sampling
error
Frequency
14
150- 160- 170159
169
179
Height (cm)
180189
D8 Sample
12
10
8
6
4
2
140149
150- 160- 170159
169
179
Height (cm)
180189
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:
Sx =
Σ(xi -
(Σxi)2
n-1
n
Where:
Sx is the standard deviation of sample
Σ stands for ‘sum of’
xi stands for the individual measurements in
the sample
n is the number of individuals in the sample
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
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
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 )
t=
(s1)2
n1
+
(s2)2
n2
Where:
x1 is the mean of sample 1
s1 is the standard deviation of sample 1
n1 is the number of individuals in sample 1
x2 is the mean of sample 2
s2 is the standard deviation of sample 2
n2 is the number of individuals in sample 2
Worked Example: Random samples were taken of pupils in
C1 and D8
Their recorded heights are shown below…
Students in C1
Student
Height
(cm)
Students in D8
145
149
152
153
154
148
153
157
161
162
154
158
160
166
166
162
163
167
172
172
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
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
n1
D8: (s2)2
n2
= 10.862 ÷ 15 = 7.86
= 11.742 ÷ 15 = 9.19
Step 5: Calculate
(s1)2
n1
(s1)2
n1
+
(s2)2
=
+
(s2)2
n2
(7.86 + 9.19) =
4.13
n2
Step 6: Calculate t
(Step 2 divided by Step 5)
x2 – x1
t=
(s1)2
n1
+
(s2)2
n2
=
6.67
4.13
= 1.62
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
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