Download Measures of central tendency and dispersion

Measures of central tendency
and dispersion
Measures of central tendency
• Mean
• Median
• Mode
• ie finding a ‘typical’
value from the middle
of the data.
You need to be able to:
• Explain how to calculate the mean, median
and mode
• State the strengths and weaknesses of mean,
median and mode
• This could include saying which one you would
use for some data e.g. 2, 2, 3, 2, 3, 2, 3, 2, 97 would you use mean or median here?
Advantages and disadvantages


Mean
More sensitive than the
median, because it makes
use of all the values of the
data.
It can be misrepresentative
if there is an extreme
value.
Median
It is not affected by
It is less sensitive than the
extreme scores, so can give mean, as it does not take
a representative value.
into account all of the
values.
Mode
It is useful when the data
are in categories, such as
the number of babies who
are securely attached.
It is not a useful way of
describing data when there
are several modes.
Measures of Dispersion
• Measures of ‘spread’
• This looks at how
‘spread out’ the data
are.
• Are the scores similar to
each other (closely
clustered), or quite
spread out?
Range and standard deviation
• The range is the difference between the highest
and lowest numbers. What is the range of …
• 3, 5, 8, 8, 9, 10, 12, 12, 13, 15
• Mean = 9.5
range = 12 (3 to 15)
• 1, 5, 8, 8, 9, 10, 12, 12, 13, 17
• Mean = 9.5
range = 16 (1 to 17)
•
Example from Cara Flanagan, Research Methods for AQA A Psychology (2005) Nelson Thornes p 15
Standard deviation
• Standard deviation tells
us the average distance
of each score from the
mean.
• 68% of normally
distributed data is
within 1 sd each side of
the mean
• 95% within 2 sd
• Almost all is within 3 sd
Example
• Mean IQ = 100, sd = 15
• What is the IQ of 68%
of population (ie what is
the range of possible
IQs)?
• Between what IQ scores
would 95% of people
be?
Another example
• Sol scores 61% in the
test. His mum says
that’s rubbish. Sol
points out that the
mean score in class was
50%, with an sd of 5.
Did he do well?
• What if the sd was only
2?
• What if sd was 15?
Advantages and disadvantages
Advantages
Disadvantages
Range
Quick and easy to calculate
Affected by extreme values
(outliers)
Does not take into account
all the values
Standard deviation
More precise measure of
dispersion because all
values are taken into
account
Much harder to calculate
than the range
I used Cara Flanagan’s (2005) Research Methods for AQA A Psychology Nelson Thornes in preparing these slides.
Standard Deviation
• Standard deviation (SD) is a statistical measure of
the amount the results vary from the mean.
• There are 2 formulas that can be used to work
out the standard deviation:
• Formula 1
S= √∑d²
• Formula 2:
n
S= √∑d²
n-1
Formula 1 is used to calculate
the SD where the whole
population has been used.
Formula 2 is used to calculate the
SD where part the population has
been used. This is the formula used
most often.
Standard Deviation
• Example - Test scores from 10 students:
– 85, 86, 94, 95, 96, 107, 108, 108, 109, 112




Calculate the mean of the data

Mean = 100
Square the differences between each score and mean:
Test score
d
85
-15
86
-14
94
-6
...
..
109
9
112
12
2
=
 Sum of d 900
d2
225
196
36
..
81
144
(85-100 = -15 -152 = 225)
N = number of scores = 10, so n-1 = 9
Substitute into the formula and we get 10, because 900
divided by 9 is 100 and the square root of 100 is 10
Standard Deviation
•
Example 2: Calculate the standard deviation of the following scores:
12, 10, 8, 4, 18, 8
–
–
Mean = 10
Subtract each score from the mean to give deviation of score from mean
•
–
Square each deviation
•
–
–
–
–
(-2, 0, 2, 6, -8, 2)
(4, 0, 4, 36, 64, 4)
Add the squares of the deviation together = 112
Count the number of scores = 6, and then subtract 1
Divide the sum of the squares of the deviation by the number of scores minus 1 = 112/5 = 22.4
(called the variance)
Square root the variance = standard deviation = 4.73
Remember…
If we were just to look at the mean score without any further information, this
data would be misleading.
THEREFORE, YOU NEVER REPORT THE MEAN WITHOUT THE
STANDARD DEVIATION.
Large standard deviations suggest that there is a large variance in the scores,
and possibly 1 or 2 scores are distorting the mean
Small standard deviations suggest that there are no large variations in the
scores distorting the mean.
Task
• Scores showing the number of seconds taken by
participants to complete a task without alcohol
(condition 1) and with alcohol (condition 2)
Condition (1)
3
4
4
5
5
5
6
6
7
Condition (2)
3
6
6
9
9
9
11
13
15
Calculate the:
•Mean
•Mode
•Median
•Standard dev
For both conditions
Task Answers
• Calculate the mode, median, mean and
standard deviation for each condition
Mode
– Condition 1
– Condition 2
5
9
Median Mean
5
9
5
9
Std Dev
1.22 secs
3.71 secs
• So what do the measures of central tendency
and dispersion tell us?
Task Interpretation
• The means tell us…
– Participants who had consumed alcohol (condition 2) took
nearly twice as long to complete the task as the
participants who had not consumed alcohol (condition 1)
• The standard deviations tell us…
– Spread of scores in condition 2 was much greater than the
spread in condition 1.
– Scores in condition 1 are clustered around the mean,
whilst those in condition 2 are more widely spread out
Task Conclusions
• What do the findings indicate generally?
– Alcohol increases the average time taken to
perform a task
– It exaggerates differences in task performance
• slows some people down a lot and others hardly at all
Task
• Complete hand outs to ensure that you are
confident in working out the measures of
central tendency and the standard deviation