Quantitative Methods
Part 3
T- Statistics
Standard Deviation
Measures the spread of scores within the
data set
◦ Population standard deviation is used when
you are only interested in your own data
◦ Sample standard deviation is used when you
want to generalise for the rest of the
population
Z - Scores
A specific method for describing a
specific location within a distribution
◦ Used to determine precise location of an in
individual score
◦ Used to compare relative positions of 2 or
more scores
Normally Distributed (Bell shaped)
Distribution of the Means
• Frequency Distribution of 4 scores (2, 4, 6,8)
X
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1
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3
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X
5
6
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7
• It is flat and not bell shaped
• Mean of population is (2+4+6+8)/4 = 5
8
9
Distribution of the Means
• Take all possible samples of pairs of scores (2,4,6,8)
• Use random sampling and replace each individual
sample into data set
• Calculate average of all sample pairs
2+6 /2 = 4
6+2 /2 = 4
2+8 /2 = 5
8+2 /2 = 5
2+2 /2 = 2
2+4 /2 = 3
4+2 /2 = 3
0
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2
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4
4+4 /2 = 4
4+6 /2 = 5
6+4 /2 = 5
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6
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Central Limit Theorem
“For any population with a mean μ and standard
deviation σ , the distribution of sample means for
sample size n will have a mean of μ and standard
deviation of σ/√n and will approach a normal
distribution as n gets very large.”
How big should the sample size be? n=30
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Standard Error
σ/√n is used to calculate the Standard Error of the sample
mean
Sample data = X
The mean of each sample =
Then the standard error becomes
It identifies how much the observed sample mean
differs from the un-measurable population mean μ.
So to be more confident that our sample mean is a
good measure of the population mean, then the
standard error should be small. One way we can ensure
this is to take large samples.
Example
Z=
- μ/ σ
The population of SATs scores is normal with μ= 500, σ
=100. What is the chance that a sample of n=25 students
has a mean score =540? Since the distribution is normal,
we can use the z-score
First calculate Standard Error
◦ 100/5 = 20
Then Z-Score
◦ 540-500/20 =2
z-value is 2, therefore around 98% of the sample means are
below this and only 2% are above. So we conclude that the
chance of getting a sample mean of 540 is 2%, so we are
98% confident that this sample mean, if recorded in an
experiment is false.
T - Statistics
So far we’ve looked at mean and sd of
populations and our calculations have had
parameters
But how do we deduce something about
the population beyond our sample?
We can use T-Statistic
T - Statistics
Remember SD from last week?
Great for population of N but not for
sample of n
Why n -1?
Because we can only freely choose n-1
(Degree of freedom = df)
T - Statistics
Standard Error
Z-Score redone to show above
=
To T-Statistic, we substitute σ (SD of
population) with s (SD of sample)
But what about μ ?
Hypothesis Testing
Sample of computer game players n =16
Intervention = inclusion of rich graphical
elements
Level has 2 rooms
◦ Room A = lots of visuals
◦ Room B = very bland
Put them in level 60 minutes
Record how long they spend in B
Results
Average time spent in B = 39 minutes
Observed “sum of squares” for the
sample is SS = 540.
A
B
Stage1: Formulation of Hypothesis
: “null hypothesis”, that the visuals have
no effect on the behaviour.
: “alternate hypothesis”, that the visuals
do have an effect on the players’ behaviour.
If visuals have no effect, how long on
average should they be in room B?
Null hypothesis is crucial; here we can
infer that μ = 30 and get rid of the
population one
Stage 2: Locate the critical region
We use the T-table to help us locate this,
enabling us to reject or accept the null
hypothesis. To get we need:
◦ Number of degree of freedom (df) 16 -1 =15
◦ Level of significance of confidence
◦ Locate in T-table (2tails)= critical value of t=2.131, t=2.131
Stage 3: Calculate statistics
Calculate sample sd
Sample Standard Error
=6
= 6 / 4 =1.5
T-Statistic
=6
The μ 30 came from the null hypothesis if
visuals had no effect, then the player
would spend 30 minutes in both rooms A
and B.
Stage 4: Decision
Can we reject the , that the visuals
have no effect on the behaviour?
◦ T = 6 which is well beyond the value of 2.313
which indicates where chance kicks in.
So yes we can safely reject it and say it
does affect behaviour
Which room do they prefer?
◦ They spent on average 39 minutes in Room B
which is bland
Workshop
Work on Workshop 6 activities
Your journal (Homework)
Your Literature Review
(Complete/update)
References
Dr C. Price’s notes 2010
Gravetter, F. and Wallnau, L. (2003) Statistics for the Behavioral
Sciences, New York: West Publishing Company
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