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
0
1
2
X
3
4
X
5
6
X
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
1
X
2
X
X
3
X
X
X
4
4+4 /2 = 4
4+6 /2 = 5
6+4 /2 = 5
X
X
X
X
5
X
X
X
6
X
X
7
X
8
9
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

X
0
1
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
2
3
4
5
6
7
8
9
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