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One Tailed and Two Tailed tests
One tailed tests: Based on a uni-directional hypothesis
Example: Effect of training on problems using PowerPoint
Population figures for usability of PP are known
Hypothesis: Training will decrease number of problems
with PP
Two tailed tests: Based on a bi-directional hypothesis
Hypothesis: Training will change the number of problems
with PP
If we know the population mean
Sampling Distribution
Population for usability of Powerpoint
1400
Identify region
Unidirectional
hypothesis: .05 level
1200
Frequency
1000
800
Bidirectional
hypothesis: .05 level
600
400
Std. Dev = .45
200
Mean = 5.65
N = 10000.00
0
Mean Usability Index
• What does it mean if our significance level is
.05?
 For a uni-directional hypothesis
 For a bi-directional hypothesis
PowerPoint example:
• Unidirectional
 If we set significance level at .05 level,
• 5% of the time we will higher mean by chance
• 95% of the time the higher mean mean will be real
• Bidirectional
 If we set significance level at .05 level
• 2.5 % of the time we will find higher mean by chance
• 2.5% of the time we will find lower mean by chance
• 95% of time difference will be real
Changing significance levels
•What happens if we decrease our significance
level from .01 to .05
Probability of finding differences that don’t exist
goes up (criteria becomes more lenient)
•What happens if we increase our significance
from .01 to .001
Probability of not finding differences that exist goes
up (criteria becomes more conservative)
• PowerPoint example:
 If we set significance level at .05 level,
• 5% of the time we will find a difference by chance
• 95% of the time the difference will be real
 If we set significance level at .01 level
• 1% of the time we will find a difference by chance
• 99% of time difference will be real
• For usability, if you are set out to find
problems: setting lenient criteria might work
better (you will identify more problems)
• Effect of decreasing significance level from .01
to .05
 Probability of finding differences that don’t exist goes
up (criteria becomes more lenient)
 Also called Type I error (Alpha)
• Effect of increasing significance from .01 to .001
 Probability of not finding differences that exist goes up
(criteria becomes more conservative)
 Also called Type II error (Beta)
Degree of Freedom
• The number of independent pieces of information
remaining after estimating one or more
parameters
• Example: List= 1, 2, 3, 4
Average= 2.5
• For average to remain the same three of the
numbers can be anything you want, fourth is
fixed
• New List = 1, 5, 2.5, __ Average = 2.5
Major Points
• T tests: are differences significant?
• One sample t tests, comparing one mean to
population
• Within subjects test: Comparing mean in
condition 1 to mean in condition 2
• Between Subjects test: Comparing mean in
condition 1 to mean in condition 2
Effect of training on Powerpoint use
• Does training lead to lesser problems
with PP?
• 9 subjects were trained on the use of PP.
• Then designed a presentation with PP.
 No of problems they had was DV
Powerpoint study data
21
• Mean = 23.89
24
21
• SD = 4.20
26
32
27
21
25
18
Mean
SD
23.89
4.20
Results of Powerpoint study.
• Results
 Mean number of problems = 23.89
• Assume we know that without training
the mean would be 30, but not the
standard deviation
Population mean = 30
• Is 23.89 enough larger than 30 to
conclude that video affected results?
Sampling Distribution of the
Mean
• We need to know what kinds of sample
means to expect if training has no effect.
 i. e. What kinds of means if m = 23.89
 This is the sampling distribution of the mean.
Sampling Distribution of the
Mean--cont.
• The sampling distribution of the mean
depends on
 Mean of sampled population
 St. dev. of sampled population
 Size of sample
Sampling Distribution
Number of problems with Powerpoint Use
1400
1200
Frequency
1000
800
600
400
Std. Dev = .45
200
Mean = 5.65
N = 10000.00
0
Mean Number of problems
Cont.
Sampling Distribution of the
mean--cont.
• Shape of the sampled population
 Approaches normal
 Rate of approach depends on sample size
 Also depends on the shape of the population
distribution
Implications of the Central
Limit Theorem
• Given a population with mean = m and
standard deviation = s, the sampling
distribution of the mean (the distribution
of sample means) has a mean = m, and a
standard deviation = s /n.
• The distribution approaches normal as n,
the sample size, increases.
Demonstration
• Let population be very skewed
• Draw samples of 3 and calculate means
• Draw samples of 10 and calculate means
• Plot means
• Note changes in means, standard
deviations, and shapes
Cont.
Parent Population
Skewed Population
3000
Frequency
2000
1000
Std. Dev = 2.43
Mean = 3.0
N = 10000.00
0
.0
20
.0
18
.0
16
.0
14
.0
12
.0
10
0
8.
0
6.
0
4.
0
2.
0
0.
X
Cont.
Sampling Distribution n = 3
Sampling Distribution
Sample size = n = 3
Frequency
2000
1000
Std. Dev = 1.40
Mean = 2.99
N = 10000.00
0
0
.0
13 0
.0
12 0
.0
11 0
.0
10
00
9.
00
8.
00
7.
00
6.
00
5.
00
4.
00
3.
00
2.
00
1.
00
0.
Sample Mean
Cont.
Sampling Distribution n = 10
Sampling Distribution
Sample size = n = 10
1600
1400
Frequency
1200
1000
800
600
400
Std. Dev = .77
200
Mean = 2.99
N = 10000.00
0
50
6.
00
6.
50
5.
00
5.
50
4.
00
4.
50
3.
00
3.
50
2.
00
2.
50
1.
00
1.
Sample Mean
Cont.
Demonstration--cont.
• Means have stayed at 3.00 throughout-except for minor sampling error
• Standard deviations have decreased
appropriately
• Shapes have become more normal--see
superimposed normal distribution for
reference
One sample t test cont.
• Assume mean of population known, but
standard deviation (SD) not known
• Substitute sample SD for population SD
(standard error)
• Gives you the t statistics
• Compare t to tabled values which show critical
values of t
t Test for One Mean
• Get mean difference between sample and
population mean
• Use sample SD as variance metric = 4.40
X  m 30  23.89 6.11
t


 1.48
s
4.40
1.46
n
9
Degrees of Freedom
• Skewness of sampling distribution of
variance decreases as n increases
• t will differ from z less as sample size
increases
• Therefore need to adjust t accordingly
• df = n - 1
• t based on df
Looking up critical t (Table
E.6)
Two-Tailed Significance Level
df
4
5
6
7
8
9
.10
1.812
1.753
1.725
1.708
1.697
1.660
.05
2.228
2.131
2.086
2.060
2.042
1.984
.02
2.764
2.602
2.528
2.485
2.457
2.364
.01
3.169
2.947
2.845
2.787
2.750
2.626
Conclusions
• Critical t= n = 9, t.05 = 2.62 (two tail
significance)
• If t > 2.62, reject H0
• Conclude that training leads to less
problems
Factors Affecting t
• Difference between sample and
population means
• Magnitude of sample variance
• Sample size
Factors Affecting Decision
• Significance level a
• One-tailed versus two-tailed test