Statistics for the Behavioral Sciences (5th ed.)
Gravetter & Wallnau
Chapter 4
Variability
University of Guelph
Psychology 3320 — Dr. K. Hennig
Winter 2003 Term
1
Chapter in outline
1)
2)
3)
4)
Individual Differences in Attachment
Quality
Factors that Influence Attachment
Security
Fathers as Attachment Objects
Attachment and Later Development
2
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0
7
8 9 10 11 12 13 14 15
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*
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0
1
1
2
2
3 4
5
6
*
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*
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3 4
5
6
7
Honours-No
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Honours-Yes
8 9 10 11 12 13 143 15
Measures of Variability
Range
Interquartile Range
Sum of Squares
(Sample) Variance
(Sample) Standard Deviation
4
5
Standard deviation and samples
Goal of inferential stats is to generalize to
populations from samples
 Representativeness? But, samples tend to
be less variable (e.g., tall basketball
players) - thus a biased estimate of
variance
 Need to correct for the bias by making an
adjustment to derive a more accurate
estimate of the population variability
 Variance = mean squared deviation = sum
of squared deviations/number of scores 6

Calculating sd and variance: 3 steps
(M = 6.8 females)
X
X-M
(Step 1)
(X-M)2
(Step 2)
3
-3.8
14.4
4
-2.8
7.8
9
2.2
4.8
Step 3: SS =  (X-M)2
7
Calculating variance and sd (contd.)
Step 3: SS =  (X-M)2 - Definition formula
(sum of squared deviations)
Alternatively: SS = X2 - (X)2/n
-computational formula
Now correct for the bias with an
adjustment, sample variance =
s2 = SS/n - 1 (sample variance) and
( X  X )
2
s
 s
n 1
2
8
Thus… (text, p. 118)
Computational formula
X
(X)2
1
1
6
36
4
16
3
9
(35)
SS  211
7
8
64
 211 175  36
7
49
6
36
 X  35
 X  211
2
n7
9
2
Degrees of freedom - two points:
Population
= 4
SS = 17
1)
the sample SS ≤ population SS, always
–
2)

Sample of
n = 3 scores
[8, 3, 4]
M=5
SS = 14
the difference between the sample mean and
the population mean is the sampling error
you need to know the mean of the sample
to compute the SS; thus one variable is
dependent on the rest - df of a sample is
n-1 (i.e., the adjustment)
df (defn) - the number of independent
10
scores.
Note
Note. an average (mean) = sum/number
 thus, variance is the average deviation
from the mean

– mean squared deviation = sum of squared
deviations/
SS
SS 
N
– but to calculate sample variance:
SS
SS
s 

n  1 df
2
11
Biased and unbiased statistics
Table 4.1
Sample 1
Sample 2
Population
= 4
2=14
Sample 3
Sample 4
Sample 5
Sample 6

63/9 = 7 but 126/9= 14
Sample
Mean
s^2
(n)
s^2
(n-1)
1
0.0
0.0
0.0
2
1.5
2.25
4.5
3
4.5
20.25
40.5
4
1.5
2.25
4.5
…9
9.0
0
0.0
total
= 36
63
126
12
Transformation rules
1)
2)
Adding a constant to each score will not
change the sd
Multiplying each score by a constant
causes the standard deviation to be
multiplied by the same constant
13
Variance and inferential stats
(seeing patterns)
conclusion: the greater the variability the
more difficult it is to see a pattern
 variance in a sample is classified as error
variance (i.e., static noise)
14
 “one suit and lots of bad tailors”

Statistics for the Behavioral Sciences (5th ed.)
Gravetter & Wallnau
Chapter 5
z-Scores
University of Guelph
Psychology 3320 — Dr. K. Hennig
Winter 2003 Term
15
Intro to z-scores
Mean & sd as methods of describing entire
distribution of scores
 We shift to describing individual scores
within the distribution - uses the mean
and sd (as “location markers”)
 “Hang a left (sign is -) at the mean and go
down two standard deviations (number) ”
 2nd purpose for z-scores is to standardize
an entire distribution

16
z-scores and location in a distribution
•Every X has a z-score location
•In a population:
-2
-1
0
+1
 ---->
+2
17
The z-score formula
z
X 

A distribution of scores has a  =50 and a
standard deviation of  = 8
 if X = 58, then z = ___ ?

18
X to z-score transformation:
Standardization
80





90
100
110
-2
-1
0
z
1
2
X 
shape stays the same
in a z-score distribution is always 0
the standard deviation is always 1
procedure:
Bob got a 70% in Biology and a 60% in Chemistry - for
which should he receive a better grade?

19
Looking ahead to inferential statistics
Population
 = 400
 = 20
Treatment
Sample of
n =1
Treated
Sample
Is treated sample different from the original
population?
 Compute z-score of sample; e.g., if X is
extreme (z=2.5), then there is a difference

20
Statistics for the Behavioral Sciences (5th ed.)
Gravetter & Wallnau
Chapter 6
Probability
University of Guelph
Psychology 3320 — Dr. K. Hennig
Winter 2003 Term
21
Example
Jar = population of 3 checker,
1 red dotted, 3 yellow dotted,
3 tiled marbles
 if you know the population you know the
probability of picking a n =1 tiled sample

– 3/10 (almost a 30% chance)
but we don’t know the population (reality)
22
 inferential statistics works backwards

Population
Sample
23
Introduction to probability
probability of A = number of outcomes A/
total number of possible outcomes
 p(spade) = 13/52 = ¼ (or 25%)
 p (red Ace) = ?
 random sample:

– each individual in the population has an equal
chance (no selection bias)
– if sample > 1, then there must be constant
probability for each and every selection
 e.g., p(jack) if first draw was not a jack?
 sampling with replacement
24
*
** *
** *
*0 *1 *2 *3 *4
0&1
2,3,4
25
“God loves a normal curve”
2.28%
= 68
= 6
74
80
13.59%
34.13%
What is the
probability of picking
a 6’ 8” (80”) tall
person from the
population?
or p(X>80)
= 80-68/6 = +2.0
p(z>2,0) = ?
26
Unit normal table (Fig. 6.6)
(A)
z
.01
(B)
(C)
(D)
.504
.496
.004
.02
.508
.492
.008
B
C
D
27
Finding scores corresponding to
specific proportions or ps
X
z-score
unit
normal
table
proportions
or ps
28
Binomial distribution
probability of A (heads) = p(A)
 probability of B (tails) = p(B)
 p + q = 1.00

1st
toss
2nd
toss
0
0
1
1
0
1
0
1
0
1
1
2
p=
.50
.25
0
-With more tosses -> normal &
mean increases (M=3 with 6 tosses)
1
2
29
The normal approximation to the
binomial distribution
With increases in n the distribution
approaches a normal curve
 Given 10 tosses the expectation is to
obtain around 5 heads; unlikely to get
values far from 5
 Samples with n>10 (the criteria)
 Mean:  = pn
(e.g., p (heads given 2 tosses) = ½(2)=1
 standard deviation:  = npq

30
Example 6.4a (text)
A PSYC dept. is ¾ female. If a random
sample of 48 students is selected, what is
p(14 males)? (i.e., 12 males)
X   X  pn
 pn=¼(48)=12=
z

 qn=3/4(48)=36=

npq
 p(X= 14) = are under curve 13.5-14.5

14.5  12
13.5  12
z
 .50 z 
 .83
3
3
31
Example 6.14a (cond.)
14.5  12
13.5  12
z
 .50 z 
 .83
3
3
12 14 X values
.50 .83 z-scores
32
Looking ahead to inferential statistics
33