Transcript Chapter 5: z

Chapter 5: z-scores – Location of
Scores and Standardized
Distributions
z-Scores and Location
• By itself, a raw score or X value provides very
little information about how that particular score
compares with other values in the distribution.
– A score of X = 53, for example, may be a
relatively low score, or an average score, or an
extremely high score depending on the mean and
standard deviation for the distribution from which
the score was obtained.
– If the raw score is transformed into a z-score,
however, the value of the z-score tells exactly
where the score is located relative to all the other
scores in the distribution.
z-Scores and Location (cont'd.)
• The process of changing an X value into a zscore involves creating a signed number, called
a z-score, such that:
– The sign of the z-score (+ or –)
identifies
whether the X value is located above the
mean (positive) or below the mean (negative).
– The numerical value of the z-score
corresponds to the number of standard
deviations between X and the mean of the
distribution.
z-Scores and Location (cont'd.)
• Thus, a score that is located two standard
deviations above the mean will have a z-score of
+2.00. And, a z-score of +2.00 always indicates
a location above the mean by two standard
deviations.
Transforming Back and Forth Between
X and z
• The basic z-score definition is usually sufficient
to complete most z-score transformations.
However, the definition can be written in
mathematical notation to create a formula for
computing the z-score for any value of X.
X– μ
z = ────
σ
Transforming Back and Forth Between
X and z (cont'd.)
• Also, the terms in the formula can be regrouped
to create an equation for computing the value of
X corresponding to any specific z-score.
X = μ + zσ
Ex 5.2-5.3 - (p.143-144)
• μ = 100, σ = 10, X = 130  z = ?
z = (130 – 100) / 10 = 30/10 = +3
• μ = 86, σ = 7, X = 95  z = ?
z = (95 – 86) / 7 = 9/7 = 1.286
• μ = 60, σ = 5, z = -3  X = ?
X = μ + zσ = 60 + (-3)*5 = 60-15 = 45
Ex 5.4-5.5 (p. 144-145)
• μ = 65, X = 59, z = -2  σ = ?
X = μ + zσ  σ = (X- μ)/z = (59-65)/(-2) = 3
see Fig 5.4 (p.145)
• σ = 4, X = 33, z = +1.5  μ = ?
μ = X – zσ = 33 – 1.5*4 = 27
#3-#4 (p.145)
• #3) μ = 50, X = 42, z = -2  σ = ?
X = μ + zσ  σ = (X- μ)/z = (42-50)/(-2)
= (-8)/(-2) = 4
• #4) σ = 12, X = 56, z = -0.25  μ = ?
μ = X – zσ = 56 – (-0.25)* 12 = 56 + 3 = 59
z-scores and Locations
• In addition to knowing the basic definition of a zscore and the formula for a z-score, it is useful to
be able to visualize z-scores as locations in a
distribution.
• Remember, z = 0 is in the center (at the mean),
and the extreme tails correspond to z-scores of
approximately –2.00 on the left and +2.00 on the
right.
• Although more extreme z-score values are
possible, most of the distribution is contained
between z = –2.00 and z = +2.00. (95%)
z-scores and Locations (cont'd.)
• The fact that z-scores identify exact locations
within a distribution means that z-scores can be
used as descriptive statistics and as inferential
statistics.
– As descriptive statistics, z-scores describe
exactly where each individual is located.
– As inferential statistics, z-scores determine
whether a specific sample is representative of its
population, or is extreme and unrepresentative.
(i.e. estimation, hypothesis testing)
z-Scores as a Standardized
Distribution
• When an entire distribution of X values is
transformed into z-scores, the resulting
distribution of z-scores will always have a mean
of zero and a standard deviation of one.
• The transformation does not change the shape
of the original distribution and it does not change
the location of any individual score relative to
others in the distribution.
X ~ N ( , )
 z ~ N (  0,   1)
z-score: Standardized distribution
z ~ N (   0,   1)
Ex. 5.6 (p.146)
• N=6
• X = 0, 6, 5, 2, 3, 2
μ = ΣX/6 = 3, σ = 2
z = -1.5, 1.5, 1, -0.5, 0, -0.5
Σz = 0,
SS = Σz2 = 2.25+2.25+1+0.25+0+0.25 = 6
σ2 = SS/N = 6/6 = 1
σ = √1 = 1
Example: p. 149
• Dave’s test score:
psychology exam: X = 60 (μ=50, σ=10)
biology exam: X = 56 (μ=48, σ=4)
which is better?
• X  z and compare the z-scores
psychology exam: z = (60-50)/10 = 1
biology exam: z = (56-48)/4 = 2
• So his biology test is getting a better score!!
#3 (p. 150)
• English exam: μ=70, σ=4
• History exam: μ=60, σ=20
• Both got X = 78, which exam got higher grade?
(Ans)
• English: z = (78-70)/4 = 2  better
• History: z = (78-60)/20 = 0.9
• If History exam’s σ=10, then....
• If History exam’s σ=9, then....
• If History exam’s σ=8, then....
z-Scores as a Standardized
Distribution (cont'd.)
• The advantage of standardizing distributions is
that two (or more) different distributions can be
made the same.
– For example, one distribution has μ = 100 and
σ = 10, and another distribution has μ = 40 and σ
= 6.
– When these distribution are transformed to zscores, both will have μ = 0 and σ = 1.
z-Scores as a Standardized
Distribution (cont'd.)
• Because z-score distributions all have the same
mean and standard deviation, individual scores
from different distributions can be directly
compared.
• A z-score of +1.00 specifies the same location in
all z-score distributions.
z-Scores and Samples
• It is also possible to calculate z-scores for
samples.
• The definition of a z-score is the same for either
a sample or a population, and the formulas are
also the same except that the sample mean and
standard deviation are used in place of the
population mean and standard deviation.
z-Scores and Samples (cont'd.)
• Thus, for a score from a sample,
X–M
z = ─────
s
• Using z-scores to standardize a sample also has
the same effect as standardizing a population.
• Specifically, the mean of the z-scores will be
zero and the standard deviation of the z-scores
will be equal to 1.00 provided the standard
deviation is computed using the sample formula
(dividing n – 1 instead of n).
Ex. 5.8 (p.153)
• M = 40, s = 10, X = 35  z = ?
z = (35-40)/10 = (-5)/10 = -0.5
• M = 40, s = 10, z=2  X = ?
X = M + zs = 40 + 2*10 = 60
Ex 5.9 (p.154)
•
•
•
•
•
•
•
•
n = 5, X = 0, 2, 4, 4, 5  M = 3, s = 2
X = 0  z = (0-3)/2 = -1.5
X = 2  z = (2-3)/2 = - 0.5
X = 4  z = (4-3)/2 = 0.5
X = 5  z = (5-3)/2 = 1
Mz = 0  Σz = -1.5-0.5+0.5+0.5+1 = 0
SS = Σz2 = 2.25+0.25+0.25+0.25+1 = 4
sz 2 = SS/ (n-1) = 4 / (5-1) = 1
Other Standardized Distributions
Based on z-Scores
• Although transforming X values into z-scores
creates a standardized distribution, many people
find z-scores burdensome because they consist
of many decimal values and negative numbers.
• Therefore, it is often more convenient to
standardize a distribution into numerical values
that are simpler than z-scores.
Other Standardized Distributions
Based on z-Scores (cont'd.)
• To create a simpler standardized distribution,
you first select the mean and standard deviation
that you would like for the new distribution.
• Then, z-scores are used to identify each
individual's position in the original distribution
and to compute the individual's position in the
new distribution.
Ex 5.7 (p.151)
• μ=57, σ=14 (burdensome)  μ=50, σ=10
(simplified) for X = 64, X=43
• 1st : μ=57, σ=14
Maria: z = (64-57)/14 = 0.5
Joe: z = (43-57)/14 = -1
• next: z X’ with μ=50, σ=10
Maria: X’ = μ+zσ = 50+ 0.5*10 = 55
Joe: X’ = μ+zσ = 50- 1*10 = 40
• check Fig 5.8 (p. 152)
Other Standardized Distributions
Based on z-Scores (cont'd.)
• Suppose, for example, that you want to
standardize a distribution so that the new mean
is μ = 50 and the new standard deviation is σ =
10.
• An individual with z = –1.00 in the original
distribution would be assigned a score of X = 40
(below μ by one standard deviation) in the
standardized distribution.
• Repeating this process for each individual score
allows you to transform an entire distribution into
a new, standardized distribution.
Ex 5.10 (p.155-156)
• the effect of new growth hormone (injected)
• test on rats (weight)
• the distribution of weights is normal: μ=400 g,
σ=20 g
• select 1 rat and experiment: X = 418 g
z = (418-400)/20 = 0.9
∆ = 18 g is not significant enough to conclude
that injection of the new growth hormone has
any noticeable effect (i.e. ~ regular rats without
injection)
Ex 5.10 (p.155-156)
• μ=400 g, σ=20 g
• If X = 450 g
z = (450-400)/20 = 2.5
∆ = 50 g is large enough to conclude that
injection of the new growth hormone has
significant effect (i.e. gaining more weights than
regular rats without injection)