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 z-score 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 – μz = ────

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 z-score 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.

)1,0(~ Nz),(~ NX

z-score: Standardized distribution

)1,0(~ Nz

Ex. 5.6 (p.146)

• N = 6• X = 0, 6, 5, 2, 3, 2μ = 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 z-scores, 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 gz = (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 gz = (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)