Round 0.0475 to three decimal places

The Standard NormalDistribution and z Scores

Flock of birds flying around Jama Masjid Mosque

Keren Su/Corbis

Chapter Learning Objectives

After reading this chapter, you should be able to do the following:

1. Identify the characteristics of the standard normal distribution.

2. Demonstrate the use of the z transformation.

3. Determine the percent of a population above a point, below a point, andbetween two points on the horizontal axis of a normal distribution.

4. Calculate z scores using Excel.

5. Describe alternative standard scores.

6. Demonstrate the use of the modified standard score.

Introduction

The data that describe characteristics of groups come from either samples or populations,explained in the first two chapters. By way of reminder, recall that populations include allpossible members of any specified group. All university students, all psychology majors, allresidents of Orange County, and all left-handed male tennis players in their 20s are eachdescriptions of a population. We rely on Greek letters, such as µ for the mean and σ for thestandard deviation, to distinguish population parameters from the statistics that describesamples. (The word parameter indicates a characteristic of a population.) Remove one ormore individuals from any population, and the resulting group is a sample.

As we were describing populations, we noted that some are “normally distributed.” Thesecharacteristics indicate normality: (a) data distributions are symmetrical, (b) all the measuresof central tendency have very similar values, and (c) the value of the standard deviation isabout one-sixth of the range.

Data normality does not simply mean that the frequency distribution will appear as a bell-shaped curve; it means that predictable proportions of the entire population will occur inspecified regions of the distribution, and this holds for all normal data distributions. Forexample, the region under a normal curve from the mean of the population to one standarddeviation below the mean always includes 34.13% of the area under the curve. Because normal distributions are symmetrical, from the mean to one standard deviation above themean also includes 34.13%, so from +1σ or −1σ includes about 68.26% of the area under thecurve in any normally distributed population. As long as the data are normally distributed,those percentages hold true. Since many mental characteristics are normally distributed,researchers can know a good deal about such a characteristic without actually gathering thedata and doing the analysis. Whether the characteristic is intelligence, achievementmotivation, anxiety, or any other normally distributed characteristics, the proportion of thedistribution within +1 or −1 standard deviation from the mean will be the same:

· If a particular intelligence scale has µ = 100 and σ = 15, about 68% of any generalpopulation will have intelligence scores between 85 and 115.

· Likewise, if an achievement motivation scale has µ = 40 and σ = 8, about 2/3 of anypopulation will have achievement motivation scores from 32 to 48.

· And for an anxiety measure with µ = 25 and σ= 5, about 68% of any generalpopulation will have scores between 20 and 30.

The consistency in the way so many characteristics are distributed affords a good deal ofinterpretive power. Anyone who needs information about the likelihood of individualsscoring in certain areas of a distribution has an advantage when data are normally distributed.In addition to the 68% of any general population likely to score between +1σ and −1σ,

· from µ to +2σ is about 47.72% of the population, so about 95% (2 × 47.72) of thepeople in any general population will have intelligence scores between 70 (100 − 30)and 130 (100 + 30).

· from +3σ (49.87%) to −3σ includes nearly everyone in any normally distributedpopulation (2 × 49.87 = 99.74).

These observations emphasize that, sometimes, isolated bits of data can be quite informative.When a 12-year-old with an intelligence score of 170 pops up on YouTube, it is immediatelyapparent that this is a very unusual child. An intelligence score of that magnitude is about4.667σ (170 − 100 = 70; 70 ÷ 15 = 4.667) beyond the mean of the general population. If from+3σ to −3σ includes more than 99% of the population, from +4.667σ to −4.667σ mustinclude all but the utmost extreme scores. We obtain an even better context for how common(or uncommon) particular measures may be when we can determine the precise probability oftheir occurrence.

This is a self-assessment and will not affect your grade. You may only take this pre-test once.

Test Ch 3: The Standard Normal Distribution and z Scores

Top of Form

1. The z transformation changes any raw score into a z score so that it fits the standard normal distribution.

· a. TRUE

· b. FALSE

2. Calculating z scores doesn’t alter the distribution; it just makes them fit a distribution where the mean is 0 and the standard deviation is 1.0.

· a. FALSE

· b. TRUE

3. We can apply the z transformation to sample data even when there is reason to believe that the population from which the sample was drawn is not normally distributed.

· a. TRUE

· b. FALSE

4. Raw scores can be determined if the mean and standard deviation are available.

· a. FALSE

· b. TRUE

5. Individual scores in the standard normal distribution are called z scores.

· a. TRUE

· b. FALSE

Finish

Bottom of Form

3.1 A Primer in Probability

Scholars, data analysts, and in fact people on the whole are rarely interested in outcomes thatoccur every time. If everyone had an intelligence score of 170, no one would pay anyattention to someone with such a score. The fact that we know it to be uncommon is whatpiques our curiosity.

If we are not interested in events that always occur, neither do we closely follow events thatnever occur. If no one had ever had an intelligence score of 170, probably no one wouldwonder about what such a score means for the person who has it. The things that occur someof the time, however, intrigue us. The “some of the time” indicates that the event has some probability, or likelihood, of occurrence.

· What is the probability that those newlyweds will divorce?

· How likely is Germany to win the World Cup?

· What is the probability that an earthquake will occur on a particular day for someonewho lives near the San Andreas Fault?

· What is the probability of an IRS audit for one taxpayer?

Major-league baseball game

Joseph Sohm/Visions of America/Corbis

The probability that something willoccur, such as how likely it is thatour favorite baseball team will winthe World Series, intrigues us and isan important component in thedecision-making process.

Because all of the items listed have happened in thepast and because their occurrence is important to atleast someone, people are interested in theprobability of those occurrences whether or notthey use the language of probability. When statednumerically, probability values range from 0 to 1.0.Something with a probability of zero (p = 0) neveroccurs. On the other hand, p = 1.0 indicates that theevent occurs every time, and p = 0.5 indicates thatthe event occurs 50% of the time.

As that last point indicates, percentages can beconverted to probability values. Dividing the percentage of times an event occurs by 100indicates the associated probability of the event.

Returning to the intelligence scores, we see thatbecause about 68% of the population has intelligence scores between 85 and 115, theprobability (p) that someone selected at random from the general population will have a scoresomewhere between 85 and 115 is 0.68 (68.26/100, if the result is rounded to two decimalplaces).

What is the probability that someone selected at random from the general population willhave an intelligence score of 100 or lower? Because 100 is the mean for intelligence scores,and because 50% of the population occur at the mean or below, p = 0.5.

What is the probability that someone selected at random will have an intelligence scorehigher than 115? First, we noted earlier that 34.13% of the population falls between themean, µ, and one standard deviation above the mean at σ = +1.0 in any normally distributedpopulation. In terms of intelligence score values, that is the region between scores of 100 and115. Since 50% of any normally distributed population will occur at the mean and above, ifwe subtract from 50% that portion between the mean and one standard deviation above themean, the remainder will be the portion of the distribution above 115: 50% − 34.13% =15.87%; that is, 15.87% of all intelligence scores in a normally distributed population willoccur above 115. Dividing by 100 (15.87/100 = 0.1587) and rounding the result to twodecimal places produces the probability p = 0.16.

By the same logic, because a score of 85 is one standard deviation below the mean, theprobability p = 0.16 means that someone selected at random from the population will scorebelow 85. If we combine the two outcomes, the probability is p = 0.32 that someone from thepopulation will score either below 85 or above 115.

Consider the number line shown in Figure 3.1.

Figure 3.1: Standard deviations for intelligence scores

The number line shows the portion of scores that fall within two standarddeviations above and below the mean. If M = 100, we can know the probabilityof someone scoring below 85 or above 115.

A straight line with intervals of 70, 85, 100, 115, and 130, situated below the statistical symbols −2σ, −1σ, M, +1σ, +2σ representing two standard deviations above and below the mean (M) and one standard deviation above and below the mean. Above the straight line are four curved lines stretching between 70 and 85, (−2σ; 16%), 85 and 100 (−1σ; 34%), 100 and 115 (+1σ; 34%), and 115 and 130 (+2σ; 16%).

If this number line represents all intelligence scores ranging from two standard deviationsbelow to two standard deviations above the mean, we can see the percentages of thepopulation that will have scores in the designated areas. Using the percentages and dividingby 100 indicates the probability of a score in any of the designated areas.

Recall that the lowest probability for any value is zero (p = 0). If p = 0, then the event oroutcome never occurs. There is no such thing as a negative probability.

3.2 The Standard Normal Distribution
Not all populations are normally distributed. Home sales are usually reported in terms of the medianprice of a home, and salary data are likewise reported as median values. Those cases use the mediansbecause the related populations are very unlikely to be normally distributed and, as a measure ofcentral tendency, medians are less affected by extreme values than are means. A few very high salariesor home values create positive skew in the resulting distribution. In contrast, when it comes to, say,mental characteristics such as intelligence, achievement motivation, problem-solving ability, verbalaptitude, reading comprehension, and so on, population data are often normally distributed.

Granddaughter and grandfather playing chess

Hello Lovely/Corbis

When evaluating information aboutpeople’s characteristics, keep in mindthat data are often normally distributed.

Although there are many normal distributions all havingthe same proportions, each has different descriptivevalues. An intelligence test might have µ = 100 and σ =15 points. A nationally administered reading test mighthave a mean of 60 and a standard deviation of 8. Thesedifferent parameters can make it difficult to compare oneindividual’s performance across multiple measures. Asone author noted regarding scores from the WechslerIntelligence Test for Children (WISC), “A raw score of 5on one [sub]test will not have the same meaning as a rawscore 5 on another [sub]test” (Brock, 2010).

One way to resolve this interpretation problem is toconvert the scores from different distributions into acommon metric, or measurement system. If researchersalter scores from different distributions so that they both fit the same distribution, they can comparescores directly. A researcher can compare them directly to determine, for example, on which test anindividual scored highest. Such comparisons are one of the purposes of the standard normaldistribution.

The standard normal distribution looks like all other normal distributions—from the mean to +1standard deviation includes 34.13% of the distribution, for example. What separates it from the othersis that in the standard normal distribution, the mean is always 0, and the standard deviation is always1.0 (Figure 3.2). Other distributions may have fixed values for their means and standard deviations, buthere µ is always 0 and σ is always 1.0.

Figure 3.2: The standard normal distribution
In the standard normal distribution, the mean is always 0, and the standard deviation isalways 1.0.

Depiction of a bell curve showing the standard normal distribution, with a line bisecting the mean (0), the highest point in the curve. The horizontal axis is numbered from -3 to +3. The standard deviation is 1.0.

The Standard Normal, or z, Distribution
Although various normal distributions have different means and standard deviations, they all mirroreach other in terms of how much of their populations occur in particular regions. The standard normaldistribution’s advantage is that the proportions of the whole that occur in the various regions of thedistribution have been calculated. That means that if data from any normal distribution are made toconform to the standard normal distribution, we can answer questions about what is likely to occur invirtually any area of the distribution, such as how likely it is to score 2.5 standard deviations below themean on a particular test, or what percentage of the entire population will likely occur between twospecified points. All such questions can be answered when adapting normal data to the characteristicsof the standard normal distribution.

Individual scores in the standard normal distribution are called z scores, which is why the standardnormal distribution is often called “the z distribution.” The formula used to turn scores from anynormal distribution into scores that conform to the standard normal distribution is the ztransformation:

Formula 3.1

z=x−Ms

where z is a score in the standard normal distribution, x is the score from the original distribution(often called a “raw” score), M is the mean of the scores before the original distribution, and s is thestandard deviation of the scores from the original distribution.

Because normality is characteristic of only very large groups, samples will rarely be normal. However,we can apply the z transformation to sample data when there is reason to believe that the populationfrom which the sample was drawn is normally distributed. This is what Formula 3 reflects. The M and s indicate that the data involved are sample data. In those situations where an analyst has access topopulation data—a social worker has all the data for those served by Head Start in a particular county,for example—µ replaces M and σ replaces s in the formula. With either sample or population data, thetransformation is from data that can have any mean and standard deviation to a distribution where themean will always equal 0 and the standard deviation will always equal 1.0.

To turn raw scores into z scores, perform the following steps:

1. Determine the mean and standard deviation for the data set.

2. Subtract the mean of the data set from each score to be transformed.

3. Divide the difference by the standard deviation of the data set.

For example, consider a psychologist interested in the level of apathy among potential voters regardingmental health issues that affect the community. Scores on the S ummary o f WH o’s A pathetic T est (theSoWHAT for short), an apathy measure, are gathered for 10 registered voters:

5, 6, 9, 11, 15, 15, 17, 20, 22, 25

What’s the z score for someone who has an apathy score of 11?

· Verify that for these 10 scores, M = 14.5 and s = 6.737.

· The z score equivalent for an apathy score of 11 is

z=x−Ms=11−14.56.737=−0.5195

An apathy score of 11 translates into a z score of −0.5195. Because the mean of the z distribution is 0and the standard deviation in the z distribution is 1.0, where would a score of −0.5195 occur on thehorizontal axis of the data distribution? It would be a little over half a standard deviation below themean, right? Figure 3.3 shows the z distribution and the point about where a raw score of 11 occurs inthis distribution once it is transformed into a z score.

It is important to know that the z transformation does not make data normal. Calculating z scores doesnot alter the distribution; it just makes them fit a distribution where the mean is 0 and the standarddeviation is 1.0. Evaluating skew and kurtosis must allow the analyst to assume that the data arenormal before using the z transformation.

With a mean of 0 in the standard normal distribution, half of all z scores—all the scores below themean—are going to be negative. A raw score of 11 from the SoWHAT data is lower than the mean,which was M = 14.5, so it has a negative z value (−0.5195).

Try It!: #1
How many standard deviations from themean of the distribution is a z score of 1.5?

Besides indicating by its sign whether the zscore is above or below the mean, the value ofthe z score indicates how far from the mean the zscore is in standard deviations. If a score had a zvalue of 1.0, it would indicate that the score isone standard deviation above the mean. The zscore for the raw score of 11 was −0.5195,indicating that it is just over half a standarddeviation below the mean. This ease of interpretation is one of the great values of z scores: the sign ofthe score indicates whether the associated raw score was above or below the mean, and the value ofthe score indicates how far from the mean the raw score falls, in standard deviation units (Fischer andMilfont, 2010).

Figure 3.3: Location of a score on the z distribution
Half of all z scores will fall below the mean, resulting in a negative value. A score of z= −0.5195 is slightly less than one-half a standard deviation below the mean.

A normal distribution showing the location of z scores along the horizontal axis ranging from −3z standard deviations below the mean (0) to +3z deviations above the mean. A score of z =−0.5195 is located about halfway between −1z and 0.

Comparing Scores from Different Instruments
Consider another application of the standard normal distribution. A counselor has intelligence andreading scores for the same person and wishes to know on which measure the individual scored higher.Table 3.1 shows the data for the two tests. On the intelligence test, the individual scored 105, and onthe reading test, the individual scored 62.

Table 3.1: Reading and intelligence test results
Test

Mean

Standard deviation

Intelligence

100

15

Reading

60

8

If the counselor transforms both scores to make them fit the standard normal distribution, they can becompared directly.

The z for the intelligence score is

z=x−Ms=105−10015=0.333

The z for the reading test score is

z=x−Ms=62−608=0.250

The intelligence score of 105 and the reading score of 62 are difficult to compare because they belongto different distributions with different means and standard deviations. When both are transformed tofit the standard normal distribution, an analyst can directly compare scores. The larger z value forintelligence makes it clear that individual scored higher in intelligence than in reading.

Expanding the Use of the z Distribution
Because the standard normal distribution is a normal distribution, we know that predictableproportions of its population will occur in specific areas. As we noted earlier, however, thoseproportions are known in great detail for the z distribution because this population is so often used toanswer detailed questions about the likelihood of particular outcomes. Table 3.2 indicates how muchof the entire population is above or below all of the most commonly occurring values of z. So, bytransforming scores from other distributions to fit the z distribution, we can use what we know aboutthis population to answer questions about scores from any normal distribution.

Not all tables for z values are alike. Probably as a matter of the developer’s preference, some tablesindicate the percentage of the population below a point. Some indicate the percentage between a pointand the mean of the distribution. Some indicate the probability of scoring in a particular area, and soon. This particular table indicates the proportion of the population between the specified value of z andthe mean of the distribution. (Table 3.2 is listed as Table B.1 in Appendix B.)

Table 3.2: The z table
0.00

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.0

0.0000

0.0040

0.0080

0.0120

0.0160

0.0199

0.0239

0.0279

0.0319

0.0359

0.1

0.0398

0.0438

0.0478

0.0517

0.0557

0.0596

0.0636

0.0675

0.0714

0.0753

0.2

0.0793

0.0832

0.0871

0.0910

0.0948

0.0987

0.1026

0.1064

0.1103

0.1141

0.3

0.1179

0.1217

0.1255

0.1293

0.1331

0.1368

0.1406

0.1443

0.1480

0.1517

0.4

0.1554

0.1591

0.1628

0.1664

0.1700

0.1736

0.1772

0.1808

0.1844

0.1879

0.5

0.1915

0.1950

0.1985

0.2019

0.2054

0.2088

0.2123

0.2157

0.2190

0.2224

0.6

0.2257

0.2291

0.2324

0.2357

0.2389

0.2422

0.2454

0.2486

0.2517

0.2549

0.7

0.2580

0.2611

0.2642

0.2673

0.2704

0.2734

0.2764

0.2794

0.2823

0.2852

0.8

0.2881

0.2910

0.2939

0.2967

0.2995

0.3023

0.3051

0.3078

0.3106

0.3133

0.9

0.3159

0.3186

0.3212

0.3238

0.3264

0.3289

0.3315

0.3340

0.3365

0.3389

1.0

0.3413

0.3438

0.3461

0.3485

0.3508

0.3531

0.3554

0.3577

0.3599

0.3621

1.1

0.3643

0.3665

0.3686

0.3708

0.3729

0.3749

0.3770

0.3790

0.3810

0.3830

1.2

0.3849

0.3869

0.3888

0.3907

0.3925

0.3944

0.3962

0.3980

0.3997

0.4015

1.3

0.4032

0.4049

0.4066

0.4082

0.4099

0.4115

0.4131

0.4147

0.4162

0.4177

1.4

0.4192

0.4207

0.4222

0.4236

0.4251

0.4265

0.4279

0.4292

0.4306

0.4319

1.5

0.4332

0.4345

0.4357

0.4370

0.4382

0.4394

0.4406

0.4418

0.4429

0.4441

1.6

0.4452

0.4463

0.4474

0.4484

0.4495

0.4505

0.4515

0.4525

0.4535

0.4545

1.7

0.4554

0.4564

0.4573

0.4582

0.4591

0.4599

0.4608

0.4616

0.4625

0.4633

1.8

0.4641

0.4649

0.4656

0.4664

0.4671

0.4678

0.4686

0.4693

0.4699

0.4706

1.9

0.4713

0.4719

0.4726

0.4732

0.4738

0.4744

0.4750

0.4756

0.4761

0.4767

2.0

0.4772

0.4778

0.4783

0.4788

0.4793

0.4798

0.4803

0.4808

0.4812

0.4817

2.1

0.4821

0.4826

0.4830

0.4834

0.4838

0.4842

0.4846

0.4850

0.4854

0.4857

2.2

0.4861

0.4864

0.4868

0.4871

0.4875

0.4878

0.4881

0.4884

0.4887

0.4890

2.3

0.4893

0.4896

0.4898

0.4901

0.4904

0.4906

0.4909

0.4911

0.4913

0.4916

2.4

0.4918

0.4920

0.4922

0.4925

0.4927

0.4929

0.4931

0.4932

0.4934

0.4936

2.5

0.4938

0.4940

0.4941

0.4943

0.4945

0.4946

0.4948

0.4949

0.4951

0.4952

2.6

0.4953

0.4955

0.4956

0.4957

0.4959

0.4960

0.4961

0.4962

0.4963

0.4964

2.7

0.4965

0.4966

0.4967

0.4968

0.4969

0.4970

0.4971

0.4972

0.4973

0.4974

2.8

0.4974

0.4975

0.4976

0.4977

0.4977

0.4978

0.4979

0.4979

0.4980

0.4981

2.9

0.4981

0.4982

0.4982

0.4983

0.4984

0.4984

0.4985

0.4985

0.4986

0.4986

3.0

0.4987

0.4987

0.4987

0.4988

0.4988

0.4989

0.4989

0.4989

0.4990

0.4990

Source: StatSoft. (2011). Electronic Statistics Textbook. Tulsa, OK: StatSoft. Retrieved from http://www.statsoft.com/textbook/distribution-tables/#z

The z value calculation for a SoWHAT score of 11 rounded to 4 decimal values for the sake of theillustration. The table rounds z values to just two decimals, so from this point forward, round z valuesto two decimals when using the table. Rounding makes the z value for a raw score of 11 = −0.52.

To interpret the z score, read the whole numbers and the tenths (the tenths are the first value to theright of the decimal) vertically down the left margin of the table. For the hundredths (the second valueto the right of the decimal), move from left to right across the columns at the top of the table.

1. Read down the left margin to the line indicating 0.5.

2. Read across the top to the column indicating 0.02.

3. The table value where row and column intersect is 0.1985. This value is the proportion (out ofa total of 1.0) of any normally distributed population that will occur between z = 0.52 and thepopulation’s mean.

4. To determine the percentage of the distribution between z = −0.52 and the mean, multiply thetable value by 100: 100 × 0.1985 = 19.85% of the distribution is between −0.52 and thepopulation mean.

Note that all the z values in Table 3.2 are positive. Our z score from the SoWHAT score was actuallynegative (z = −0.52). Since the mean of the standard normal distribution is z = 0, the z value for anyscore below the mean will be negative. However, the negative values pose no problem because allnormal distributions are symmetrical, so the proportion of a normal population between z = −0.52 andthe mean will be the same as that between z = 0.52 and the mean. We simply look up the proportionfor the appropriate value of z, remembering that when z is negative, it is a proportion to the left of themean rather than to the right.

Try It!: #2
Table 3.2 has table values only for positive z scores. How do we interpret the valuewhen z turns out to be negative?

To state all this as a principle, because normaldistributions are symmetrical, z scores with thesame absolute value (the same numbers withoutregard to the sign) include the same proportionsbetween their values and the mean of thedistribution. For this reason, the z table indicatesonly the proportions for half the distribution. Inthe case of Table 3.2, that half is the positive(right) half of the distribution.

Because 50% of the distribution occurs eitherside of the mean, if 19.85% of the distribution is from a z = −0.52 back to the mean, the balance of theleft (negative) half of the distribution must occur below a z score of −0.52. That proportion is 50 −19.85 = 30.15%, as the number line illustrates:

https://media.thuze.com/MediaService/MediaService.svc/constellation/book/AUPSY325.16.1/%7bimages%7dch03_01.jpg

Working in the other direction: if the question is what percentage of the population will score 11 orlower on the SoWHAT, the answer is 50 − 19.85 = 30.15%.

If instead someone asks what the probability of scoring at or below 11 (30.15%) is, we must turn thepercentage back into a probability: 30.15 / 100 = 0.3015, or p = 0.3015 of scoring at or below 11.

Try It!: #3
What is the largest possible value for z?

Note that the language above is “11 or lower,”and “at or below.” The characteristics of thenormal curve allow us to determine thepercentage between points, but not at a discretepoint. Technically, a particular point has nowidth and so no associated percentage.

Converting z Scores to Percentage
Now that we have learned how to transform scores from other distributions to fit the z distribution, wewill take a further look at how we can convert scores on opposite sides of the mean and scores with thesame sign to percentages.

Two Scores on Opposite Sides of the Mean
If 5 and 25 are the most extreme apathy scores gathered in the sample of SoWHAT scores, we mightask what percentage of the entire distribution will score between 5 and 25. Because those were thelowest and highest scores, the answer should be 100%, correct? Remember that the collected data werea sample:

5, 6, 9, 11, 15, 15, 17, 20, 22, 25

Although everyone in the sample scored between 5 and 25, it is entirely possible, even probable, thatsomeone in the larger population will have a more extreme score. Using the z distribution, we candetermine how probable by following these steps:

1. Convert both 5 and 25 into z scores.

2. Determine the table values for both z scores.

3. Turn the table values into percentages.

4. Add the percentages together.

The z score formula is

z=x−Ms

Allowing that the subscript to each z indicates the raw score and that M = 14.5 and s = 6.737 from thesample data produces the following calculations:

z5=5−14.56.737=−1.410,

for which the table value is 0.4207,

which corresponds to a percentage of 42.07% (0.4207 × 100).

z25=25−14.56.737=1.559=1.56,

which has a table value of 0.4406.

Expressed as a percentage, the value is 44.06% (0.4406 × 100).

Adding the two percentages together to determine the total percentage between them produces thefollowing:

42.07 + 44.06 = 86.13% from 5 to 25.

Clearly, these scores do not equal 100%. The results indicate that in the population for which thesedata are a sample, about 13.87% (100 − 86.13) will score either lower than 5 or higher than 25. Figure3.4 indicates this result.

Figure 3.4: Areas under the normal curve below z = −1.41 andbeyond z = 1.56
In this distribution, z values that fall below −1.41 or above +1.56 (raw scores below 5 orabove 25) are considered extreme scores, comprising only about 13.87% of thepopulation.

A normal curve, with z value intervals ranging from −3z to +3z indicated along the horizontal axis. A lighter-shaded region to either size of the mean (M = 0) shows where the majority of scores will fall (42% below the mean; z = −1.41; and 44% above the mean, z = +1.56). Darker regions to the left of −1.41 and to the right of +1.56 indicate extreme scores.

The answer to this problem underscores two important concepts. First, remember that we are dealingwith sample data, and the sample will never exactly duplicate a population. The second, more subtlepoint reveals that there is no point at which we can be confident that no one will produce a moreextreme score. The curve represents this fact by extending the tails (the endpoints of the curve)outward in either direction along the horizontal axis. Although the gap between tail and axis narrowsconstantly, the tails never touch the axis (the 50-cent word is that the tails are “asymptotic” to thehorizontal axis). The application means a value of z will never account for 100% of the distribution.

z Scores with the Same Sign
The previous example raised the question about the percentage of the distribution between z scores onopposite sides of the mean—two z scores where one was positive (z = 1.56) and the other negative (z =−1.41). Perhaps the researcher has a question about the percentage of the distribution betweenSoWHAT scores of 15 and 20. When M = 14.5, both of these raw scores are higher than the mean andboth will result in positive z values. When two z scores have the same sign, determining the percentageof the distribution between them requires that we complete the following steps:

1. Calculate z scores for the raw scores.

2. Determine the table values for each z.

3. Subtract the smaller proportion from the larger.

4. Convert the result into a percentage by multiplying by 100.

z=x−Ms

z15=15−14.56.737=0.0742,

or 0.07, for which the table value is 0.0279.

The 0.0279 is the proportion of the distribution from z = 0.07 and the mean of the distribution. For araw score of 20,

z20=20−14.56.737=0.8164,

or 0.82, which corresponds to p = 0.2939.

This is the proportion of the distribution between z = 0.82 and the mean of the distribution.

When the z scores are on opposite sides of the mean, as they were in our first example, determining theproportion of the distribution between them was a simple matter of adding the two table values. Whenboth z scores are on the same side of the distribution, however, their table values overlap. To determinethe proportion between two values of z with the same sign, take the proportion between the larger(absolute) value and the mean minus the proportion from the smaller (absolute) value to the mean:0.2939 − 0.0279 = 0.2660. Multiplying that by 100 produces the percentage: 100 × 0.2660 = 26.6% ofthe distribution will score between 15 and 20.

Figure 3.5 illustrates this result.

Figure 3.5: Areas under the curve between z = 0.07 and z = 0.82
The percentage of scores between two z values with the same sign is determined bycalculating the difference between the smaller z score table value and the larger one,then multiplying the result by 100.

A normal curve with z value intervals ranging from -3z to +3z indicated along the horizontal axis. Two positive z values of 0.07 and 0.82 are situated in their respective places below the curve, with the figure 26.6% shown below the two values in the center.

When trying to answer a question about the percentage of the distribution in a particular area, drawinga simple diagram like Figure 3.5 helps make the question less abstract.

Apply It! Attention to Detail
Group of people working on a circuit board

gerenme/iStock/Thinkstock

A psychological services company administers a test thatmeasures the respondent’s attention to detail. The company’sclients are employers in a variety of organizations that requirepeople with good analytical skills. Respondents who score inthe lowest ranges of the scale are indifferent to potentiallyimportant details. Those who score in the highest ranges tendto fixate on details that may be unimportant to an outcome.Individuals who meet the qualification on this particular testscore in the range from 3.80 to 4.30. Data for those who havetaken the test in the past indicate that M = 4.00 and s = 0.120.For researchers, the initial question is, “Of those who take thetest, what proportion are rejected because either they areinattentive to important details or they become focused on thewrong details?” In terms of the z distribution, the equivalentquestions are the following:

a. What proportion of those who took the test in the past failed to meet the minimumqualification for attention to relevant detail? In other words, what proportion scoredlower than 3.80?

b. What proportion of test-takers scored higher than 4.30?

Regarding question (a), to determine the value of z, the following apply:

x = 3.80

M = 4.00

s = 0.12

Since

z=x−Ms=(3.80−4.00)/0.120

z3.80 = −1.67

The z score table (Table 3.2) indicates that a proportion of 0.4525 of the entire population willfall between this z score and the mean of the distribution. However, the researchers’ interest isin the proportion below this point. Therefore,

0.5 − 0.4525 = 0.0475

In other words, a proportion of 0.0475 occurs below x = 3.80. Stated as a percentage, 4.75% ofthe candidates will score below 3.80 on the test.

For the proportion above 4.30,

z4.30 = (4.30 − 4.00) / 0.12 = 2.5

Table 3.2 indicates that

· this z score corresponds to a proportion of 0.4938, indicating that, as a percentage,49.38% of the population occurs between a score of 4.30 and the mean of thedistribution, and

· the percentage above this point will be 50 − 49.38 = 0.62, or 0.62%, of those who takethe test score at 4.30 or beyond.

Apply It! boxes written by Shawn Murphy

Comparing Data from Different Tests
Earlier chapters discussed how test scores from two different instruments with different means andstandard deviations can be compared. Perhaps a juvenile gang member under court-ordered counselingis required to complete two different assessments: one measuring aggression and one social alienation.The gang member scores 39 on the aggression test and 15 on the alienation test. Table 3.3 shows themeans and standard deviations of the two tests.

Table 3.3: Test results for aggression and social alienation
Test

Mean

Standard deviation

Aggression measure

32.554

5.824

Social alienation

12.917

2.674

In both cases, the gang member scored higher than average on both aggression and social alienation.For which measure is the score the most extreme?

Because the two tests have different means and standard deviations, comparing the raw scores directlyis not helpful. However, employing the z transformation allows both scores to fit a distribution wherethe mean is 0 and the standard deviation is 1.0. The raw scores may not reveal much, the z scores canbe directly compared. Recall that

z=x−Ms

Calculating z for the aggression score produces:

z39=39−32.5545.824=1.107

Then calculate the z for social alienation:

z15=15−12.9172.674=0.779

Young man being arrested.

Doug Menuez/Photodisc/Thinkstock

Using z scores enables researchers tobetter understand test results measuringaggression and social alienation injuvenile gang members.

Interpreting Multiple z Values
Since the question is which of the juvenile’s two testscores is the more extreme, we have no need for tablevalues—only the value of z. As both z values arepositive, the z for aggression is more extreme than thatfor social alienation. Performing the z transformationallows us to note that the aggression value is 1.107standard deviations from the mean of the distribution.Alienation, meanwhile, is just 0.779 standard deviationsfrom its mean. Practically, the results show thisindividual is more aggressive than alienated. As long asraw scores, means, and standard deviations are available,researchers can use z to make direct comparison of verydifferent qualities, in this case, aggression and socialalienation in the same individual.

Another Comparison
Psychologist Lewis Terman developed the Stanford-Binet test, which measures children’s intelligence.Suppose a psychologist is similarly interested in giftedness among children. Because unusual verbalability often seems to accompany superior intelligence in gifted children, the psychologist measuresboth characteristics for a group of subjects. One particular subject scores 140 on intelligence and 55.0on verbal ability. Table 3.4 lists the descriptive data for each test.

Table 3.4: Test results for intelligence and verbal ability
Test

Mean

Standard deviation

Intelligence

100

15

Verbal ability measure

40

5.451

As in the previous example, the researcher must convert scores into z scores before they can bedirectly compared.

For the intelligence score, the z score is calculated as:

z=x−Ms

z140=140−10015=2.667

For the verbal ability measure, the z score is calculated as:

z=x−Ms

z55=55−405.451=2.752

The z scores indicate that both test scores are about the same distance from their respective means.This makes it more difficult to glance at the raw scores and know which is higher. But because bothhave been transformed into z scores, the two measures now belong to a common distribution, and theresearcher can see that the verbal ability measure is slightly higher than the intelligence score.

Determining How Much of the Distribution Occurs Under Particular Areas of theCurve
If we draw a distribution and clarify what is at issue, questions about how much of the distribution isabove a point, below a point, or between two points do not require researchers to observe formal rules.For the sake of order and clarity, however, the flowchart in Figure 3.6 provides some direction foranswering different questions a researcher might ask about proportions within a distribution.

Figure 3.6: Flowchart to address questions pertaining to adistribution
Use the steps illustrated in the flowchart to resolve questions about the proportionswithin a population.

A flowchart showing four possible paths a researcher might follow to answer questions about a given population, based on whether a distribution is above, below, between, or above and below two points on a curve.

Try It!: #4
Figuratively speaking, how does the ztransformation allow you to compareapples to oranges?

The list of steps must seem like a great deal toremember. In fact, the better course whenconfronted with a z score problem is to sketchout a distribution to produce something likeFigures 3.3 and 3.4. The visual displays helpclarify the question and suggest the steps neededto answer it.

The Normal Curve
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3.3 z Scores, Percentile Ranks, and Other Standard Scores
Our task to this point has been to transform raw scores into z scores and then to percentagesor proportions of the distribution in specified areas. If the percentages are already available,but neither the raw data nor the related descriptive statistics are, Table 3.2 (the z table) allowsus to work backward to determine the z value—even without the mean and standarddeviation for the data.

Let us assume that published data indicate that only 1% of the population has intelligencescores above 140. What z score does this represent?

1. Because Table 3.2 lists proportions, the first step is to turn the percentage into aproportion: 1% is 1/100, which is the same as a proportion of 0.01.

2. Recall that Table 3.2 indicates the proportion of a normal population between a particular value of z and the mean for half (0.5) of the distribution. Therefore, we needa z value which includes all but that most extreme 0.01, which will be the z value for aproportion of 0.50 − 0.01 = 0.49. A z value for a proportion of 0.49, will be the valuethat includes the 49% of the distribution, which means it excludes the highest 1% ofthe distribution.

3. Table 3.2 does not list a proportion of exactly 0.49, but it does list 0.4901, which isvery close. Reading leftward from the proportion to the margin and also vertically tothe column heading, the associated z value for 0.4901 is 2.33. If data were gathered forintelligence scores, a z = 2.33 excludes close to the top 0.01 or 1%.

To state this more directly, when viewed as z scores, any intelligence score where z > 2.33 issomewhere among the top 1% of all intelligence scores. Figure 3.7 illustrates this proportion.

Figure 3.7: The value of z associated with a particularproportion
A normal distribution curve that shows where the highest 1% of scores fallwithin a given population.

A normal distribution curve, with z values along the horizontal axis ranging from -3z to +3z. A vertical line indicates the z value of 2.33 just to the right of +2z. The shaded area between z = 2.33 and +3z represents a proportion of 0.01 (1% of the population).

Converting z Scores to Percentile Ranks
Chapter 2 introduced percentile scores. Recall that percentiles indicate the point belowwhich a specified percentage of the group occurs. For example, 73% of the distributionoccurs at or below the point defined by the 73rd percentile, and so on. Because researcherscan use the table values associated with z scores to determine the percentage of thedistribution occurring below a point, it is not difficult to take one more step and turn thatpercentage into a percentile score. For example, because

· z = 1.0 includes 34.13% between that point and the mean and

· that part of the distribution from the mean downward is 50%, then

· 34.13% + 50% = 84.13% of scores are at or below z = 1.0; therefore, z = 1.0 occurs atthe 84th percentile.

Although percentile scores can be easily determined from the table values that are associatedwith z scores, note an important difference between percentile scores and z scores. The zscore is one of several standard scores. Standard scores are all equal-interval scores—theinterval between consecutive integers is constant, which means that in terms of data scale,standard scores are interval scale. The increase in whatever is measured from z = −1.5 to z =−1.0 is the same as it is from z = 0.3 to z = 0.8. The increase is 0.5 in either case.

This interval scale does not apply to percentile scores. Because these scores indicate thepercentage of scores below a point rather than reflecting a direct measure of somecharacteristic, the distances between consecutive scores differ widely in various parts of thedistribution. Most of the data in any normal distribution are in the middle portion, wherescores have the greatest frequency. The frequency with which scores occur diminishes asscores become more distant from the mean, something reflected in the curves in frequencydistributions that are vertically highest in the middle and then decline as they extend outwardto the two tails. Note the comparison between percentiles and z scores in Figure 3.8.

Figure 3.8: z scores and percentile scores
A comparison of z scores and percentiles for a normal distribution shows thatthe majority of scores are found within the 50th percentile. Meanwhile, thefrequency of scores above the 99th and below the 1st percentiles is low in anormal distribution.

A normal distribution curve showing both z scores, ranging from -3 to +3, and percentile scores along the horizontal axis. The scores align along the horizontal axis as follows: <1 percentile immediately under -3 standard deviations, 2nd percentile immediately under -2 standard deviations, 16th percentile under -1 standard deviations, 50th percentile under 0, 84th percentile under +1 standard deviation, 98th percentile under +2 standard deviations, and >99th percentile under +3 standard deviations.

As a result of high frequency in the middle of the distribution, in any normal distribution thedifference between consecutive percentile scores is always much smaller near the middle ofthe distribution (between the 50th and 51st percentiles, for example) than betweenconsecutive percentile scores in the tails (between the 10th and 11th, or the 90th and 91stpercentiles, for example). This characteristic has important implications. The differencebetween scoring at the 50th and 51st percentile score on something like the Beck DepressionInventory is almost inconsequential compared to the difference between the 90th and 91stpercentile, a much greater difference. Percentile scores are ordinal scale, whereas z scores areinterval scale.

Converting z to Other Standard Scores
Part of the appeal of the z score is that it enables researchers to readily determine relativeperformance. A positive z value indicates that the individual has scored in the upper half ofthe distribution. Someone who scores one standard deviation beyond the mean, as we notedearlier, has scored at the 84th percentile, and so on. The z scores belong to a family ofmeasures called “standard scores.” They have in common these characteristics: a) a fixedmean and standard deviation and b) equal intervals between consecutive data points.

Another standard score is the t score. It is used in the place of z scores when those reportingthem prefer not to report negative scores, which of course are half of all possible z values.After calculating the z value, a researcher can easily change it to a t score. In fact this is truefor any score that has a fixed mean and standard deviation, whether it is a standard score like t or, for example, a Graduate Record Exam (GRE) score (see Table 3.5), which also has afixed mean and standard deviation.

Table 3.5: Comparison of t scores and GRE scores
Mean

Standard deviation

t scores

50

10

Graduate Record Exam

500

100

Try It!: #5
What makes a score a standard score?

Either score can be derived from z. Toconvert from z to t, for example, simplymultiply z by 10 and add 50. So, if z = 1.75,then

t = 10 × 1.75 = 17.5 + 50 = 67.5

z = 1.75 is the same as t = 67.5.

For GRE, we would multiply z by 100 and add 500:

GRE = 100 × 1.75 = 175.0 + 500 = 675

z = 1.75 is the same as GRE= 675 (and as t = 67.5).

Although more common in educational than in psychological testing and research, normalcurve equivalent scores (NCE) and “stanine” scores (standard nine-point scale) are alsoexamples of standard scores. Like z and t, each is equal-interval, and both have fixed meansand standard deviations.

3.4 Using Excel to Perform the z Score Transformation

Students taking a test in a classroom.

monkeybusinessimages/iStock/Thinkstock

Evaluating achievement motivationscores can give researchers valuableinformation about the relationshipbetween poverty and achievement inschools.

The z score transformation is a fairly simple formula. Asa result, to program it into Excel and transform an entiredata set into z scores is not difficult. In fact, theapplication offers several ways to do this, but thischapter will explore just one. It involves programmingthe z score transformation formula directly into the datasheet.

A researcher interested in the relationship betweenpoverty and achievement motivation among secondary-school-aged young people gathers data from a group ofstudents whose families qualify for free and reduced-price lunches at school. The achievement motivationscores are as follows:

4, 5, 7, 7, 8, 9, 9, 9, 10, 13

To use Excel to transform those data into their z score equivalents, follow these steps:

1. List the data in Excel in Column B, with the label “Ach Mot” in B1.

2. Enter the 10 scores into cells B2 to B11.

3. In cell B12, enter the formula =average(B2:B11). (Note: Virtually all spreadsheets, includingExcel, have shortcuts for the more common calculations, such as means and standard deviations.A user can enter the formula, as we have done here, or use a shortcut. Shortcut procedures vary,however, depending on the operating system and the version of Excel. Excel for Mac, forexample, allows users to enter the data, position the cursor where they desire the statistic’s valueto appear, and then double-click the name of the desired statistic under the Formula tab.)

c. The equal sign indicates to Excel that a formula follows.

c. The command average will provide the arithmetic mean.

c. When several cells are to be included in the function, they are placed in parentheses ( ).When the cells are consecutive, the colon (:) indicates that all cells from B2 to B11 are tobe included in the function.

1. Press Enter.

1. In cell A12, enter the label “mean =.” The value in cell B12 will be 8.1, the mean of theachievement motivation scores.

1. In cell B13, enter the formula =stdev(B2:B11). Note that stdev is the Excel abbreviation for“sample standard deviation.” For Mac users, the abbreviation is stdev.s.

1. Press Enter.

1. In cell A13, enter the label std dev =. The value in cell B13 will be 2.558211, the standarddeviation of the scores.

1. In cell C1, enter the label equiv z.

1. In cell C2, enter the formula =(B2_8.1)/2.558 and press Enter. Consistent with the z scoretransformation, this formula subtracts the mean from the raw score in cell B2 and then dividesthe result by the standard deviation, 2.558, which we rounded to three decimals.

1. Repeat that operation for all the other scores as shown next:

k. With the cursor in cell C2, click and drag the cursor down from C2 to C11 so that cells C2to C11 are highlighted.

k. In the Editing section at the top of the page near the right side is a Fill command with adown-arrow at the left (for Macs, the command is on the left side, below the Home tab).Click the down-arrow to the right of the Fill command, then click Down. This action willrepeat the result in C2 for the other nine cells, adjusting for the different test scores in eachcell.

Figure 3.9 shows how the spreadsheet will look after Step 11, with the z score equivalents of all theoriginal achievement motivation scores displayed to the right of the original scores.

Figure 3.9: Raw scores transformed to z scores in Excel

Excel converts raw scores to z scores using a simple formula.

Screen capture of an Excel spreadsheet, with achievement motivation raw scores listed vertically in the B column. The C column lists the equivalent z scores for each raw value. The mean and standard deviation for this sample are also indicated below the two columns.

Source: Microsoft Excel. Used with permission from Microsoft.

Using Excel to Perform the z Score Transformation

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3.5 Using z Scores to Determine Other Measures
Occasionally, a researcher has access to z scores and the mean and standard deviation but notthe original raw scores. Formula 3.1 used the raw score (x), the mean (M), and the standarddeviation (s) to determine the value of z, but actually, any three of the values in the formulacan be used to determine the value of the fourth. Just as Formula 3.1 uses x, M, and s todetermine z, we could use z, M, and s to derive x. Altering Formula 3.1 to determine thevalue of something other than z involves a little algebra but is not difficult.

Determining the Raw Score
To determine the raw score, follow these steps:

1. Because z = (x − M)/s, swap the terms before and after the equal sign so that (x − M)/s= z.

2. To eliminate the s in the denominator of the first term, multiply both sides by s so thatit disappears from the first term and emerges in the second: x − M = sz.

3. To isolate x, add M to both sides of the equation so that x = sz + M.

Returning to the Excel problem, if the z scores and descriptive statistics are available, we candetermine the raw score for which z = −1.603 as follows:

If M = 8.10, s = 2.558, and x = z × s + M, substituting the values we have produces x = (−1.603)(2.558) + 8.10 = 3.9995, which rounds to 4.0.

Checking the earlier data reveals that 4 was indeed the raw score for which z = −1.603.

Determining the Standard Deviation
If the raw scores, the mean, and z are available, but s is lacking, z = (x − M)/s, so (x − M)/s = z. Taking the reciprocal of each half of the equation—which means inverting the term so that(x − M)/s becomes s/(x − M) and z/1 becomes 1/z, giving us s/(x − M) = 1/z. Multiplying bothsides by (x − M) yields s = (x − M).

Using the data from the Excel problem again, for the first participant, x = 4, M = 8.10, and z =−1.603. According to the adjusted formula,

s = (x − M)/z

therefore, substituting the values we have produces the following:

s=4−8.1−1.603=2.5577

which rounds to 2.558, the standard deviation value for the original data set.

Determining the Mean
It would be very unusual for a researcher to have the z scores, the standard deviation, and theoriginal achievement motivation score but not have the mean. However, just to complete theset, the mean can be determined from the other three values as follows:

Because

z = (x − M)/s

if both halves of the equation are multiplied by s, then s appears in the first termand disappears from the second. The result is

sz = x − M

If M is then added to both sides, M appears in the first term and is eliminated in thesecond. The result is

sz + M = x.

If sz is then subtracted from both sides, it is eliminated from the first term andadded to the second. The result is

M = x − sz

For the first participant, the achievement motivation score x = 4.0, the z score =−1.603, and s = 2.558. The mean for the test can be determined as follows:

M = x − sz, or

M = 4 − (2.558 × −1.603) = 8.10, which was the original mean.

Maintaining Fixed Means and Standard Deviation
One of the characteristics of widely used standardized tests is that their mean and standarddeviation values remain the same over time. The major intelligence tests, for example, have afixed mean of 100 and a standard deviation of 15, even though the Stanford-Binet andWechsler tests have been revised several times. When a test is revised and updated, do themeans and standard deviations likewise change? In fact, they do. Flynn and Weiss (2007)documented significant increases in intelligence scores over a 70-year period, but to make thescores comparable over time, psychologists use what are called modified standard scores.The modified standard score allows those working with the test to gather data that have anymean and any standard deviation and then adjust them so that they conform to predeterminedvalues. This process follows these steps:

1. Gather data with the new instrument.

2. Determine the equivalent z scores for test-takers’ raw scores.

3. Apply the formula.

Formula 3.2 is used to modify a score so that the mean and standard deviation for thepopulation of scores take on specified values.

Formula 3.2

MSS = (sspec)(z)+ Mspec

where

MSS = the modified standard score,

sspec = the specified standard deviation, and

Mspec = the specified mean.

Note that this formula is the same used to transform z scores into t scores. By way of anexample, perhaps a psychologist has developed what she has labeled the Brief IntelligenceTest (BIT). To compare results to tests her colleagues have used traditionally, she wants theBIT’s descriptive characteristics to conform to those of the more established tests. For eightparticipants, the BIT scores are as follows:

22, 25, 26, 29, 29, 32, 32, 35

Of course, no one norms an intelligence test on only eight people. The potential for what wewill later call sampling error is too great. Still, to illustrate the process, we will assume thesample scores are valid.

Verify that M = 28.75 and s = 4.268.

For the participant with an intelligence score of 22, the corresponding z value is

z=x−Ms=22−28.754.268=−1.582

To determine that participant’s score on an instrument with a mean of 100 and a standarddeviation of 15, the psychologist will apply the formula:

MSS = (sspec)(z) + Mspec

= (15)(=1.582) + 100

= 76.276

Although the original BIT score was 22, using the z transformation and modified standardscore procedures makes the BIT score conform to the mean and standard deviation of a moreestablished test. Among scores for which the mean is 100 and the standard deviation is 15,the modified standard score for the BIT score of 22 is 76.276.

Writing Up Statistics
Although z scores are an important part of data analysis, like the raw scores that researchersgather in their work, z scores often do not appear in research reports. Reports list the meansand standard deviations, but often omit the raw scores and their z score transformation. In astudy of the weight-gain side effect that anti-psychotic drugs might have on adolescents,Overbeek (2012), however, used a combination of height and weight to determine body-mass-index (BMI) scores for each subject, and then transformed the BMI scores into easier-to-interpret z scores. A z score near 0 indicated that given the subject’s height, weight wasprobably appropriate. A positive z score indicated that the individual might be overweight,negative z scores indicated underweight, and so on. Overbeek also used z scores to index theweight-gain data over the course of the study.

Brown (2012), too, used z scores. In his study, they offered a way to counter the effect thatgrade inflation has on college students’ class rankings. He posited that class rankings are lessinformative than they once were because weaker students in departments where courseworkis easier are ranked ahead of students who have a higher level of academic aptitude butcompete in more demanding programs. Brown’s solution was to use the z transformationwithin departments to indicate how much above or below the mean students were in theirindividual programs.

Summary and Resources
Chapter Summary
Normal distributions are unimodal and symmetrical, and their standard deviations tend to beabout one-sixth of the range. Although not all data are normally distributed, many of themental characteristics that psychologists and social scientists measure are normal. Becausethe proportions of the population that occur in specified ranges remains constant in normallydistributed populations, we can have some confidence about how scores will be arrayed evenbefore we view a display of the data.

What the standard normal distribution, or z distribution, does is capitalize on the consistencyin normally distributed populations by offering one distribution by which all other normalpopulations can be referenced. In this distribution, where the mean is always 0 and thestandard deviation is 1.0 (Objective 1), table values indicate the proportions of the populationlikely to occur anywhere along its range. By transforming raw scores (Objective 2) from anynormal population so that they fit this z distribution, we can take advantage of how well thecharacteristics of this distribution are known and answer important questions about data fromany population (Objective 3) in terms of z:

· For example, when someone scores at a particular level, we can ask what proportion ofthe entire population is likely to score below (or above) that point.

· When most of the people in a particular group score between two points, we can askwhat proportion of the entire population will score between (or outside) those points.

Because the z score transformation is a relatively simple formula, programming Excel toproduce the z equivalents for any set of scores (Objective 4) is simple and can be helpful withlarge data sets.

The z is one of several standard scores in fairly common use. Whether z or some other, allstandard scores indicate how distant one individual’s score is from the mean of thedistribution. Rather than providing an absolute measure of some characteristic, standardscores are normative, meaning that they indicate the level of what is measured relative toothers in the same population. Those who prefer not to deal in negative values (whichcharacterize half of the z distribution) can employ t scores. In all material respects, t is thesame as z, except that the mean is 50 and the standard deviation is 10.

The modified standard score (Objective 6) enhances standard scores’ ability to communicatean individual’s standing relative to a population. Researchers often use standard scores toreport the data from standardized tests, but these tests are revised from time to time, whichcan affect the test means and standard deviations. To ensure stability, the modified standardscore uses the z transformation as a way to maintain constant descriptive characteristics, evenas the instrument used to measure it, or even the characteristic measured, changes with time.

In the incremental nature of statistics books, each chapter prefaces the next. Chapters 1–3 area prelude to Chapter 4. With all our effort to label, display, and describe data sets, the focusin the discussion of z scores and the other topics has been primarily about analyzing theperformance of individuals. Behavioral scientists, however, are generally much moreinterested in asking questions about groups. Analyzing how those in a sample compare tothose in the entire population is the focus of Chapter 4. It will do so by expanding discussionof the z distribution.

The math and the logic involved in Chapter 4 will be much the same. If the discussion in thischapter makes sense, the material in Chapter 4 will not be difficult. Still, it is a good idea toreview the Chapter 3 material and recalculate the sample problems, as repetition has value.

Chapter 3 Flashcards
Key Terms
modified standard scores

percentile

probability

standard normal distribution

standard scores

t score

z score

z transformation

Review Questions
Answers to the odd-numbered questions are provided in Appendix A.

1. A researcher is interested in people’s resistance to change. For the dogmatism scale(DS), data for 10 participants are as follows:

28, 28, 29, 29, 32, 33, 35, 36, 39, 42

a. What is the z score for someone who has a DS score of 28?

a. Will a z score for a raw score of 35 be positive or negative? How do you know?

a. How many standard deviations is a score of 28 from the mean?

a. What will be the z value of a raw score of 33.1?

a. Since there are no scores below 28, does it make any sense to calculate z for a rawscore of 25, for example? Shouldn’t such a score have a zero probability ofoccurring?

1. Examining the relationship between recreational activity and level of optimism amongsenior citizens, a psychologist develops the Recreation Activity Test (RAT). Scores for8 participants are as follows:

11, 11, 14, 14, 14, 17, 18, 22

b. What is the z score for someone with RAT = 23?

b. Why isn’t 0 the answer to 2a, since none of the participants scored 23?

b. What is the z score for someone with RAT = 15.125?

b. Explain the answer to 2c.

b. How does z allow one to compare tests with entirely different means and standarddeviations?

1. One participant has RAT = 11. The same individual is administered a ConsistentApproval Test (CAT) and scores 45. The CAT data, including that participant’s score,are as follows:

42, 45, 48, 49, 55, 58, 62, 64

c. Which score is higher, the RAT or the CAT?

c. Why isn’t the answer to 3a automatically CAT, since it has the higher mean value?

1. Researchers developed the ANxious, Gnawing Stress Test (ANGST) to measure emotional stability among law-enforcement professionals. A random sample of policepatrol officers yielded the following scores:

54, 58, 61, 64, 75, 81, 82, 85

d. What proportion of the population will score 81 or higher?

d. What proportion will score 60 or higher?

d. If x > 75 is the cutoff for “highly stressed,” what is the probability that someone,selected at random, will be highly stressed?

d. What is the probability of scoring between 60 and 81?

1. Using the data in Question 4, what percentage of the population will score lower than55?

1. What is the t equivalent to a z score for someone with an ANGST score of 64?

f. What is the mean of the t distribution?

f. Why is t sometimes preferred over z?

f. If z = 2.5, what is t?

1. Refer to the data in Questions 3 and 4: For an individual who scores 60 on RAT and 78on ANGST, which is the higher score?

1. In any standard normal distribution, determine the following:

h. What percentage of scores will occur below z = 0?

h. What is the probability of a positive value of z?

h. What percentage of scores will occur between ±1.96 z?

1. If someone scores z = 1.0, what is the corresponding percentile rank?

1. What percentile rank is z = 0? What measure of central tendency represents the 50thpercentile?

1. A psychologist wishes to maintain a mean of 25 and a standard deviation of 5 for a testdeveloped to measure compulsive behavior. On a revised test, the scores are asfollows:

14, 17, 19, 19, 22, 27, 28, 29

k. What is the modified standard score for the person who scored 17 on the revisedinstrument?

k. What is the probability of scoring 17 or lower according to the eight scores?

1. Given the data in Question 11,

l. What is the z equivalent of a raw score of 28?

l. What is the probability of scoring somewhere from 14 to 29?

1. Draw a normal distribution and identify where z = −1.17 and z = +2.53 are located.What percentage of the population occurs between these two z values?

Answers to Try It! Questions
1. A z of 1.5 indicates that the associated raw score is 1.5 standard deviations (thedenominator in z) from the mean.

2. Because the standard normal distribution (the z distribution) is normal, the distributionis symmetrical. The proportion of the distribution between a value of z and the meanwill be the same for a negative z as it is for a positive z with the same numerical value.

3. Do not refer to Table 3.2 for help with this one. The table’s highest score is z = 3.09,but in fact z has no upper limit. In theory, the tails in the z distribution never actuallytouch the horizontal axis of the graph, which means that there exists always at least thepossibility of scores higher (or lower) than any already measured.

4. One of the values of the z transformation is that scores that have any descriptivecharacteristics can be recalibrated so that they fit a distribution where the mean is 0 andthe standard deviation is 1.0. By doing so, scores from any variety of sources can becompared directly after converting them to z scores. The only requirement is that theybe normally distributed.

5. Standard scores are equal-interval scores with a fixed mean and standard deviation,thereby allowing the magnitude of the score to indicate how an individual compares toall others for whom scores are available.

Previous section

Next section

4

Applying z to Groups

Crowd of people in Hong Kong, some of whom are holding umbrellas and flags

Victor Faile/Corbis

Chapter Learning Objectives:

After reading this chapter, you should be able to do the following:

1. Describe the distribution of sample means.

2. Explain the central limit theorem.

3. Analyze the relationship between sample size and confidence in normality.

4. Calculate and explain z test results.

5. Explain statistical significance.

6. Calculate and explain confidence intervals.

7. Explain how decision errors can affect statistical analysis.

8. Calculate the z test using Excel.

Introduction

As we noted at the end of Chapter 3, researchers are generally more interested in groups thanin individuals. Individuals can be highly variable, and what occurs with one is not necessarilya good indicator of what to expect from someone else. What occurs in groups, on the otherhand, can be very helpful in understanding the nature of the entire population. A Googlesearch indicates that the suicide rate is higher among dentists than it is among those of manyother professions. If we wanted to experiment with some therapy designed to relievedepressive symptoms among dentists, we would be more confident observing how a group of50 dentists responds than in examining results from just one. This chapter will use thematerial from the first three chapters to begin analyzing people in groups.

Noting that many of the characteristics that interest behavioral scientists are normallydistributed in a population implies that some characteristics are not. Since samples can neverexactly emulate their populations, it may not be clear in the midst of a particular study whendata are normally distributed. This uncertainty potentially poses a problem: we may wish touse the z transformation and Table B.1 of Appendix B in our analysis, but Table B.1 is basedon the normality assumption. If the data are not normal, where does that leave the relatedanalysis?

This is a self-assessment and will not affect your grade. You may only take this pre-test once.

Test Ch 4: Applying z to Groups

Top of Form

1. The potential for sampling error diminishes as the size of the sample grows.

· a. TRUE

· b. FALSE

2. Populations based on samples are more inclined to normality than populations based on individual scores.

· a. FALSE

· b. TRUE

3. Determining normality requires all of the scores in a population.

· a. TRUE

· b. FALSE

4. The z test produces a z value based on individual scores rather than on sample means.

· a. TRUE

· b. FALSE

5. A statistically significant result is one that is unlikely to have occurred by chance.

· a. FALSE

· b. TRUE

Finish

Bottom of Form

4.1 Distribution of Sample Means
Sign on sidewalk displaying American flag and the words “vote here”

iStockphoto/Thinkstock

A population is allmembers of a definedgroup, such as all voters ina county.

What options do researchers have if they are suspiciousabout data normality? One important answer is the distribution of sample means, so named because the scoresthat constitute the distribution are the means of samplesrather than individual scores.

Note that the descriptor population means all possiblemembers of a defined group. Recall that the frequencydistribution—the bell-shaped curve representing thepopulation—was a figure based on the individual measuressampled one subject at a time. In discussing the frequencydistribution, we assumed that we would measure eachindividual on some trait, and then plot each individual score.Instead of selecting each individual in a population one at atime, suppose a researcher

1. selects a group with a specified size;

2. calculates the sample mean (M) for each group;

3. plots the value of M (rather than the value of eachscore) in a frequency distribution;

4. and continues doing this until the population is exhausted.

How would plotting group scores rather than individual ones affect the distribution? Wouldthe end result still be a population? The answer to the second question is yes: because everymember is included, it is still a population. Whether a population is measured individually oras members of a group is incidental, as long as all are included.

Perhaps researchers are interested in language development among young children and wishto measure mean length of utterance (MLU) in a county population. Whether the researchersmeasure and plot MLU for each child in a county’s Head Start program or plot the meanMLU for every group of 25 in the program, the result is population data for Head Startlearners for that county.

The Central Limit Theorem
The answer to the question “how would the distribution be affected?” is a little moreinvolved, but it is important to nearly everything we do in statistical analysis. It involveswhat is called the central limit theorem:

If a population is sampled an infinite number of times using sample size n and themean (M) of each sample is determined, then the multiple M measures will take onthe characteristics of a normal distribution, whether or not the original populationof individuals is normal.

Take a minute to absorb this. A population of an infinite number of sample means drawnfrom one population will reflect a normal distribution whatever the nature of the originaldistribution. A healthy skepticism prompts at least two questions: 1) How would we provewhether this is true since no one can gather an infinite number of samples? and 2) Why doessampling in groups rather than as individuals affect normality?

Although prove is too strong a word, we can at least provide evidence for the effect of thecentral limit theorem using an example. Perhaps a psychologist is working with 10 people ontheir resistance to change, their level of dogmatism. Technically, because 10 constitutes theentire group, the population is N = 10. Recall that N refers to the number in a population.Even with a small population we cannot have an infinite number of samples, of course, butfor the sake of the illustration we will assume that

· dogmatism scores are available for each of the 10 people;

· the data are interval scale;

· the scores range from 1 to 10; and

· each person receives a different score.

So with N = 10, the scores are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Figure 4.1 depicts a frequencydistribution of those 10 scores.

The distribution in Figure 4.1 is not normal. With R = 10 − 1 = 9 and s = 3.028 (a calculationworth checking), the distribution is extremely platykurtic (i.e., flatter than normal); the rangeis less than 3 times the value of the standard deviation rather than the approximately 6 timesassociated with normal distributions. There is either no mode or there are 10 modes, neitherof which suggests normality. We can illustrate the workings of the central limit theorem witha procedure Diekhoff (1992) used. We will use samples of n = 2, and make the examplemanageable by using one sample for each possible combination of scores in samples of n = 2from the population, rather than an infinite number of samples.

Figure 4.1: A frequency distribution for the scores 1through 10: Each score occurring once
A frequency distribution of ten scores, each with a different value. This type ofdistribution, which is not normal, is highly platykurtic.

Chart showing a distribution that is completely flat, with values numbering 1–10 along both x- and y-axes. All of the boxes along the chart corresponding with the value of 1 on the x-axis are shaded.

Table 4.1 lists all the possible combinations of two scores from values 1–10. Ninetycombinations of the 10 dogmatism scores are possible. The larger the sample size, the morereadily it demonstrates the tendency toward normality, but all combinations of (for example)three scores would result in a very large table.

Table 4.1: All possible combinations of the integers 1–10

1, 2

2, 1

3, 1

4, 1

5, 1

6, 1

7, 1

8, 1

9, 1

10, 1

1, 3

2, 3

3, 2

4, 2

5, 2

6, 2

7, 2

8, 2

9, 2

10, 2

1, 4

2, 4

3, 4

4, 3

5, 3

6, 3

7, 3

8, 3

9, 3

10, 3

1, 5

2, 5

3, 5

4, 5

5, 4

6, 4

7, 4

8, 4

9, 4

10, 4

1, 6

2, 6

3, 6

4, 6

5, 6

6, 5

7, 5

8, 5

9, 5

10, 5

1, 7

2, 7

3, 7

4, 7

5, 7

6, 7

7, 6

8, 6

9, 6

10, 6

1, 8

2, 8

3, 8

4, 8

5, 8

6, 8

7, 8

8, 7

9, 7

10, 7

1, 9

2, 9

3, 9

4, 9

5, 9

6, 9

7, 9

8, 9

9, 8

10, 8

1, 10

2, 10

3, 10

4, 10

5, 10

6, 10

7, 10

8, 10

9, 10

10, 9

For each possible pair of scores, if we calculate a mean and plot the value in a frequencydistribution as a test of the central limit theorem, the result is Figure 4.2. Because the entiredistribution is based on sample means, Figure 4.2 is a distribution of sample means. Strictlyspeaking, this distribution is not normal, but although based on precisely the same data,Figure 4.2’s distribution is a good deal more normal than the distribution in Figure 4.1.

Figure 4.2: A frequency distribution of the means of allpossible pairs of scores 1 through 10
A distribution of the means of each possible pair of scores with values between1 and 10. This distribution is not normal, but has more normality than thedistribution shown in Figure 4.1.

A bar graph with values of 1–10 on the y-axis. The x-axis is numbered 1.5 to 9.5 in increments of 0.5. The greatest frequency of scores (highest bar) occurs at 5.5 and the lowest frequency occurs at 1.5, 2.0, 9.0, and 9.5.

Mean of the Distribution of Sample Means
The symbol used for a population mean to this point, µ, is actually the symbol for apopulation mean formed from one score at a time. To distinguish between the mean of thepopulation of individual scores and the mean of the population of sample means, we’llsubscript µ with an M: µM. This symbol indicates a population mean (µ) based on samplemeans (M).

Try It!: #1
Why is there less variability in thedistribution of sample means than in adistribution of individual scores?

With a distribution of just 90 sample means,Figure 4.2 shows nothing like an infinitenumber, of course, but it is instructivenevertheless. The mean of the scores 1through 10 is 5.5 (µ = 5.5). Study Figure4.2 for a moment. What is the mean of thatdistribution? The mean of our distributionof sample means is also 5.5: µM = 5.5. Thepoint is this: When the same data are usedto create two distributions—one apopulation based on individual scores andthe other a distribution of sample means—the two population means will have the samevalue, or, symbolically stated: µ = µM.

Describing the distribution as “normal” is a stretch, but Figure 4.2 is certainly more normalthan Figure 4.1. For one thing, rather than the perfectly flat distribution that occurs when allthe scores have the same frequency, some scores have greater frequency than others. Thesample means near the middle of the distribution in Figure 4.2 occur more frequently than thesample means at either the extreme right or left.

Why are extreme scores less likely than scores near the middle of the distribution? It isbecause many combinations of scores can produce the mean values in the middle of thedistribution, but comparatively few combinations can produce the values in the tails of thedistribution. With repetitive sampling, the mean scores that can be produced by multiplecombinations increase in frequency and the more extreme scores occur only occasionally,which the next section illustrates.

Variability in the Distribution of Sample Means
In the original distribution of 10 scores (Figure 4.1), what is the probability that someonecould randomly select one score (x) that happens to have a value of 1? Because there are 10scores, and just one score of 1, the probability is p = 1/10 = 0.1, right? By the same token,what is the probability of selecting x = 10? It is the same, p = 0.1.

Moving to the distribution based on 90 scores (Figure 4.2), what is the probability ofselecting a sample of n = 2 that will have M = 1.0? Is there any probability of selecting twoscores out of the 10 that will have M = 1.0? Because there is only one value of 1, there is noway to select two values with M = 1.0. As soon as a score of 1 is averaged with any otherscore in the group, all of which are greater than one, the result is M > 1. That is why thelowest possible mean score in Figure 4.2 is 1.5, which can only occur when 1 and 2 are in thesame sample.

The same thing occurs in the upper end of the distribution. The probability of selecting agroup of n = 2 with M = 10 is also zero (p = 0) because all other scores have lower valuesthan 10. For the 90 possible combinations, the highest possible mean score is 9.5, which canoccur only when the 10 and the 9 happen to be in the same sample.

The point is that variability in group scores is always less than the variability in individualscores. A related point is that the impact of the most extreme scores in a distributiondiminishes when they are included in samples with less extreme scores. Applied, theseprinciples mean that a researcher examining, for example, problem-solving ability among agroup of subjects can afford to be less concerned about the impact of one extremely low orone extremely high score as the size of the group increases. Larger group sizes minimize theeffect of extreme scores.

Standard Error of the Mean
Recall that the sigma, σ, indicates a population’s standard deviation. Specifically, σ indicatesa standard deviation from a population of individual scores. The symbol for the standarddeviation in a distribution of sample means is σM and as the subscript M suggests, it measuresvariability among the sample means. The formal name for σM is the standard error of themean.

In the language of statistics, error as in standard error of the mean refers to unexplainedvariability. As we move through the different procedures, we will calculate other standarderrors, which all have this in common: all are measures of unexplained data variability.

Earlier, we noted that whether charting the distribution of individual scores or the distributionof sample means, the means of the two distributions will always be equal: µ = µM. Is itlogical to expect the same from the measures of variability; in other words, will σ = σM? Thefact that the distribution of individual scores always has more variability than the distributionof sample means answers this question. Symbolically speaking, σ > σM, something that thedogmatism data show.

The standard deviation of the 10 original scores (1, 2, 3, 4, 5, 6, 7, 8, 9, 10) is σ = 2.872. Notethat this instance and the calculation of the standard error of the mean just below deal withpopulations and the formula must, therefore, employ N, rather than n − 1. Elsewhere in thebook, however, the formula will always be n − 1.

The standard error of the mean can be calculated by taking the standard deviation of themean scores of each of those 90 samples which constituted the distribution of sample means.The calculation is a little laborious, and happily, not a pattern that must be followed later, butthe value is σM = 1.915.

So, as predicted, σ has a larger value than σM. The smaller value for σM reflects the way lessextreme scores moderate the more extreme scores when they occur in the same sample.

Sampling Error
Although the standard error of the mean does not per se refer to a mistake, another kind oferror, sampling error, does. In inferential statistics, samples are important for what theyreveal about populations. However, information from the sample is helpful for drawinginferences only when the sample accurately represents the population. The degree to whichthe sample does not represent the population is the degree of sampling error.

Samples reflect the population with the greatest fidelity when two prerequisites are met: 1)the sample must be relatively large, and 2) the sample must be based on random selection.

The safety of large samples is explained by the law of large numbers. According to thismathematical principle, as a proportion of the whole, errors diminish as the number of datapoints increases. The potential for serious sampling error diminishes as the size of the samplegrows. The text earlier referred to this principle in noting that the distorting effect of extremescores diminishes as sample size grows.

Random selection refers to a situation where every member of the population has an equalprobability of being selected. Random selection contrasts with what are called conveniencesamples, samples that are used intact because they are handy. A sociology professor who usesthe students in a particular section of his class is relying on a type of convenience sampleknown as a nonrandom sample.

A random sample of n = 5 could be created from the 10 people being treated for dogmaticbehavior by assigning each person a number, placing the 10 numbers into a paper bag,shaking the bag well, and without looking, drawing out five numbers.

The result would be a randomly selected sample. When randomly selected, samples differfrom populations only by chance. They will differ, of course, but the differences are less andless important as sample size grows.

A map of the electoral college from the 1936 election between F. D. Roosevelt and Alf Landon.

National Atlas

Systematic sampling error providesresults drastically different from theactual outcome. Such a samplingerror occurred before the 1936presidential election, when a studypredicted a win by Alf Landon. Theelection results, displayed in themap, were drastically different.

If the sample should fail to capture some importantcharacteristic of the population other than its size,the problem is sampling error. The importantcharacteristic might be the mean, for example, andwhen M ≠ µ, sampling error has occurred. In fact,some sampling error always occurs because asample can never exactly duplicate all thedescriptive characteristics of the population, butsampling error will usually be minor if samples arerelatively large and randomly selected.

Statistical analysis procedures tolerate minor,random sampling error, but systematic samplingerror is another matter. Systematic sampling erroroccurs when the same mistake is made time aftertime.

In 1936, the publishers of Literary Digest, aprominent publication of the time, decided topredict the outcome of that year’s presidential election in the United States. To ensure thatsample size would not pose a problem, they sent out millions of postcards to registeredvoters. Literary Digest seemed to at least have met the requirement for a relatively largesample, because the Harris and Gallup polling organizations typically obtain very accurateresults with a few thousand, and sometimes just a few hundred, responses. Unfortunately, thepublishers decided to rely on telephone books and automobile registrations to locate pollrecipients. Consider the historical setting. At the height of the Great Depression, twoindicators of relative prosperity identified voters: a telephone in the home and a currentlyregistered car. The study proved disastrous for the magazine’s reputation. Poll resultsindicated that Alf Landon would win, but of course Franklin Roosevelt was elected in alandslide to a second term, carrying every state in the union except Maine and Vermont.

The problem was systematic sampling error. The voters were consistently and nonrandomlyselected from groups not representative of the entire population. If they had been randomlyselected, chances are that with the large sample size, the study would have predicted the election results accurately, but the sample size alone was not enough to compensate for theerror.

4.2 The z Test
The distribution of sample means is a distribution based not on individual scores but on themeans of samples of the same size repeatedly drawn from a population. The central limittheorem assures that such a distribution will be normal. Consequently, if the z score formulafrom Chapter 3 is adjusted to accommodate groups rather than individual scores, Table B.1answers all the same questions about groups that it did about individuals in Chapter 3.

Recall that the z score formula (3.1) had the following form:

z=x−Ms

If the following substitutions are made:

· M for x, so that the focus is on a sample mean rather than on an individual score;

· µM for M to shift from the sample mean to the mean of the distribution of samplemeans;

· σM for σ so that the measure of variability is for the distribution rather than the sample;

then the result is the z test:

Formula 4.1

z=M−μMσM

The z test produces a z value for groups rather than individual scores. Just as it did forindividual scores, z indicates how distant a particular sample mean is from the mean of thedistribution of sample means.

Note the similarities in Formulas 3.1 and 4.1:

· Both formulas produce values of z.

· Both numerators call for subtractions that result in difference scores.

· Both denominators measure data variability.

Calculating the z Test
When calculating z scores, as shown in Chapter 3, everything that is needed (x, M, and s) canbe determined from the sample:

z=x−Ms

Values needed for the z test, however, are often not as easy to determine. Because µM = µ,o