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3The Standard Normal Distribution and z Scores

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, and between 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.

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Introduction

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 all pos- sible members of any specified group. All university students, all psychology majors, all resi- dents of Orange County, and all left-handed male tennis players in their 20s are each descrip- tions of a population. We rely on Greek letters, such as µ for the mean and σ for the standard deviation, to distinguish population parameters from the statistics that describe samples. (The word parameter indicates a characteristic of a population.) Remove one or more indi- viduals from any population, and the resulting group is a sample.

As we were describing populations, we noted that some are “normally distributed.” These characteristics indicate normality: (a) data distributions are symmetrical, (b) all the mea- sures of central tendency have very similar values, and (c) the value of the standard deviation is about 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 in specified regions of the distribution, and this holds for all normal data distributions. For exam- ple, the region under a normal curve from the mean of the population to one standard devia- tion 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 the mean also includes 34.13%, so from 11σ or 21σ includes about 68.26% of the area under the curve in any normally distributed population. As long as the data are normally distributed, those percentages hold true. Since many mental characteristics are normally distributed, research- ers can know a good deal about such a characteristic without actually gathering the data and doing the analysis. Whether the characteristic is intelligence, achievement motivation, anxiety, or any other normally distributed characteristics, the proportion of the distribution within 11 or 21 standard deviation from the mean will be the same:

• If a particular intelligence scale has µ 5 100 and σ 5 15, about 68% of any general population will have intelligence scores between 85 and 115.

• Likewise, if an achievement motivation scale has µ 5 40 and σ 5 8, about 2/3 of any population will have achievement motivation scores from 32 to 48.

• And for an anxiety measure with µ 5 25 and σ5 5, about 68% of any general popu- lation will have scores between 20 and 30.

The consistency in the way so many characteristics are distributed affords a good deal of interpretive power. Anyone who needs information about the likelihood of individuals scor- ing 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 11σ and 21σ,

• from µ to 12σ is about 47.72% of the population, so about 95% (2 3 47.72) of the people in any general population will have intelligence scores between 70 (100 2 30) and 130 (100 1 30).

• from 13σ (49.87%) to 23σ includes nearly everyone in any normally distributed population (2 3 49.87 5 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 immediately

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Section 3.1 A Primer in Probability

apparent that this is a very unusual child. An intelligence score of that magnitude is about 4.667σ (170 2 100 5 70; 70 4 15 5 4.667) beyond the mean of the general population. If from 13σ to 23σ includes more than 99% of the population, from 14.667σ to 24.667σ must include 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 of their occurrence.

3.1 A Primer in Probability Scholars, data analysts, and in fact people on the whole are rarely interested in outcomes that occur every time. If everyone had an intelligence score of 170, no one would pay any attention to someone with such a score. The fact that we know it to be uncommon is what piques our curiosity.

If we are not interested in events that always occur, neither do we closely follow events that never occur. If no one had ever had an intelligence score of 170, probably no one would won- der about what such a score means for the person who has it. The things that occur some of 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 some-

one who lives near the San Andreas Fault? • What is the probability of an IRS audit for one taxpayer?

Because all of the items listed have happened in the past and because their occurrence is important to at least someone, people are interested in the probability of those occurrences whether or not they use the language of probability. When stated numerically, probabil- ity values range from 0 to 1.0. Something with a probability of zero (p 5 0) never occurs. On the other hand, p 5 1.0 indicates that the event occurs every time, and p 5 0.5 indicates that the event occurs 50% of the time.

As that last point indicates, percentages can be con- verted to probability values. Dividing the percent- age of times an event occurs by 100 indicates the associated probability of the event.

Returning to the intelligence scores, we see that because about 68% of the population has intelligence scores between 85 and 115, the probability (p) that someone selected at ran- dom from the general population will have a score somewhere between 85 and 115 is 0.68 (68.26/100, if the result is rounded to two decimal places).

Joseph Sohm/Visions of America/Corbis

The probability that something will occur, such as how likely it is that our favorite baseball team will win the World Series, intrigues us and is an important component in the decision- making process.

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70 85 100 115 130

Intelligence scores

–2σ –1σ +1σ +2σM

16% 16%34% 34%

Section 3.2 The Standard Normal Distribution

What is the probability that someone selected at random from the general population will have 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 5 0.5.

What is the probability that someone selected at random will have an intelligence score higher than 115? First, we noted earlier that 34.13% of the population falls between the mean, µ, and one standard deviation above the mean at σ 5 11.0 in any normally distributed population. In terms of intelligence score values, that is the region between scores of 100 and 115. Since 50% of any normally distributed population will occur at the mean and above, if we subtract from 50% that portion between the mean and one standard deviation above the mean, the remainder will be the portion of the distribution above 115: 50% 2 34.13% 5 15.87%; that is, 15.87% of all intelligence scores in a normally distributed population will occur above 115. Dividing by 100 (15.87/100 5 0.1587) and rounding the result to two decimal places produces the probability p 5 0.16.

By the same logic, because a score of 85 is one standard deviation below the mean, the prob- ability p 5 0.16 means that someone selected at random from the population will score below 85. If we combine the two outcomes, the probability is p 5 0.32 that someone from the popu- lation 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 standard deviations above and below the mean. If M 5 100, we can know the probability of someone scoring below 85 or above 115.

70 85 100 115 130

Intelligence scores

–2σ –1σ +1σ +2σM

16% 16%34% 34%

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

Recall that the lowest probability for any value is zero (p 5 0). If p 5 0, then the event or out- come 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 median price of a home, and salary data are likewise reported as median values. Those cases

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–3 –2 –1 0 +1 +2 +3

The Mean = 0 The Standard Deviation = 1.0

Section 3.2 The Standard Normal Distribution

use the medians because the related populations are very unlikely to be normally distributed and, as a measure of central tendency, medians are less affected by extreme values than are means. A few very high salaries or home values create positive skew in the resulting distribution. In contrast, when it comes to, say, mental characteristics such as intel- ligence, achievement motivation, problem-solving ability, verbal aptitude, reading comprehension, and so on, population data are often normally distributed.

Although there are many normal distributions all having the same proportions, each has different descriptive values. An intelligence test might have µ 5 100 and σ 5 15 points. A nationally adminis- tered reading test might have a mean of 60 and a standard deviation of 8. These different parameters can make it difficult to compare one individual’s performance across multiple measures. As one author noted regarding scores from the Wechsler Intelligence Test for Children (WISC), “A raw score of 5 on one [sub]test will not have the same meaning as a raw score 5 on another [sub]test” (Brock, 2010).

One way to resolve this interpretation problem is to convert the scores from different distribu- tions into a common metric, or measurement system. If researchers alter scores from different distributions so that they both fit the same distribution, they can compare scores directly. A researcher can compare them directly to determine, for example, on which test an individual scored highest. Such comparisons are one of the purposes of the standard normal distribution.

The standard normal distribution looks like all other normal distributions—from the mean to 11 standard deviation includes 34.13% of the distribution, for example. What separates it from the others is that in the standard normal distribution, the mean is always 0, and the standard deviation is always 1.0 (Figure 3.2). Other distributions may have fixed values for their means and standard deviations, but here 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 is always 1.0.

–3 –2 –1 0 +1 +2 +3

The Mean = 0 The Standard Deviation = 1.0

Hello Lovely/Corbis

When evaluating information about people’s characteristics, keep in mind that data are often normally distributed.

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Section 3.2

The Standard Normal, or z, Distribution Although various normal distributions have different means and standard deviations, they all mirror each other in terms of how much of their populations occur in particular regions. The standard normal distribution’s advantage is that the proportions of the whole that occur in the various regions of the distribution have been calculated. That means that if data from any nor- mal distribution are made to conform to the standard normal distribution, we can answer ques- tions about what is likely to occur in virtually any area of the distribution, such as how likely it is to score 2.5 standard deviations below the mean on a particular test, or what percentage of the entire population will likely occur between two specified points. All such questions can be answered when adapting normal data to the characteristics of the standard normal distribution.

Individual scores in the standard normal distribution are called z scores, which is why the standard normal distribution is often called “the z distribution.” The formula used to turn scores from any normal distribution into scores that conform to the standard normal distri- bution is the z transformation:

Formula 3.1

z 5 x 2 M

where z is a score in the standard normal distribution, x is the score from the original distri- bution (often called a “raw” score), M is the mean of the scores before the original distribu- tion, and s is the standard 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 population from 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 situ- ations where an analyst has access to population 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, the transformation is from data that can have any mean and standard deviation to a distribution where the mean 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 regarding mental health issues that affect the community. Scores on the Summary of WHo’s Apa- thetic Test (the SoWHAT 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 5 14.5 and s 5 6.737.

The Standard Normal Distribution

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–3z –2z –1z 0 +1z +2z +3z

z = – 0.5195

Section 3.2

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

z 5 x 2 M

s 5 11 2 14.5

6.737 5 20.5195

An apathy score of 11 translates into a z score of 20.5195. Because the mean of the z dis- tribution is 0 and the standard deviation in the z distribution is 1.0, where would a score of 20.5195 occur on the horizontal axis of the data distribution? It would be a little over half a standard deviation below the mean, right? Figure 3.3 shows the z distribution and the point about where a raw score of 11 occurs in this 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 does not alter the distribution; it just makes them fit a distribution where the mean is 0 and the standard deviation is 1.0. Evaluating skew and kurtosis must allow the analyst to assume that the data are normal before using the z transformation.

With a mean of 0 in the standard normal distribution, half of all z scores—all the scores below the mean—are going to be negative. A raw score of 11 from the SoWHAT data is lower than the mean, which was M 5 14.5, so it has a negative z value (20.5195).

Besides indicating by its sign whether the z score is above or below the mean, the value of the z score indicates how far from the mean the z score is in standard deviations. If a score had a z value of 1.0, it would indicate that the score is one standard deviation above the mean. The z score for the raw score of 11 was 20.5195, indicating that it is just over half a standard deviation below the mean. This ease of interpretation is one of the great values of z scores: the sign of the score indi- cates whether the associated raw score was above or below the mean, and the value of the score indicates how far from the mean the raw score falls, in standard deviation units (Fischer and Milfont, 2010).

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

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 5 20.5195 is slightly less than one-half a standard deviation below the mean.

–3z –2z –1z 0 +1z +2z +3z

z = – 0.5195

The Standard Normal Distribution

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Section 3.2

Comparing Scores from Different Instruments Consider another application of the standard normal distribution. A counselor has intelli- gence and reading 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 on the 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 be compared directly.

The z for the intelligence score is

z 5 x 2 M

s 5 105 2 100

15 5 0.333

The z for the reading test score is

z 5 x 2 M

s 5 62 2 60

8 5 0.250

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

Expanding the Use of the z Distribution Because the standard normal distribution is a normal distribution, we know that predictable proportions of its population will occur in specific areas. As we noted earlier, however, those proportions are known in great detail for the z distribution because this population is so often used to answer detailed questions about the likelihood of particular outcomes. Table 3.2 indicates how much of the entire population is above or below all of the most commonly occurring values of z. So, by transforming scores from other distributions to fit the z distribu- tion, we can use what we know about this 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 tables indicate the percentage of the population below a point. Some indicate the percentage between a point and the mean of the distribution. Some indicate the probability of scoring in a particular area, and so on. This particular table indicates the proportion of the popula- tion between the specified value of z and the mean of the distribution. (Table 3.2 is listed as Table B.1 in Appendix B.)

The Standard Normal Distribution

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Section 3.2

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 Standard Normal Distribution

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Section 3.2

The z value calculation for a SoWHAT score of 11 rounded to 4 decimal values for the sake of the illustration. The table rounds z values to just two decimals, so from this point forward, round z values to two decimals when using the table. Rounding makes the z value for a raw score of 11 5 20.52.

To interpret the z score, read the whole numbers and the tenths (the tenths are the first value to the right of the decimal) vertically down the left margin of the table. For the hundredths (the second value to 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 of a total of 1.0) of any normally distributed population that will occur between z 5 0.52 and the population’s mean.

4. To determine the percentage of the distribution between z 5 20.52 and the mean, multiply the table value by 100: 100 3 0.1985 5 19.85% of the distribution is between 20.52 and the population mean.

Note that all the z values in Table 3.2 are positive. Our z score from the SoWHAT score was actually negative (z 5 20.52). Since the mean of the standard normal distribution is z 5 0, the z value for any score below the mean will be negative. However, the negative values pose no problem because all normal distributions are symmetrical, so the proportion of a normal population between z 5 20.52 and the mean will be the same as that between z 5 0.52 and the mean. We simply look up the proportion for the appropriate value of z, remembering that when z is negative, it is a proportion to the left of the mean rather than to the right.

To state all this as a principle, because normal distributions are symmetrical, z scores with the same absolute value (the same numbers without regard to the sign) include the same propor- tions between their values and the mean of the distribution. For this reason, the z table indi- cates only the proportions for half the distribution. In the case of Table 3.2, that half is the positive (right) half of the distribution.

Because 50% of the distribution occurs either side of the mean, if 19.85% of the distribution is from a z 5 20.52 back to the mean, the balance of the left (negative) half of the distribution must occur below a z score of 20.52. That proportion is 50 2 19.85 5 30.15%, as the number line illustrates:

50%

30.15 | 19.85% | z 5 20.52 0

Working in the other direction: if the question is what percentage of the population will score 11 or lower on the SoWHAT, the answer is 50 2 19.85 5 30.15%.

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

The Standard Normal Distribution

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Section 3.2

If instead someone asks what the probability of scoring at or below 11 (30.15%) is, we must turn the percentage back into a probability: 30.15 / 100 5 0.3015, or p 5 0.3015 of scoring at or below 11.

Note that the language above is “11 or lower,” and “at or below.” The characteristics of the normal curve allow us to determine the percentage between points, but not at a discrete point. Technically, a particular point has no width 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 distri- bution, we will take a further look at how we can convert scores on opposite sides of the mean and scores with the same 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 might ask what percentage of the entire distribution will score between 5 and 25. Because those were the lowest and highest scores, the answer should be 100%, correct? Remember that the collected data were a 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 prob- able, that someone in the larger population will have a more extreme score. Using the z distri- bution, we can determine 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 5 x 2 M

Allowing that the subscript to each z indicates the raw score and that M 5 14.5 and s 5 6.737 from the sample data produces the following calculations:

z5 5 5 2 14.5

6.737 5 21.410, for which the table value is 0.4207,

which corresponds to a percentage of 42.07% (0.4207 3 100).

z25 5 25 2 14.5

6.737 5 1.559 5 1.56, which has a table value of 0.4406.

Expressed as a percentage, the value is 44.06% (0.4406 3 100).

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

The Standard Normal Distribution

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–3zz values: –2z –1z 0 +1z +2z +3z

About 42%

–1.41 1.56

About 44%

Section 3.2

Adding the two percentages together to determine the total percentage between them pro- duces the following:

42.07 1 44.06 5 86.13% from 5 to 25.

Clearly, these scores do not equal 100%. The results indicate that in the population for which these data are a sample, about 13.87% (100 2 86.13) will score either lower than 5 or higher than 25. Figure 3.4 indicates this result.

Figure 3.4: Areas under the normal curve below z 5 21.41 and beyond z = 1.56

In this distribution, z values that fall below 21.41 or above 11.56 (raw scores below 5 or above 25) are considered extreme scores, comprising only about 13.87% of the population.

–3zz values: –2z –1z 0 +1z +2z +3z

About 42%

–1.41 1.56

About 44%

The answer to this problem underscores two important concepts. First, remember that we are dealing with sample data, and the sample will never exactly duplicate a population. The second, more subtle point reveals that there is no point at which we can be confident that no one will produce a more extreme 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 narrows constantly, the tails never touch the axis (the 50-cent word is that the tails are “asymptotic” to the horizontal 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 on opposite sides of the mean—two z scores where one was positive (z 5 1.56) and the other negative (z 5 21.41). Perhaps the researcher has a question about the percentage of the distribution between SoWHAT scores of 15 and 20. When M 5 14.5, both of these raw scores are higher than the mean and both will result in positive z values. When two z scores have the same sign, determining the percentage of the distribution between them requires that we complete the following steps:

The Standard Normal Distribution

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–3z –2z –1z

26.6%

0 +1z +2z +3z

z = 0.07 0.82

Section 3.2

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 5 x 2 M

z15 5 15 2 14.5

6.737 5 0.0742, or 0.07, for which the table value is 0.0279.

The 0.0279 is the proportion of the distribution from z 5 0.07 and the mean of the distribu- tion. For a raw score of 20,

z20 5 20 2 14.5

6.737 5 0.8164, or 0.82, which corresponds to p 5 0.2939.

This is the proportion of the distribution between z 5 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, deter- mining the proportion of the distribution between them was a simple matter of adding the two table values. When both z scores are on the same side of the distribution, however, their table values overlap. To determine the 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 2 0.0279 5 0.2660. Multiplying that by 100 pro- duces the percentage: 100 3 0.2660 5 26.6% of the 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 by calculating the difference between the smaller z score table value and the larger one, then multiplying the result by 100.

–3z –2z –1z

26.6%

0 +1z +2z +3z

z = 0.07 0.82

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

The Standard Normal Distribution

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Section 3.2

Apply It! Attention to Detail

A psychological services company administers a test that mea- sures the respondent’s attention to detail. The company’s clients are employers in a variety of organizations that require people with good analytical skills. Respondents who score in the low- est ranges of the scale are indifferent to potentially important details. Those who score in the highest ranges tend to fixate on details that may be unimportant to an outcome. Individuals who meet the qualification on this particular test score in the range from 3.80 to 4.30. Data for those who have taken the test in the past indicate that M 5 4.00 and s 5 0.120. For researchers, the initial question is, “Of those who take the test, what proportion are rejected because either they are inattentive to important details or they become focused on the wrong details?” In terms of the z distribution, the equivalent questions are the following:

(a) What proportion of those who took the test in the past failed to meet the minimum qualification for attention to relevant detail? In other words, what proportion scored lower 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 5 3.80

M 5 4.00

s 5 0.12

Since z 5 x 2 M

s 5 (3.80 2 4.00) / 0.120

z3.80 5 21.67

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

0.5 2 0.4525 5 0.0475

In other words, a proportion of 0.0475 occurs below x 5 3.80. Stated as a percentage, 4.75% of the candidates will score below 3.80 on the test.

For the proportion above 4.30,

z4.30 5 (4.30 2 4.00) / 0.12 5 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 the distribu- tion, and

• the percentage above this point will be 50 2 49.38 5 0.62, or 0.62%, of those who take the test score at 4.30 or beyond.

gerenme/iStock/Thinkstock

Apply It! boxes written by Shawn Murphy

The Standard Normal Distribution

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Section 3.2

Comparing Data from Different Tests Earlier chapters discussed how test scores from two different instruments with different means and standard deviations can be compared. Perhaps a juvenile gang member under court-ordered counseling is 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 the means 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 directly is not helpful. However, employing the z transformation allows both scores to fit a distribution where the mean is 0 and the standard deviation is 1.0. The raw scores may not reveal much, the z scores can be directly compared. Recall that

z 5 x 2 M

Calculating z for the aggression score produces:

z39 5 39 2 32.554

5.824 5 1.107

Then calculate the z for social alienation:

z15 5 15 2 12.917

2.674 5 0.779

Interpreting Multiple z Values Since the question is which of the juvenile’s two test scores is the more extreme, we have no need for table values—only the value of z. As both z values are positive, the z for aggression is more extreme than that for social alienation. Performing the z transfor- mation allows us to note that the aggression value is 1.107 standard deviations from the mean of the distribution. Alienation, meanwhile, is just 0.779 standard deviations from its mean. Practically, the results show this individual is more aggressive than alienated. As long as raw scores, means, and stan- dard deviations are available, researchers can use

Doug Menuez/Photodisc/Thinkstock

Using z scores enables researchers to better understand test results measuring aggression and social alienation in juvenile gang members.

The Standard Normal Distribution

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Section 3.2

z to make direct comparison of very different qualities, in this case, aggression and social alienation in the same individual.

Another Comparison Psychologist Lewis Terman developed the Stanford-Binet test, which measures children’s intel- ligence. Suppose a psychologist is similarly interested in giftedness among children. Because unusual verbal ability often seems to accompany superior intelligence in gifted children, the psychologist measures both characteristics for a group of subjects. One particular subject scores 140 on intelligence and 55.0 on 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 be directly compared.

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

z 5 x 2 M

z140 5 140 2 100

15 5 2.667

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

z 5 x 2 M

z55 5 55 2 40

5.451 5 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 both have been transformed into z scores, the two measures now belong to a com- mon distribution, and the researcher 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 the Curve If we draw a distribution and clarify what is at issue, questions about how much of the distri- bution is above 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 for answering different questions a researcher might ask about pro- portions within a distribution.

The Standard Normal Distribution

tan82773_03_ch03_061-090.indd 76 3/3/16 9:59 AM

Scores on opposite sides of M

Scores on same side of M

Calculate both zs and locate

table values

Calculate both zs and locate

table values

Subtract both values from .050 and sum

Subtract the smaller value from the larger

and subtract the result from 1.0

Positive z? Add the table value to 0.50

Negative z? Subtract the table value from 0.50

Scores on opposite sides of the mean

Calculate both zs and locate table values

Subtract the smaller value from the larger

Sum the table values

Calculate both zs and locate table values

Scores on the same side of the mean

Below a score

Determine the z value

Above a score

Determine the z value

Locate the proportion in Table 3.2Locate the proportion in Table 3.2

Positive z? Subtract the table value from 0.50

Negative z? Add the table value to 0.50

Questions about the proportions of a population

Between two scores Above and below two scores

Section 3.2

The list of steps must seem like a great deal to remem- ber. In fact, the better course when confronted with a z score problem is to sketch out a distribution to pro- duce something like Figures 3.3 and 3.4. The visual displays help clarify the question and suggest the steps needed to answer it.

Figure 3.6: Flowchart to address questions pertaining to a distribution

Use the steps illustrated in the flowchart to resolve questions about the proportions within a population.

Scores on opposite sides of M

Scores on same side of M

Calculate both zs and locate

table values

Calculate both zs and locate

table values

Subtract both values from .050 and sum

Subtract the smaller value from the larger

and subtract the result from 1.0

Positive z? Add the table value to 0.50

Negative z? Subtract the table value from 0.50

Scores on opposite sides of the mean

Calculate both zs and locate table values

Subtract the smaller value from the larger

Sum the table values

Calculate both zs and locate table values

Scores on the same side of the mean

Below a score

Determine the z value

Above a score

Determine the z value

Locate the proportion in Table 3.2Locate the proportion in Table 3.2

Positive z? Subtract the table value from 0.50

Negative z? Add the table value to 0.50

Questions about the proportions of a population

Between two scores Above and below two scores

Try It!: #4 Figuratively speaking, how does the z transformation allow you to compare apples to oranges?

The Standard Normal Distribution

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–3z –2z –1z 0 +1z +2z +3z

z = 2.33

proportion = 0.01

Section 3.3 z Scores, Percentile Ranks, and Other Standard Scores

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 percent- ages or proportions of the distribution in specified areas. If the percentages are already avail- able, but neither the raw data nor the related descriptive statistics are, Table 3.2 (the z table) allows us to work backward to determine the z value—even without the mean and standard deviation for the data.

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

1. Because Table 3.2 lists proportions, the first step is to turn the percentage into a proportion: 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 need a z value which includes all but that most extreme 0.01, which will be the z value for a proportion of 0.50 2 0.01 5 0.49. A z value for a proportion of 0.49, will be the value that includes the 49% of the distribution, which means it excludes the highest 1% of the distribution.

3. Table 3.2 does not list a proportion of exactly 0.49, but it does list 0.4901, which is very close. Reading leftward from the proportion to the margin and also vertically to the column heading, the associated z value for 0.4901 is 2.33. If data were gathered for intelligence scores, a z 5 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 is somewhere among the top 1% of all intelligence scores. Figure 3.7 illustrates this proportion.

Figure 3.7: The value of z associated with a particular proportion

A normal distribution curve that shows where the highest 1% of scores fall within a given population.

–3z –2z –1z 0 +1z +2z +3z

z = 2.33

proportion = 0.01

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–3 –2 –1 0 +1 +2 +3 Percentiles

z scores

<1st 2nd 16th 50th 84th 98th >99th

Section 3.3 z Scores, Percentile Ranks, and Other Standard Scores

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

• z 5 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% 1 50% 5 84.13% of scores are at or below z 5 1.0; therefore, z 5 1.0 occurs

at the 84th percentile.

Although percentile scores can be easily determined from the table values that are associ- ated with z scores, note an important difference between percentile scores and z scores. The z score is one of several standard scores. Standard scores are all equal-interval scores—the interval 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 5 21.5 to z 5 21.0 is the same as it is from z 5 0.3 to z 5 0.8. The increase is 0.5 in either case.

This interval scale does not apply to percentile scores. Because these scores indicate the per- centage of scores below a point rather than reflecting a direct measure of some characteristic, the distances between consecutive scores differ widely in various parts of the distribution. Most of the data in any normal distribution are in the middle portion, where scores have the greatest frequency. The frequency with which scores occur diminishes as scores become more distant from the mean, something reflected in the curves in frequency distributions that are vertically highest in the middle and then decline as they extend outward to 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 that the majority of scores are found within the 50th percentile. Meanwhile, the frequency of scores above the 99th and below the 1st percentiles is low in a normal distribution.

–3 –2 –1 0 +1 +2 +3 Percentiles

z scores

<1st 2nd 16th 50th 84th 98th >99th

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Section 3.3

As a result of high frequency in the middle of the distribution, in any normal distribution the difference between consecutive percentile scores is always much smaller near the middle of the distribution (between the 50th and 51st percentiles, for example) than between consecu- tive percentile scores in the tails (between the 10th and 11th, or the 90th and 91st percen- tiles, for example). This characteristic has important implications. The difference between scoring at the 50th and 51st percentile score on something like the Beck Depression Inven- tory is almost inconsequential compared to the difference between the 90th and 91st per- centile, a much greater difference. Percentile scores are ordinal scale, whereas z scores are interval scale.

Converting z to Other Standard Scores Part of the appeal of the z score is that it enables researchers to readily determine rela- tive performance. A positive z value indicates that the individual has scored in the upper half of the distribution. Someone who scores one standard deviation beyond the mean, as we noted earlier, has scored at the 84th percentile, and so on. The z scores belong to a family of measures called “standard scores.” They have in common these characteristics: a) a fixed mean 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 reporting them 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 true for 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 a fixed 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

Either score can be derived from z. To convert from z to t, for example, simply multiply z by 10 and add 50. So, if z 5 1.75, then

t 5 10 3 1.75 5 17.5 1 50 5 67.5

z 5 1.75 is the same as t 5 67.5.

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

GRE 5 100 3 1.75 5 175.0 1 500 5 675

z 5 1.75 is the same as GRE5 675 (and as t 5 67.5).

z Scores, Percentile Ranks, and Other Standard Scores

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Section 3.4 Using Excel to Perform the z Score Transformation

Although more common in educational than in psycho- logical testing and research, normal curve equivalent scores (NCE) and “stanine” scores (standard nine-point scale) are also examples of standard scores. Like z and t, each is equal-interval, and both have fixed means and standard deviations.

3.4 Using Excel to Perform the z Score Transformation The z score transformation is a fairly simple for- mula. As a result, to program it into Excel and trans- form an entire data set into z scores is not difficult. In fact, the application offers several ways to do this, but this chapter will explore just one. It involves programming the z score transformation formula directly into the data sheet.

A researcher interested in the relationship between poverty and achievement motivation among secondary-school-aged young people gathers data from a group of students whose families qualify for free and reduced-price lunches at school. The achievement motivation scores 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,

including Excel, 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, for example, allows users to enter the data, position the cursor where they desire the statistic’s value to appear, and then dou- ble-click the name of the desired statistic under the Formula tab.)

a. The equal sign indicates to Excel that a formula follows. b. The command average will provide the arithmetic mean. c. When several cells are to be included in the function, they are placed in paren-

theses ( ). When the cells are consecutive, the colon (:) indicates that all cells from B2 to B11 are to be included in the function.

4. Press Enter. 5. In cell A12, enter the label “mean 5.” The value in cell B12 will be 8.1, the mean of

the achievement motivation scores.

monkeybusinessimages/iStock/Thinkstock

Evaluating achievement motivation scores can give researchers valuable information about the relationship between poverty and achievement in schools.

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

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Section 3.4 Using Excel to Perform the z Score Transformation

6. 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.

7. Press Enter. 8. In cell A13, enter the label std dev =. The value in cell B13 will be 2.558211, the

standard deviation of the scores. 9. In cell C1, enter the label equiv z.

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

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

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

b. In the Editing section at the top of the page near the right side is a Fill command with a down-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 will repeat the result in C2 for the other nine cells, adjusting for the different test scores in each cell.

Figure 3.9 shows how the spreadsheet will look after Step 11, with the z score equivalents of all the original 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.

Source: Microsoft Excel. Used with permission from Microsoft.

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Section 3.5 Using z Scores to Determine Other Measures

3.5 Using z Scores to Determine Other Measures Occasionally, a researcher has access to z scores and the mean and standard deviation but not the original raw scores. Formula 3.1 used the raw score (x), the mean (M), and the standard deviation (s) to determine the value of z, but actually, any three of the val- ues in the formula can be used to determine the value of the fourth. Just as Formula 3.1 uses x, M, and s to determine z, we could use z, M, and s to derive x. Altering Formula 3.1 to determine the value 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 5 (x 2 M)/s, swap the terms before and after the equal sign so that (x 2 M)/s 5 z.

2. To eliminate the s in the denominator of the first term, multiply both sides by s so that it disappears from the first term and emerges in the second: x 2 M 5 sz.

3. To isolate x, add M to both sides of the equation so that x 5 sz 1 M.

Returning to the Excel problem, if the z scores and descriptive statistics are available, we can determine the raw score for which z 5 21.603 as follows:

If M 5 8.10, s 5 2.558, and x 5 z 3 s 1 M, substituting the values we have produces x 5 (21.603)(2.558) 1 8.10 5 3.9995, which rounds to 4.0.

Checking the earlier data reveals that 4 was indeed the raw score for which z 5 21.603.

Determining the Standard Deviation If the raw scores, the mean, and z are available, but s is lacking, z 5 (x 2 M)/s, so (x 2 M)/s 5 z. Taking the reciprocal of each half of the equation—which means inverting the term so that (x 2 M)/s becomes s/(x 2 M) and z/1 becomes 1/z, giving us s/(x 2 M) 5 1/z. Multiplying both sides by (x 2 M) yields s 5 (x 2 M).

Using the data from the Excel problem again, for the first participant, x 5 4, M 5 8.10, and z 5 21.603. According to the adjusted formula,

s 5 (x 2 M)/z

therefore, substituting the values we have produces the following:

s 5 4 2 8.1 21.603 5 2.5577

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

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Section 3.5 Using z Scores to Determine Other Measures

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

Because

z 5 (x 2 M)/s

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

sz 5 x 2 M

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

sz 1 M 5 x.

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

M 5 x 2 sz

For the first participant, the achievement motivation score x 5 4.0, the z score 5 21.603, and s 5 2.558. The mean for the test can be determined as follows:

M 5 x 2 sz, or

M 5 4 2 (2.558 3 21.603) 5 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 standard devia- tion values remain the same over time. The major intelligence tests, for example, have a fixed mean of 100 and a standard deviation of 15, even though the Stanford-Binet and Wechsler tests have been revised several times. When a test is revised and updated, do the means and standard devia- tions likewise change? In fact, they do. Flynn and Weiss (2007) documented significant increases in intelligence scores over a 70-year period, but to make the scores comparable over time, psy- chologists use what are called modified standard scores. The modified standard score allows those working with the test to gather data that have any mean and any standard deviation and then adjust them so that they conform to predetermined values. 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.

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Section 3.5 Using z Scores to Determine Other Measures

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

Formula 3.2

MSS 5 (sspec)(z)1 Mspec

where

MSS 5 the modified standard score,

sspec 5 the specified standard deviation, and

Mspec 5 the specified mean.

Note that this formula is the same used to transform z scores into t scores. By way of an example, perhaps a psychologist has developed what she has labeled the Brief Intelligence Test (BIT). To compare results to tests her colleagues have used traditionally, she wants the BIT’s descriptive characteristics to conform to those of the more established tests. For eight participants, 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 we will later call sampling error is too great. Still, to illustrate the process, we will assume the sample scores are valid.

Verify that M 5 28.75 and s 5 4.268.

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

z 5 x 2 M

s 5 22 2 28.75

4.268 5 21.582

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

MSS 5 (sspec)(z) 1 Mspec 5 (15)(51.582) 1 100

5 76.276

Although the original BIT score was 22, using the z transformation and modified standard score procedures makes the BIT score conform to the mean and standard deviation of a more established 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.

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Summary and Resources

Writing Up Statistics Although z scores are an important part of data analysis, like the raw scores that researchers gather in their work, z scores often do not appear in research reports. Reports list the means and standard deviations, but often omit the raw scores and their z score transformation. In a study 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 was probably 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 the weight-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 that grade inflation has on college students’ class rankings. He posited that class rankings are less informative than they once were because weaker students in departments where coursework is easier are ranked ahead of students who have a higher level of academic aptitude but com- pete in more demanding programs. Brown’s solution was to use the z transformation within departments to indicate how much above or below the mean students were in their indi- vidual programs.

Summary and Resources

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

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

• For example, when someone scores at a particular level, we can ask what proportion of the 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 ask what proportion of the entire population will score between (or outside) those points.

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

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Summary and Resources

modified standard scores Standard scores adjusted to reflect a specified mean and standard deviation.

percentile A value below which a certain percentage of all scores in a distribution may be found; 37% of all scores occur below the 37th percentile.

probability The measure of the likelihood that an event will occur. The values range from p 5 0, for events that never occur, to p 5 1.0, for events that occur every time.

standard normal distribution A normal distribution in which µ 5 0 and s 5 1.0.

standard scores Normally distributed, equal-interval scores that have a fixed mean and standard deviation.

t score A standard score based on a normal distribution in which M 5 50 and s 5 10. They are sometimes preferred to z scores because they rarely involve negative values.

z score The score that results when scores from any source are made to conform to the characteristics of the standard normal, or z, distribution. The z distribution has M 5 0 and s 5 1.

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

Key Terms

The z is one of several standard scores in fairly common use. Whether z or some other, all standard scores indicate how distant one individual’s score is from the mean of the distribu- tion. Rather than providing an absolute measure of some characteristic, standard scores are normative, meaning that they indicate the level of what is measured relative to others in the same population. Those who prefer not to deal in negative values (which characterize half of the z distribution) can employ t scores. In all material respects, t is the same 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 communi- cate an individual’s standing relative to a population. Researchers often use standard scores to report the data from standardized tests, but these tests are revised from time to time, which can affect the test means and standard deviations. To ensure stability, the modified standard score uses the z transformation as a way to maintain constant descriptive char- acteristics, even as 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 are a prelude to Chapter 4. With all our effort to label, display, and describe data sets, the focus in the discussion of z scores and the other topics has been primarily about analyzing the performance of individuals. Behavioral scientists, however, are generally much more interested in asking questions about groups. Analyzing how those in a sample compare to those in the entire population is the focus of Chapter 4. It will do so by expanding discussion of the z distribution.

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

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Summary and Resources

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? b. Will a z score for a raw score of 35 be positive or negative? How do you know? c. How many standard deviations is a score of 28 from the mean? d. What will be the z value of a raw score of 33.1? e. Since there are no scores below 28, does it make any sense to calculate z for

a raw score of 25, for example? Shouldn’t such a score have a zero probability of occurring?

2. Examining the relationship between recreational activity and level of optimism among senior citizens, a psychologist develops the Recreation Activity Test (RAT). Scores for 8 participants are as follows:

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

a. What is the z score for someone with RAT 5 23? b. Why isn’t 0 the answer to 2a, since none of the participants scored 23? c. What is the z score for someone with RAT 5 15.125? d. Explain the answer to 2c. e. How does z allow one to compare tests with entirely different means and standard

deviations?

3. One participant has RAT 5 11. The same individual is administered a Consistent Approval Test (CAT) and scores 45. The CAT data, including that participant’s score, are as follows:

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

a. Which score is higher, the RAT or the CAT? b. Why isn’t the answer to 3a automatically CAT, since it has the higher mean value?

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

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

a. What proportion of the population will score 81 or higher? b. What proportion will score 60 or higher? c. 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?

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Summary and Resources

5. Using the data in Question 4, what percentage of the population will score lower than 55?

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

a. What is the mean of the t distribution? b. Why is t sometimes preferred over z? c. If z 5 2.5, what is t?

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

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

a. What percentage of scores will occur below z 5 0? b. What is the probability of a positive value of z? c. What percentage of scores will occur between ±1.96 z?

9. If someone scores z 5 1.0, what is the corresponding percentile rank?

10. What percentile rank is z 5 0? What measure of central tendency represents the 50th percentile?

11. A psychologist wishes to maintain a mean of 25 and a standard deviation of 5 for a test developed to measure compulsive behavior. On a revised test, the scores are as follows:

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

a. What is the modified standard score for the person who scored 17 on the revised instrument?

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

12. Given the data in Question 11,

a. What is the z equivalent of a raw score of 28? b. What is the probability of scoring somewhere from 14 to 29?

13. Draw a normal distribution and identify where z 5 21.17 and z 5 12.53 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 (the denominator in z) from the mean.

2. Because the standard normal distribution (the z distribution) is normal, the distri- bution is symmetrical. The proportion of the distribution between a value of z and

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Summary and Resources

the mean will 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 5 3.09, but in fact z has no upper limit. In theory, the tails in the z distribution never actually touch the horizontal axis of the graph, which means that there exists always at least the possibility of scores higher (or lower) than any already measured.

4. One of the values of the z transformation is that scores that have any descriptive characteristics can be recalibrated so that they fit a distribution where the mean is 0 and the standard deviation is 1.0. By doing so, scores from any variety of sources can be compared directly after converting them to z scores. The only requirement is that they be 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 to all others for whom scores are available.

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