A measure of central tendency provides information about the “middleness” of a distribution of scores, but not about the width or spread of the distribution. To assess the width of a distribution, we need a measure of variability or dispersion. A measure of variation indicates how scores are dispersed around the mean of the distribution. As an illustration, consider the two very small distributions represented in Table 5.1 . Each represents a small distribution of exam scores. Notice that the mean for each distribution is the same. If these data represented two very small classes of students, reporting that the two classes had the same mean on the exam might lead one to conclude that they performed essentially the same. Notice, however, how different the distributions are. Providing a measure of variability along with a measure of central tendency would convey the information that even though the distributions have the same mean, the spread of the distributions is very different.

measure of variation A number that indicates how dispersed scores are around the mean of the distribution.

We will discuss three measures of variability: the range, the average deviation, and the standard deviation. The range can be used with any type of data. However, the standard deviation and average deviation are appropriate only for interval and ratio data.

TABLE 5.1 Two distributions of exam scores

CLASS 1	CLASS 2
0	45
50	50
100	55
Σ = 150	Σ = 150
μ = 50	μ = 50

Range

The simplest measure of variation is the range —the difference between the lowest and the highest score in a distribution. To find the range, simply subtract the lowest score from the highest score. In our hypothetical distributions of exam scores, the range for Class 1 is 100 points, whereas the range for Class 2 is 10 points. Thus, the range provides some information concerning the difference in the spread of the distributions. In this simple measure of variation, however, only the highest and lowest scores enter the calculation, and all other scores are ignored. For example, referring back to Module 4 and the distribution of 30 exam scores in Table 4.1 , only 2 of the 30 scores would be used in calculating the range (95 − 45 = 50). Thus, the range is easily distorted by one unusually high or low score in a distribution.

range A measure of variation; the difference between the lowest and the highest scores in a distribution.

Average Deviation and Standard Deviation

More sophisticated measures of variation use all of the scores in the distribution in their calculation. The most commonly used measure of variation is the standard deviation. Most people have heard this term before and may even have calculated a standard deviation if they have taken a statistics class. However, many people who know how to calculate a standard deviation do not really appreciate the information it provides.

To begin, let's think about what the phrase standard deviation means. Other words that might be substituted for the word standard include average, normal, or usual. The word deviation means to diverge, move away from, or digress. Putting these terms together, we see that the standard deviation means the average movement away from something. But what? It is the average movement away from the center of the distribution—the mean.

The standard deviation , then, is the average distance of all of the scores in the distribution from the mean or central point of the distribution—or, as we shall see shortly, the square root of the average squared deviation from the mean. Think about how you would calculate the average distance of all of the scores from the mean of the distribution. First, you would have to determine how far each score is from the mean; this is the deviation, or difference, score. Then, you would have to average these scores. This is the basic idea behind calculating the standard deviation.

standard deviation A measure of variation; the average difference between the scores in the distribution and the mean or central point of the distribution, or more precisely, the square root of the average squared deviation from the mean.

The data from Table 4.1 are presented again in Table 5.2 . Let's use these data to calculate the average distance from the mean. We will begin with a calculation that is slightly simpler than the standard deviation, known as the average deviation . The average deviation is essentially what the name implies—the average distance of all of the scores from the mean of the distribution. Referring to Table 5.2 , you can see that we begin by determining how much each score deviates from the mean, or

average deviation An alternative measure of variation that also indicates the average difference between the scores in a distribution and the mean of the distribution.

X − μ

where

X = each individual score, and

μ = the population mean

Then we need to sum the deviation scores. Notice, however, that if we were to sum these scores, they would add to zero. Therefore, we first take the absolute value of the deviation scores (the distance from the mean, irrespective of direction), as shown in the third column of Table 5.2 . To calculate the average deviation, we sum (Σ) the absolute value of each deviation score:

∑|X−μ|∑|X−μ|

TABLE 5.2 Calculations for the sum of the absolute values of the deviation scores (μ = 74)

X	X − μ	\|X − μ\|
45	−29.00	29.00
47	−27.00	27.00
54	−20.00	20.00
56	−18.00	18.00
59	−15.00	15.00
60	−14.00	14.00
60	−14.00	14.00
63	−11.00	11.00
65	−9.00	9.00
69	−5.00	5.00
70	−4.00	4.00
74	0.00	0.00
74	0.00	0.00
74	0.00	0.00
75	1.00	1.00
76	2.00	2.00
77	3.00	3.00
78	4.00	4.00
78	4.00	4.00
80	6.00	6.00
82	8.00	8.00
82	8.00	8.00
85	11.00	11.00
86	12.00	12.00
87	13.00	13.00
90	16.00	16.00
92	18.00	18.00
93	19.00	19.00
94	20.00	20.00
95	21.00	21.00
	332.00 = Σ\|X − μ\|

Then we divide by the total number of scores (N) to find the average deviation:

AD=∑|X−μ|NAD=∑|X−μ|N

Using the data from Table 5.2 , we would calculate the average deviation as follows:

AD=∑|X−μ|N=33230=11.07AD=∑|X−μ|N=33230=11.07

Thus, for the exam score distribution, the scores fall an average of 11.07 points from the mean of 74.00.

Although the average deviation is fairly easy to compute, it is not as useful as the standard deviation because, as we will see in later modules, the standard deviation is used in many other statistical procedures.

The standard deviation is very similar to the average deviation. The only difference is that rather than taking the absolute value of the deviation scores, we use another method to “get rid of” the negative deviation scores—we square the deviation scores. This procedure is illustrated in Table 5.3 . Notice that this table is very similar to Table 5.2 . It includes the distribution of exam scores, the deviation scores, and the squared deviation scores. The formula for the standard deviation is:

σ=√∑(X−μ)2Nσ=∑(X−μ)2N

where

X = each individual score

μ = the population mean

N = the total number of scores, and

Σ = summation of

This formula represents the standard deviation for a population. The symbol for the population standard deviation is σ (pronounced “sigma”). To derive the standard deviation for a sample, the calculation is the same but the symbols differ. This will be discussed later in the module.

Notice that the formula is similar to that for the average deviation. We determine the deviation scores, square the deviation scores, sum the squared deviation scores, and divide by the number of scores in the distribution. Lastly, we take the square root of that number. Why? Squaring the deviation scores has inflated them. We now need to bring the squared deviation scores back to the same level of measurement as the mean so that the standard deviation is measured on the same scale as the mean.

TABLE 5.3 Calculations for the sum of the squared deviation scores

X	X − μ	(X − μ)2
45	−29.00	841.00
47	−27.00	729.00
54	−20.00	400.00
56	−18.00	324.00
59	−15.00	225.00
60	−14.00	196.00
60	−14.00	196.00
63	−11.00	121.0
65	−9.00	81.00
69	−5.00	25.00
70	−4.00	16.00
74	0.00	0.00
74	0.00	0.00
74	0.00	0.00
75	1.00	1.00
76	2.00	4.00
77	3.00	9.00
78	4.00	16.00
78	4.00	16.00
80	6.00	36.00
82	8.00	64.00
82	8.00	64.00
85	11.00	121.00
86	12.00	144.00
87	13.00	169.00
90	16.00	256.00
92	18.00	324.00
93	19.00	361.00
94	20.00	400.00
95	21.00	441.00
5,580.00 = Σ(X − μ)2

Now, using the sum of the squared deviation scores (5,580.00) from Table 5.3 , we can calculate the standard deviation:

σ=√∑(X−μ)2N=√5,580.0030=√186.00=13.64σ=∑(X−μ)2N=5,580.0030=186.00=13.64

Compare this number to the average deviation calculated on the same data (AD = 11.07). The standard deviation tells us that the exam scores fall an average of 13.64 points from the mean of 74.00. The standard deviation is slightly larger than the average deviation of 11.07 and will always be larger whenever both of these measures of variation are calculated on the same distribution of scores. Can you see why? It is because we are squaring the deviation scores and thus giving more weight to those that are farther from the mean of the distribution. The scores that are lowest and highest will have the largest deviation scores; squaring them exaggerates the difference. When all of the squared deviation scores are summed, these large scores will contribute disproportionately to the numerator and, even after dividing by N and taking the square root, will result in a larger number than what we see for the average deviation.

If you have taken another statistics class, you may have used the “raw-score formula” to calculate the standard deviation. The raw-score or computational formula is shown in Table 5.4 , where it is used to calculate the standard deviation for the same distribution of exam scores. The numerator represents an algebraic transformation from the original formula that is somewhat shorter to use. Although the raw-score formula is slightly easier to use, it is more difficult to equate this formula with what the standard deviation actually is—a means of determining the average deviation (or distance) from the mean for all of the scores in the distribution.

As mentioned previously, the calculation of the standard deviation for a sample (S) differs from the calculation for the standard deviation for a population (σ) only in the symbols used to represent each term. The formula for a sample is:

S=√∑(X−¯¯¯X)2NS=∑(X−X¯)2N

where

S = sample standard deviation

X = each individual score

¯¯¯X=samplemeanX¯=sample mean

N = number of scores in the distribution

Σ = summation of

TABLE 5.4 Standard deviation raw-score formula

σ=

⎷∑X2−(∑X)2NN=√169,860−(2,220)23030=√169,860−4,928,4003030=√169,860−164,28030=√5,580.0030=√186.00=13.64σ=∑X2−(∑X)2NN=169,860−(2,220)23030=169,860−4,928,4003030=169,860−164,28030=5,580.0030=186.00=13.64

Note that the only difference is in the notation for the mean (¯¯¯XX¯ rather than μ). This difference simply reflects the scientific notation for the population mean versus the sample mean. However, the calculation is exactly the same as that for σ. Thus, if we were to use the data set from Table 5.3 to calculate S, we would arrive at exactly the same answer as we did for σ, 13.64.

If, however, you are using sample data to estimate the population standard deviation, then the standard deviation formula must be slightly modified. The modification provides what is called an “unbiased estimator” of the population standard deviation based on sample data. The modified formula is:

S=√∑(X−¯¯¯X)2N−1S=∑(X−X¯)2N−1

where

s = unbiased estimator of population standard deviation

X = each individual score

¯¯¯X=samplemeanX¯=sample mean

N = number of scores in the distribution

Σ = summation of

Notice that the symbol for the unbiased estimator of the population standard deviation is s (lowercase), whereas the symbol for the sample standard deviation is S. The main difference, however, is in the denominator—dividing by N − 1 versus N. The reason is that the standard deviation within a small sample may not be representative of the population; that is, there may not be as much variability in the sample as there actually is in the population. We therefore divide by N − 1, because dividing by a smaller number increases the standard deviation and thus provides a better estimate of the population standard deviation.

We can use the formula for s to calculate the standard deviation on the same set of exam score data. Before we even begin the calculation, we know that because we are dividing by a smaller number (N − 1), s should be larger than σ and S (which were 13.64 for the same distribution of scores). Normally we would not compute both σ and s on the same distribution of scores because σ is the standard deviation for the population and s is the unbiased estimator of the population standard deviation based on sample data. I am doing so here simply to illustrate the difference in the formulas.

s=√∑(X−¯¯¯X)2N−1=√5,580.0030−1=√5,580.0029=√192.41=13.87s=∑(X−X¯)2N−1= 5,580.0030−1=5,580.0029=192.41=13.87

Note that s (13.87) is slightly larger than σ and S (13.64). The procedures for calculating σ, S, and s using Excel, SPSS, and the TI-84 calculator are shown in the Statistical Software Resource Section at the end of this chapter.

One final measure of variability is called the variance. The variance is equal to the standard deviation squared. Thus, the variance for a population would be σ2 and for a sample, S2. Because variance is not as useful a descriptive statistic as the standard deviation, we will not discuss it here. We will see, however, that it is used in more advanced statistical procedures presented later in the text.

The formulas for the average deviation, standard deviation, and variance all use the mean. Thus, it is appropriate to use these measures with interval or ratio data, but not with ordinal and nominal data.

MEASURES OF VARIATION

TYPE OF VARIATION MEASURE
	Range	Average Deviation	Standard Deviation
Definition	The difference between the lowest and highest scores in the distribution	The average distance of all of the scores from the mean of the distribution	The square root of the average squared deviation from the mean of a distribution
Use with	Primarily interval and ratio data, but can be used with any type of data	Only interval and ratio data	Only interval and ratio data
Caution	A simple measure that does not use all scores in the distribution in its calculation	A more sophisticated measure in which all scores are used, but which may not weight extreme scores adequately	The most sophisticated and most frequently used measure of variation

1.For a distribution of scores, what information does a measure of variation add that a measure of central tendency does not convey?

2.Today's weather report included information on the normal rainfall for this time of year. The amount of rain that fell today was 1.5 inches above normal. To decide whether this is an abnormally high amount of rain, you need to know that the standard deviation for rainfall is .75 inches. What would you conclude about how normal the amount of rainfall was today? Would your conclusion be different if the standard deviation were 2 inches rather than .75 inches?

3.Draw two distributions with the same mean but different standard deviations in one graph. Draw a second set of distributions on another graph with different means but the same standard deviation.

Types of Distributions

In addition to knowing the central tendency and width or spread of a distribution, it is also important to know about the shape of the distribution.

Normal Distributions

When a distribution of scores is very large, it tends to approximate a pattern called a normal distribution. When plotted as a frequency polygon, a normal distribution forms a symmetrical, bell-shaped pattern often called a normal curve (see Figure 5.1 ). We say that the pattern approximates a normal distribution because a true normal distribution is a theoretical construct not actually observed in the real world.

normal curve A symmetrical, bell-shaped frequency polygon representing a normal distribution.

The normal distribution is a theoretical frequency distribution that has certain special characteristics. First, it is bell-shaped and symmetrical—the right half is a mirror image of the left half. Second, the mean, median, and mode are equal and are located at the center of the distribution. Third, the normal distribution is unimodal—it has only one mode. Fourth, most of the observations are clustered around the center of the distribution, with far fewer observations at the ends, or “tails,” of the distribution. Lastly, when standard deviations are used on the x-axis, the percentage of scores falling between the mean and any point on the x-axis is the same for all normal curves. This important property of the normal distribution will be discussed more fully in Module 6 .

normal distribution A theoretical frequency distribution having certain special characteristics.

Kurtosis

Although we typically think of the normal distribution as being similar to the curve depicted in Figure 5.1 , there are variations in the shape of normal distributions. Kurtosis refers to how flat or peaked a normal distribution is. In other words, kurtosis refers to the degree of dispersion among the scores, or whether the distribution is tall and skinny or short and fat. The normal distribution depicted in Figure 5.1 is called mesokurtic—meso means “middle.” Mesokurtic curves have peaks of medium height and the distributions are moderate in breadth. Now look at the distributions depicted in Figure 5.2 .

kurtosis How flat or peaked a normal distribution is.

mesokurtic Normal curves that have peaks of medium height and distributions that are moderate in breadth.

The normal distribution on the left is the same mesokurtic distribution depicted in Figure 5.1 . The distribution in the middle is leptokurtic—lepto means “thin.” Leptokurtic curves are tall and thin, with only a few scores in the middle of the distribution having a high frequency. Last, see the curve on the right side of Figure 5.2 . This is a platykurtic curve—platy means “broad” or “flat.” Platykurtic curves are short and more dispersed (broader). In a platykurtic curve, there are many scores around the middle score that all have a similar frequency.

leptokurtic Normal curves that are tall and thin, with only a few scores in the middle of the distribution having a high frequency.

platykurtic Normal curves that are short and more dispersed (broader).

FIGURE 5.1 A normal distributionFIGURE 5.2 Types of distributions: Mesokurtic, Leptokurtic, and platykurtic

Positively Skewed Distributions

Most distributions do not approximate a normal or bell-shaped curve. Instead, they are skewed, or lopsided. In a skewed distribution, scores tend to cluster at one end or the other of the x-axis, with the tail of the distribution extending in the opposite direction. In a positively skewed distribution , the peak is to the left of the center point and the tail extends toward the right, or in the positive direction. (See Figure 5.3 .)

positively skewed distribution A distribution in which the peak is to the left of the center point and the tail extends toward the right or in the positive direction.

Notice that what is skewing the distribution, or throwing it off center, are the scores toward the right or positive direction. A few individuals have extremely high scores that pull the distribution in that direction. Notice also what this does to the mean, median, and mode. These three measures do not have the same value, nor are they all located at the center of the distribution as they are in a normal distribution. The mode—the score with the highest frequency—is the high point on the distribution. The median divides the distribution in half. The mean is pulled in the direction of the tail of the distribution; that is, the few extreme scores pull the mean toward them and inflate it.

Negatively Skewed Distributions

The opposite of a positively skewed distribution is a negatively skewed distribution —a distribution in which the peak is to the right of the center point and the tail extends toward the left or in the negative direction. The term negative refers to the direction of the skew. As can be seen in Figure 5.3 , in a negatively skewed distribution, the mean is pulled toward the left by the few extremely low scores in the distribution. As in all distributions, the median divides the distribution in half, and the mode is the most frequently occurring score in the distribution.

negatively skewed distribution A distribution in which the peak is to the right of the center point and the tail extends toward the left or in the negative direction.

FIGURE 5.3 Positively and negatively skewed distributions

Knowing the shape of a distribution provides valuable information concerning the distribution. For example, would a class of students prefer to have a negatively or positively skewed distribution of exam scores for an exam they have taken? Students will frequently answer that they would prefer a positively skewed distribution, because they think the term positive means good. Keep in mind that positive and negative describe the skew of the distribution, not whether the distribution is “good” or “bad.” Assuming that the exam scores span the entire possible range (say from 0 to 100), you should prefer a negatively skewed distribution—meaning that most people have high scores and only a few people have low scores.

Another example of the value of knowing the shape of a distribution is provided by Harvard paleontologist Stephen Jay Gould (1985). Gould was diagnosed in 1982 with a rare form of cancer. He immediately began researching the disease and learned that it was incurable and had a median mortality rate of only 8 months after discovery. Rather than immediately assuming that he would be dead in 8 months, Gould realized this meant that half of the patients lived longer than 8 months. As he was diagnosed with the disease in its early stages and was receiving high-quality medical treatment, he reasoned that he could expect to be in the half of the distribution that lived beyond 8 months. The other piece of information that Gould found encouraging was the shape of the distribution. Look again at the two distributions in Figure 5.3 , and decide which you would prefer in this situation. With a positively skewed distribution, the cases to the right of the median could stretch out for years; this is not true for a negatively skewed distribution. The distribution of life expectancy for Gould's disease was positively skewed, and Gould was obviously in the far right-hand tail of the distribution because he lived and remained professionally active for another 20 years.

REVIEW OF KEY TERMS

average deviation (p. 71 )

kurtosis (p. 78 )

leptokurtic (p. 78 )

measure of variation (p. 70 )

mesokurtic (p. 78 )

negatively skewed distribution (p. 79 )

normal curve (p. 78 )

normal distribution (p. 78 )

platykurtic (p. 78 )

positively skewed distribution (p. 79 )

range (p. 70 )

standard deviation (p. 71 )

MODULE EXERCISES

(Answers to odd-numbered questions appear in Appendix B .)

Calculate the range, average deviation, and standard deviation for the following five distributions (Exercises 1–5).

1.1, 2, 3, 4, 5, 6, 7, 8, 9

2.−4, −3, −2, −1, 0, 1, 2, 3, 4

3.10, 20, 30, 40, 50, 60, 70, 80, 90

4..1, .2, .3, .4, .5, .6, .7, .8, .9

5.100, 200, 300, 400, 500, 600, 700, 800, 900

6.What is the difference in calculation and use of σ, S, and s?

7.Using the data from question 1 in Module 4 , determine whether the data represent a normal or skewed distribution. If skewed, what type of skew do the data represent?

8.Describe the difference between mesokurtic, leptokurtic, and platykurtic curves.

CRITICAL THINKING CHECK ANSWERS

Critical Thinking Check 5.1

1.A measure of variation tells us about the spread of the distribution. In other words, are the scores clustered closely about the mean, or are they spread over a wide range?

2.The amount of rainfall for the indicated day is 2 standard deviations above the average. I would therefore conclude that the amount of rainfall was well above average. If the standard deviation were 2 rather than .75, then the amount of rainfall for the indicated day would be less than 1 standard deviation above average—above average, but not greatly above average.

MODULE 6

Standard Scores ( z Scores)

Learning Objectives

•Describe what a z score is.

•Know how to calculate a z score.

•Describe what a percentile rank is.

•Know how to calculate a percentile rank.

Introduction to Standard Scores ( z Scores)

The descriptive statistics and types of distributions discussed so far are valuable for describing a sample or group of scores. Sometimes, however, we want information about a single score. For example, in our exam score distribution, we may want to know how one person's exam score compares with those of others in the class. Or we may want to know how an individual's exam score in one class, say psychology, compares with the same person's exam score in another class, say English. Because the two distributions of exam scores are different (different means and standard deviations), simply comparing the raw scores on the two exams will not provide this information.

Let's say an individual who was in the psychology exam distribution used as an example in Module 3 ( Table 3.1 ) scored 86 on the exam. Remember, from Modules 4 and 5 that the exam had a mean of 74.00 with a standard deviation (S) of 13.64. Assume that the same person took an English exam and earned a score of 91 and that the English exam had a mean of 85 with a standard deviation of 9.58. On which exam did the student do better? Most people would immediately say the English exam because the score on this exam was higher. However, we are interested in how well this student did in comparison to everyone else who took the exams. In other words, did the individual do better on the psychology exam in comparison to those taking that exam than on the English exam in comparison to those taking the English exam?

Calculation of z scores

To answer this question, we need to convert the exam scores into a form we can use to make comparisons. A standard score , or z score , is a measure of how many standard deviation units an individual raw score falls from the mean of the distribution. We can convert each exam score to a z score and then compare the z scores because they will be in the same unit of measurement. We can think of z scores as a translation of raw scores into scores of the same language for comparative purposes. The formulas for z score transformations are:

z score (standard score) A number that indicates how many standard deviation units a raw score is from the mean of a distribution.

Z=X−¯¯¯XSZ=X−X¯S

and

z=X−μσz=X−μσ

where z is the symbol for the standard score. The difference between the two formulas is simply that the first is used when calculating a z score for an individual in comparison to a sample (thus, the use of the sample mean and sample standard deviation in the formula), and the second when calculating a z score for an individual in comparison to a population (thus, the use of the population mean and population standard deviation in the formula). Notice that they do exactly the same thing—indicate the number of standard deviations an individual score is from the mean of the distribution.

Conversion to a z score is a statistical technique that is appropriate for use with data measured on a ratio or interval scale of measurement (scales for which means are calculated). Let's use the formula to calculate the z scores for the previously mentioned student's psychology and English exam scores. The necessary information is summarized in Table 6.1 .

To calculate the z score for the English test, we first calculate the difference between the score and the mean and then divide by the standard deviation. We use the same process to calculate the z score for the psychology exam. These calculations are as follows:

zEnglish=X−¯¯¯XS=91−859.58=69.58=+0.626zEnglish=X−X¯S=91−859.58=69.58=+0.626

ZPsychology=X−¯¯¯XS=86−7413.64=1213.64=+0.880ZPsychology=X−X¯S=86−7413.64=1213.64=+0.880

TABLE 6.1 Raw score (x), sample mean (¯¯¯¯X)(X¯), and standard deviation (S) for English and psychology exams

	X	(¯¯¯¯X)(X¯)	S
English	91	85	9.58
Psychology	86	74	13.64

The individual's z score for the English test is 0.626 standard deviations above the mean, and the z score for the psychology test is 0.880 standard deviations above the mean. Thus, even though the student answered more questions correctly on the English exam (had a higher raw score) than on the psychology exam, the student performed better on the psychology exam relative to other students in the psychology class than on the English exam in comparison to other students in the English class.

The z scores calculated in the previous example were both positive, indicating that the individual's scores were above the mean in both distributions. When a score is below the mean, the z score is negative, indicating that the individual's score is lower than the mean of the distribution. Let's go over another example so that you can practice calculating both positive and negative z scores.

Suppose you administered a test to a large sample of people and computed the mean and standard deviation of the raw scores, with the following results:

¯¯¯X=45S=4X¯=45S=4

Suppose also that four of the individuals who took the test had the following scores:

Person	Score (X)
Rich	49
Debbie	45
Pam	41
Henry	39

Let's calculate the z score equivalents for the raw scores of these individuals, beginning with Rich.

ZRich=XRich−¯¯¯XS=49−454=44=+1ZRich=XRich−X¯S=49−454=44=+1

Notice that we substitute Rich's score (XRich) and then use the group mean (¯¯¯X)(X¯) and the group standard deviation (S). The positive sign (+) indicates that the z score is positive, or above the mean. We find that Rich's score of 49 is 1 standard deviation above the group mean of 45.

Now let's calculate Debbie's z score.

zDebbie=XDebbie−¯¯¯XS=45−454=04=0zDebbie=XDebbie−X¯S=45−454=04=0

Debbie's score is the same as the mean of the distribution. Therefore, her z score is 0, indicating that she scored neither above nor below the mean. Keep in mind that a z score of 0 does not indicate a low score; it indicates a score right at the mean or average. See if you can calculate the z scores for Pam and Henry on your own. Did you get zPam = −1 and zHenry= −1.5? Good work!

In summary, the z score tells you if an individual raw score is above the mean (a positive z score) or below the mean (a negative z score), and it tells you how many standard deviations the raw score is above or below the mean. Thus, z scores are a means of transforming raw scores to standard scores for purposes of comparison in both normal and skewed distributions. The procedure for calculating z scores using Excel is shown in the Statistical Software Resource section at the end of this chapter.

z Scores and the Standard Normal Distribution

If the distribution of scores for which you are calculating transformations (z scores) is normal (symmetrical and unimodal), then it is referred to as the standard normal distribution —a normal distribution with a mean of 0 and a standard deviation of 1. The standard normal distribution is actually a theoretical distribution defined by a specific mathematical formula. All other normal curves approximate the standard normal curve to a greater or lesser extent. The value of the standard normal curve is that it can provide information about the proportion of scores that are higher or lower than any other score in the distribution. A researcher can also determine the probability of occurrence of a score that is higher or lower than any other score in the distribution. The proportions under the standard normal curve hold only for normal distributions—not for skewed distributions. Even though z scores can be calculated on skewed distributions, the proportions under the standard normal curve do not hold for skewed distributions.

standard normal distribution A normal distribution with a mean of 0 and a standard deviation of 1.

Take a look at Figure 6.1 , which represents the area under the standard normal curve in terms of standard deviations. Based on this figure, we see that approximately 68% of the observations in the distribution fall between − 1.0 and +1.0 standard deviations from the mean. This approximate percentage holds for all data that are normally distributed. Notice also that approximately 13.5% of the observations fall between −1.0 and −2.0 and another 13.5% between +1.0 and +2.0, and that approximately 2% of the observations fall between −2.0 and −3.0 and another 2% between +2.0 and +3.0. Only .13% of the scores are beyond a z score of either ±3.0. If you sum the percentages in Figure 6.1 , you will have 100%—all of the area under the curve, representing everybody in the distribution. If you sum half of the curve, you will have 50%—half of the distribution.

FIGURE 6.1 Area under the standard normal curve

With a curve that is normal or symmetrical, the mean, median, and mode are all at the center point; thus, 50% of the scores are above this number and 50% are below this number. This property helps us determine probabilities. For example, what is the probability of randomly choosing a score that falls above the mean? The probability is equal to the proportion of scores in that area, or .50. Figure 6.1 gives us a rough estimate of the proportions under the normal curve. Luckily for us, statisticians have determined the exact proportion of scores that will fall between any two z scores—for example, between z scores of + 1.30 and +1.39. This information is provided in Table A.1 in Appendix A at the back of the text. A small portion of this table is shown in Table 6.2 .

The columns along the top of the table are labeled z, Area Between Mean and z, and Area Beyond z. The column heads also include pictorial representations. The z columns refer to the z score with which you are working. The Area Between Mean and z represents the area under the curve between the mean of the distribution (where z = 0) and the z score with which you are working—that is, the proportion of scores between the mean and the z score in column 1. The Area Beyond z is the area under the curve from the z score out to the tail end of the distribution. Notice that the entire table only goes as far as a z score of 4.00, because it is very unusual for a normally distributed population of scores to include scores larger than this. Notice also that the table provides information only about positive z scores, even though the distribution of scores actually ranges from approximately −4.00 to +4.00. Because the distribution is symmetric, the areas between the mean and z and beyond the z scores are the same whether the z score is positive or negative.

Let's use some of the examples from earlier in the module to illustrate how to use these proportions under the normal curve. Assume that the test data described earlier (with ¯¯¯XX¯ = 45 and S = 4) are normally distributed, so that the proportions under the normal curve apply. We calculated z scores for four individuals who took the test—Rich, Debbie, Pam, and Henry. Let's use Rich's z score to illustrate the use of the normal curve table. Rich had a z score equal to +1.00—1 standard deviation above the mean. I like to begin by drawing a picture representing the normal curve, and sketching in the z score with which I am working. Thus, Figure 6.2 shows a representation of the normal curve, with a line drawn at a zscore of + 1.00.

Before we look at the proportions under the normal curve, we can begin to gather information from this picture. We see that Rich's score is above the mean. Using the information from Figure 6.1 , we see that roughly 34% of the area under the curve falls between his z score and the mean of the distribution, whereas approximately 16% of the area falls beyond his z score. Using Table A.1 to get the exact proportions, we find (from the Area Beyond z column) that the proportion of scores falling above the z score of +1.0 is .1587. This number can be interpreted to mean that 15.87% of the scores were higher than Rich's score, or that the probability of randomly choosing a score with a z score greater than +1.00 is .1587. To determine the proportion of scores falling below Rich's z score, we need to use the Area Between Mean and z column and add .50 to this proportion. According to the table, the area between the mean and the z score is .3413. Why must we add .50 to this number? The table provides information about only one side of the standard normal distribution. We must add in the proportion of scores represented by the other half of the distribution, which is always .50. Look back at Figure 6.2 . Rich's score is +1.00 above the mean, which means that he did better than those between the mean and his z score (.3413) and also better than everybody below the mean (.50). Hence, the proportion of scores below Rich's score is .8413.

TABLE 6.2 A portion of the standard normal curve table

AREAS UNDER THE STANDARD NORMAL CURVE FOR VALUES OF Z

Z	AREA BETWEEN MEAN AND Z	AREA BEYOND Z
0.00	.0000	.5000
0.01	.0040	.4960
0.02	.0080	.4920
0.03	.0120	.4880
0.04	.0160	.4840
0.05	.0199	.4801
0.06	.0239	.4761
0.07	.0279	.4721
0.08	.0319	.4681
0.09	.0359	.4641
0.10	.0398	.4602
0.11	.0438	.4562
0.12	.0478	.4522
0.13	.0517	.4483
0.14	.0557	.4443
0.15	.0596	.4404
0.16	.0636	.4364
0.17	.0675	.4325
0.18	.0714	.4286
0.19	.0753	.4247
0.20	.0793	.4207
0.21	.0832	.4268
0.22	.0871	.4129
0.23	.0910	.4090
0.24	.0948	.4052
0.25	.0987	.4013
0.26	.1026	.3974
0.27	.1064	.3936
0.28	.1103	.3897
0.29	.1141	.3859
0.30	.1179	.3821
0.31	.1217	.3783
0.32	.1255	.3745
0.33	.1293	.3707
0.34	.1331	.3669
.35	.1368	.3632
0.36	.1406	.3594
0.37	.1443	.3557
0.38	.1480	.3520
0.39	.1517	.3483
0.40	.1554	.3446
0.41	.1591	.3409
0.42	.1628	.3372
0.43	.1664	.3336
0.44	.1770	.3300
0.45	.1736	.3264
0.46	.1772	.3228
0.47	.1808	.3192
0.48	.1844	.3156
0.49	.1879	.3121
0.50	.1915	.3085
0.51	.1950	.3050
0.52	.1985	.3015
0.53	.2019	.2981
0.54	.2054	.2946
0.55	.2088	.2912
0.56	.2123	.2877
0.57	.2157	.2843
0.58	.2190	.2810
0.59	.2224	.2776
0.60	.2257	.2743
0.61	.2291	.2709
0.62	.2324	.2676
0.63	.2357	.2643
0.64	.2389	.2611
0.65	.2422	.2578
0.66	.2454	.2546
0.67	.2486	.2514
0.68	.2517	.2483
0.69	.2549	.2451
0.70	.2580	.2420
0.71	.2611	.2389
0.72	.2642	.2358
0.73	.2673	.2327
0.74	.2704	.2296
0.75	.2734	.2266
0.76	.2764	.2236
0.77	.2794	.2206
0.78	.2823	.2177
0.79	.2852	.2148
0.80	.2881	.2119
0.81	.2910	.2090
0.82	.2939	.2061
0.83	.2967	.2033

SOURCE: Lehman, R. S. (1995). Statistics in the Behavioral Sciences: A Conceptual Introduction. Pacific Grove, CA: Brooks/Cole Publishing.

FIGURE 6.2 Standard normal curve with z score of +1.00 indicatedFIGURE 6.3 Standard normal curve with z scores of −1.0 and −1.5 indicated

Let's use Debbie's z score to further illustrate the use of the table. Debbie's z score was 0.00—right at the mean. We know that if she is at the mean (z = 0), then half of the distribution is below her score and half is above her score. Does this match what Table A.1 tells us? According to the table, .5000 (50%) of scores are beyond this z score, so the information in the table does agree with our reasoning.

Using the table with Pam and Henry's z scores will be slightly more difficult, because both Pam and Henry had negative z scores. Remember, Pam had a z score of −1.00, and Henry had a z score of −1.50. Let's begin by drawing a normal distribution and then marking where both Pam and Henry fall on that distribution. This information is represented in Figure 6.3 .

Before even looking at the z table, let's think about what we know from Figure 6.3 . We know that both Pam and Henry scored below the mean, that they are in the lower 50% of the class, that the proportion of people scoring higher than them is greater than .50, and that the proportion of people scoring lower than them is less than .50. Keep this overview in mind as we use Table A.1 . Using Pam's z score of −1.0, see if you can determine the proportion of scores lying above and below her score. If you determined that the proportion of scores above hers was .8413 and that the proportion below was .1587, then you were correct!

Why is the proportion above her score .8413? We begin by looking in the table at a z score of 1.0 (remember, there are no negatives in the table). The Area Between Mean and zis .3413, and then we need to add the proportion of .50 in the top half of the curve. Adding these two proportions, we get .8413. The proportion below her score is represented by the area in the tail, or the Area Beyond z of .1587. Now see if you can compute the proportions above and below Henry's z score of −1.5. Did you get .9332 above his score and .0668 below his score? Good work!

Now, let's try something slightly more difficult by determining the proportion of scores that fall between Henry's z score of −1.5 and Pam's z score of −1.0. Referring back to Figure 6.3 , you can see that we are targeting the area between the two z scores represented on the curve. Again, we use Table A.1 to provide the proportions. The area between the mean and Henry's z score of −1.5 is .4332, whereas the area between the mean and Pam's z score of −1.0 is .3413. To determine the proportion of scores that fall between the two, we subtract .3413 from .4332, obtaining a difference of .0919. This result is illustrated in Figure 6.4 .

FIGURE 6.4 Proportion of scores between z scores of −1.0 and −1.5

Percentile Ranks

The standard normal curve can also be used to determine an individual's percentile rank —the percentage of scores equal to or below the given raw score, or the percentage of scores the individual's score is higher than. To determine a percentile rank, we must first know the individual's z score. Let's say we wanted to calculate an individual's percentile rank based on this person's score on an intelligence test. The scores on the intelligence test are normally distributed, with μ = 100 and σ = 15. Let's suppose the individual scored 119. Using the zscore formula, we have:

percentile rank A score that indicates the percentage of people who scored at or below a given raw score.

z=X−μσ=119−10015=1915=1.27z=X−μσ=119−10015=1915=1.27

Looking at the Area Between Mean and z column in Table A.1 in Appendix A for a score of +1.27, we find the proportion .3980. To determine all of the area below the score, we must add .50 to .3980; the entire area below a z score of +1.27, then, is .8980. If we multiply this proportion by 100, we can describe the intelligence test score of 119 as being in the 89.80th percentile.

To practice calculating percentile ranks, see if you can calculate the percentile ranks for Rich, Debbie, Pam, and Henry from our previous example. You should arrive at the following percentile ranks.

Person	Score (X)	z Score	Percentile Rank
Rich	49	+1.0	84.13th
Debbie	45	0.0	50.00th
Pam	41	−1.0	15.87th
Henry	39	−1.5	6.68th

Students most often have trouble determining percentile ranks from negative z scores. Always draw a figure representing the normal curve with the z scores indicated; this will help you determine which column to use from the z table. When the z score is negative, the proportion of the curve representing those who scored lower than the individual is found in the Area Beyond z. We then multiply this proportion by 100 to determine the percentile rank. When the z score is positive, the proportion of the curve representing those who scored lower than the individual is found by using the Area Between Mean and z, adding .50 (the bottom half of the distribution) to this proportion, and then multiplying by 100 to determine the percentile rank.

What if we know an individual's percentile rank and want to determine this person's raw score? Let's say we know that an individual scored at the 75th percentile on the intelligence test described previously. We want to know what score has 75% of the scores below it. We begin by using Table A.1 to determine the z score for this percentile rank. If the individual is at the 75th percentile, we know the Area Between Mean and z is .25. How do we know this? The person scored higher than the 50% of people in the bottom half of the curve, and .75 − .50 = .25. Therefore, we look in the column labeled Area Between Mean and z and find the proportion that is closest to .25. The closest we come to .25 is .2486, which corresponds to a z score of 0.67.

Remember the z score formula is:

z=X−μσz=X−μσ

We know that μ = 100 and σ = 15, and now we know that z = 0.67. What we want to find is the person's raw score, X. So, let's solve the equation for X.

z=X−μσzσ=X−μzσ+μ=Xz=X−μσzσ=X−μzσ+μ=X

Substituting the values we have for μ, σ, and z,

X=zσ+μX=0.67(15)+100=10.05+100=110.05X=zσ+μX=0.67(15) + 100=10.05 + 100=110.05

As you can see, the standard normal distribution is useful for determining how a single score compares with a population or sample of scores and also for determining probabilities and percentile ranks. Knowing how to use the proportions under the standard normal curve increases the information we can derive from a single score.

You should also appreciate that if we know an individual's raw score on a variable (for example, an individual's intelligence test score), then we can calculate a z score from the raw score. Once we have the z score, we can then calculate a percentile rank or determine a proportion. On the other hand, if we are given an individual's z score, we can either calculate his or her raw score (solve for X) or a percentile rank or proportion. Lastly, if we know what an individual's percentile rank is, we can then determine his or her z score and from there determine his or her raw score. The point is that you cannot determine a percentile rank directly from an individual's raw score—we must first determine the individual's zscore.

Z SCORES AND PERCENTILE RANKS

	TYPE OF DISTRIBUTION
	Normal	Positively Skewed	Negatively Skewed
z score transformations applicable?	Yes	Yes	Yes
Percentile ranks and proportions under standard normal curve applicable?	Yes	No	No

1.Why is it not possible to use the proportions under the standard normal curve with skewed distributions?

2.Students in the Psychology Department at General State University have an average SAT score of 1,025 with a standard deviation of 175. The distribution is normal.

a.What proportion of students scored equal to or greater than 1,000?

b.What proportion of students scored equal to or greater than 1,150?

c.What proportion of students scored between 1,000 and 1,150?

d.What is the percentile rank for an individual who scored 950?

e.What score would an individual at the 75th percentile have?

3.Based on what you have learned about z scores, percentile ranks, and the use of the area under the standard normal curve, fill in the missing information in the following table representing performance on an exam that is normally distributed with ¯¯¯XX¯ = 55 and S = 6.

	X	z Score	Percentile Rank
John	63
Ray		− 1.66
Betty			72

REVIEW OF KEY TERMS

percentile rank (p. 90 )

standard normal distribution (p. 85 )

z score (standard score) (p. 83 )

MODULE EXERCISES

(Answers to odd-numbered questions appear in Appendix B )

Exercises 1–6 are based on the following data: The results of a recent survey indicate that the average new car costs $23,000 with a standard deviation of $3,500. The price of cars is normally distributed.

1.If someone bought a car for $32,000, what proportion of cars cost an equal amount or more than this?

2.If someone bought a car for $16,000, what proportion of cars cost an equal amount or more than this?

3.At what percentile rank is a car that sold for $30,000?

4.At what percentile rank is a car that sold for $12,000?

5.What proportion of cars were sold for an amount between $12,000 and $30,000?

6.For what price would a car at the 16th percentile have sold?

Exercises 7-12 are based on the following data: A survey of college students was conducted during final exam week to assess the number of hours spent studying each day. The mean number of hours was 5 with a standard deviation of 1.5 hours. The distribution was normal.

7.What proportion of students studied 7 or more hours per day?

8.What proportion of students studied 2 or more hours per day?

9.What proportion of individuals studied between 2 and 7 hours per day?

10.How many hours would an individual at the 60th percentile rank study?

11.What is the percentile rank for an individual who studied 4 hours per day?

12.What is the percentile rank for an individual who studied 7.5 hours per day?

13.Fill in the missing information in the following table representing performance on an exam that is normally distributed with ¯¯¯XX¯ = 75 and S = 9.

	X	z Score	Percentile Rank
Ken	73	—	—
Drew	—	1.55	—
Cecil	—	—	82

CRITICAL THINKING CHECK ANSWERS

Critical Thinking Check 6.1

1.The proportions hold only for normal (symmetrical) distributions in which one half of the distribution is equal to the other. If the distribution were skewed, this condition would be violated.

a..5557

b..2388

c.3169

d.33.36

e.1,142

	X	z Score	Percentile Rank
John	63	+1.33	90.82
Ray	45.05	−1.66	4.85
Betty	58.48	+0.58	72
CHAPTER THREE SUMMARY AND REVIEW Descriptive Statistics II

CHAPTER SUMMARY

In this chapter, we have discussed some additional descriptive statistics. Descriptive statistics that tell us about the spread or dispersion of the distribution are known as measures of variation (range, average deviation, and standard deviation). A distribution may be normal, positively skewed, or negatively skewed. The shape of the distribution affects the relationship among the mean, median, and mode. Finally, we discussed the calculation of z score transformations as a means of standardizing raw scores for comparative purposes. Although z scores can be used with either normal or skewed distributions, the proportions under the standard normal curve can be applied only to data that approximate a normal distribution.

Based on our discussion of these descriptive methods, you can begin to organize and summarize a large data set and also compare the scores of individuals to the entire sample or population.

CHAPTER 3 REVIEW EXERCISES

(Answers to exercises appear in Appendix B .)

Fill-in Self-Test

Answer the following questions. If you have trouble answering any of the questions, restudy the relevant material before going on to the multiple-choice self-test.

1.Measures of ______________ are numbers that indicate how dispersed scores are around the mean of the distribution.

2.An alternative measure of variation that indicates the average difference between the scores in a distribution and the mean of the distribution is the

3.When we divide the squared deviation scores by N — 1 rather than by N, we are using the ______________ of the population standard deviation.

4.σ represents the ______________ standard deviation and S represents the ______________ standard deviation.

5.A distribution in which the peak is to the left of the center point and the tail extends toward the right is a ______________ skewed distribution.

6.A number that indicates how many standard deviation units a raw score is from the mean of a distribution is a ______________.

7.The normal distribution with a mean of 0 and a standard deviation of 1 is the ______________.

Multiple-Choice Self-Test

Select the single best answer for each of the following questions. If you have trouble answering any of the questions, restudy the relevant material.

1.The calculation of the standard deviation differs from the calculation of the average deviation in that the deviation scores are

a.squared.

b.converted to absolute values.

c.squared and converted to absolute values.

d.It does not differ.

2.Imagine that distribution A contains the following scores: 11, 13, 15, 18, 20.

Imagine that distribution B contains the following scores: 13, 14, 15, 16, 17.

Distribution A has a ________ standard deviation and a ________ average deviation in comparison to distribution B.

a.larger; larger

b.smaller; smaller

c.larger; smaller

d.smaller; larger

3.Which of the following is not true?

a.All scores in the distribution are used in the calculation of the range.

b.The average deviation is a more sophisticated measure of variation than the range; however, it may not weight extreme scores adequately.

c.The standard deviation is the most sophisticated measure of variation because all scores in the distribution are used and because it weights extreme scores adequately.

d.None of the other alternatives is false.

4.If the shape of a frequency distribution is lopsided, with a long tail projecting longer to the left than to the right, how would the distribution be skewed?

a.Normally

b.Negatively

c.Positively

d.Average

5.If Jack scored 15 on a test with a mean of 20 and a standard deviation of 5 what is his z score?

a.1.5

b.− 1.0

c.0.0

d.Cannot be determined.

6.Faculty in the Physical Education department at State University consume an average of 2,000 calories per day with a standard deviation of 250 calories. The distribution is normal. What proportion of faculty consumes an amount between 1,600 and 2,400 calories?

a..4452

b..8904

c..50

d.None of the above.

7.If the average weight for women is normally distributed with a mean of 135 pounds and a standard deviation of 15 pounds, then approximately 68% of all women should weigh between ______________ and ________ pounds.

a.120; 150

b.120; 135

c.105; 165

d.Cannot say from the information given.

8.Sue's first philosophy exam score is −1 standard deviation from the mean in a normal distribution. The test has a mean of 82 and a standard deviation of 4. Sue's percentile rank would be approximately

a.78%.

b.84%.

c.16%.

d.Cannot say from the information given.

Self-Test Problems

1.Calculate the range, average deviation, and standard deviation for the following distribution:

2, 2, 3, 4, 5, 6, 7, 8, 8.

2.The results of a recent survey indicate that the average new home costs $100,000 with a standard deviation of $15,000. The price of homes is normally distributed.

a.If someone bought a home for $75,000, what proportion of homes cost an equal amount or more than this?

b.At what percentile rank is a home that sold for $112,000?

c.For what price would a home at the 20th percentile have sold?

CHAPTER THREE

Statistical Software Resources

If you need help getting started with Excel or SPSS, please see Appendix C : Getting Started with Excel and SPSS.

MODULE 5 Measures of Variation

The data we'll be using to illustrate how to calculate measures of variation are exam score data for a class of 30 students and are presented in Table 5.1 in Module 5 .

Using Excel

To begin using Excel to conduct this analysis, the data must be entered into an Excel spreadsheet. This simply involves opening Excel and entering the data into the spreadsheet.

You can see in the following image that I have entered the exam grade data from Table 5.1 into an Excel spreadsheet.

Once the data have been entered, we use the Data Analysis tool to calculate descriptive statistics. This is accomplished by clicking on the Data tab or ribbon and then clicking the Data Analysis icon on the far top right side of the window. Once the Data Analysis tab is active, a dialog box of options will appear (see next).

Select Descriptive Statistics as is indicated in the preceding box, and then click OK. This will lead to the following dialog box:

With the cursor in the Input Range box, highlight the data that you want analyzed from column A in the Excel spreadsheet so that they appear in the input range. In addition, check the Summary statistics box. Once you have done this, click OK. The summary statistics will appear in a new Worksheet, as seen next.

As you can see, there are several descriptive statistics reported, including measures of variation (range, standard deviation, and variance).

Using SPSS

As with the Excel exercise above, we will once again be using the data from Table 5.1 to calculate measures of variation. We begin by entering the data from Table 5.1 into an SPSS spreadsheet. This simply involves opening SPSS and entering the data into the spreadsheet. You can see in the following image that I have entered the exam grade data from Table 5.1 into an SPSS spreadsheet:

Notice that the variable is simply named VAR00001. To rename the variable to something appropriate for your data set, click on the Variable View tab on the bottom left of the screen. You will see the following window:

Type the name you wish to give the variable in the highlighted Name box. The variable name cannot have any spaces in it. Because these data represent exam grade data, we'll type in Examgrade. Note also that the Type of data is Numeric. Once the variable is named, highlight the Data View tab on the bottom left of the screen in order to get back to the data spreadsheet. Once you've navigated back to the data spreadsheet, click on the Analyze tab at the top of the screen and a drop-down menu with various statistical analyses will appear. Select Descriptive Statistics and then Descriptives…. The following dialog box will appear:

Examgrade will be highlighted, as above. Click on the arrow in the middle of the window and the Examgrade variable will be moved over to the Variables box. Then click on Options to receive the following dialog box:

You can see that the Mean, Standard Deviation, Minimum, and Maximum are all checked. However, you could select any of the descriptive statistics you want calculated. After making your selections, click Continue and then OK. The output will appear on a separate page as an Output file like the one below where you can see the minimum and maximum scores for this distribution along with the mean exam score of 74 and the standard deviation of 13.87. Please note that if you had more than one set of data, for example, two classes of exam scores, they could each occupy one column in your SPSS spreadsheet and you could conduct analyses on both variables at the same time. In this situation, separate descriptive statistics would be calculated for each data set.

Using the TI-84

Follow the steps below to use your TI-84 calculator to calculate the standard deviation for the data set from Table 5.1 .

TI-84 Exercise: Calculation of σ (standard deviation for population) and s (estimated population standard deviation).

1.With the calculator on, press the STAT key.

2.EDIT will be highlighted. Press the ENTER key.

3.Under L1 enter the data from Table 5.1 .

4.Press the STAT key once again and highlight CALC.

5.Number 1: 1—VAR STATS will be highlighted. Press ENTER.

6.Press ENTER once again.

Descriptive statistics for the single variable on which you entered data will be shown. The population standard deviation (σ) is indicated by the symbol σχ. The unbiased estimator of the population standard deviation (s) is indicated by the symbol Sχ.

MODULE 6 Standard Scores or z Scores

Using Excel to Determine z Scores

To illustrate how to calculate z scores, we'll use the example in Module 6 , also presented in the following table. We can see in the table the mean and standard deviation for two groups of students—one group who took an English exam and a second group who took a psychology exam.

Mean and standard deviation for English and psychology exams for two classes.

	Mean	Standard Deviation
English	85	9.58
Psychology	74	13.64

The scores for two of the students who took each of these exams follow:

Person	English Exam	Psychology Exam
Rich	91	88
Debbie	82	80

We'll use Excel to calculate the z scores for each of these individuals on the two exams. To calculate z scores using Excel, we use a function other than the Data Analysis ToolPak. Open Excel and click on the Formulas tab. You can see in the following Excel worksheet that this tab is highlighted:

We'll start with the English exam data from the preceding table to calculate Rich's z score. You can see from that table that the English exam had a mean of 85 and a standard deviation of 9.58. To calculate the first z score, click on the fx button on the far left side of the formulas ribbon. You should receive the following dialog box:

Make sure that Statistical is selected in the “Or select a category” field and then scroll down and select STANDARDIZE as in the preceding window. Finally select OK to receive the following dialog box. Enter Rich's English exam score into the X box and the mean and standard deviation where indicated. Then click OK.

Excel will give you the preceding output, where you can see the z score of +0.6263 in the A2 cell. This z score indicates that Rich scored 0.63 standard deviations above the mean on the English exam. If we want to compare this score to his performance on the psychology exam, or to Debbie's performance on the English exam, we must calculate these zscores also. Thus, to calculate Rich's z score on the psychology exam we use the same procedure as above. Use this procedure to calculate Rich's z score on the psychology exam and Debbie's z scores for both the English and psychology exams. You should receive the following results:

Rich's psychology exam z score = +1.03 Debbie's English exam z score = −.31 Debbie's psychology exam z score = + .44

Thus, we can see that although Rich's English exam score was higher than his psychology exam score (91 vs. 88), his z score on the psychology exam was larger than his z score on the English exam, indicating that he was 1.03 standard deviations above the mean on the psychology exam, but only .63 standard deviation above the mean on the English exam. Debbie, on the other hand, had a negative z score on the English exam, indicating that she scored − .31 standard deviation below the class mean. However, her psychology z score was + .44 standard deviation above the mean of the class, even though her raw score on the psychology exam was lower than her raw score on the English exam (80 vs. 82).