A colorful mound of blocks above two aligned rulers; their shared center is labeled X on the original-score ruler and z on the standardized ruler.
In the last two chapters, we’ve covered central tendency, particularly the mean, and variability, particularly focusing on standard deviation. Now we’ll see how we can combine these two statistics into a new concept: the \(z\)-score. We’ll talk about what \(z\)-scores are for and how we calculate them. Then we’ll talk more about describing distributions, particularly what we mean by standardized distributions. And lastly, we’ll see how we can begin to think about making probabilistic inferences using this new tool.
Comparing scores
To see why this is such a useful way of describing scores, suppose we have created two math tests and given them to a few people already. One test consists of a handful of fairly simple sums:
\(34 - 7\)
\(8 \times 4\)
\(37 + 64\)
\(54 \div 6\)
The second test is more difficult:
\(87 \times 13\)
\(3^2 \times 4^3\)
\(14 \times 5 + 15\)
\(8 \div 2(2 + 2)\)
If \(3x + 5y = 15\) and \(5x + 3y = 33\), what is the value of \(x + y\)?
We’re curious how long it takes people to complete each test, and whether times for the two tests are related. If somebody is quick on one test are they also quick on the other? Or does performance on one not translate to the other?
Now suppose you take each of the tests, and we record how long they each take you. You’re done with the first in 8 seconds, while the second takes you 70 seconds.
Describe the Scores
You completed one test in 8 seconds and another in 70 seconds. Would
you describe your performances as good, bad, or just okay? Why, or why
not?
Hint
What else would you want to know before comparing your performances
relative to other people?
Answer
You might have a sense of how you did from trying the questions
yourself. But the times alone don’t tell us whether either performance
was good compared with other people’s. Eight seconds might be
impressively quick on one test and nothing special on another.
You sense a problem, right? I’m asking you to compare performances, but the tests are pretty different. There’s no reason to expect that your time would be similar; indeed we should expect quite different times. So should we just give up trying to compare at all? Not so fast! We have more information. Remember, we already gave some people these tests. So let’s say we know that on average it takes people 10 seconds to complete the first test and 100 seconds to complete the second.
Describe the Scores Again
The mean completion times are 10 seconds for the easy test and 100
seconds for the hard test. Your times were 8 and 70 seconds. With this
extra information, how would you describe your performances? Why?
Hint
What do the averages tell you that the two times alone didn’t? What
do they still not tell you?
Answer
You were faster than average on both tests: 2 seconds faster on the
easy test and 30 seconds faster on the hard one. Whether one performance
is more impressive than the other is still harder to judge.
Faster than average on both tests. Not too shabby. But wait a minute. Can we be more precise? What if we have one more piece of information: the standard deviation. For the easy test the standard deviation is 1 second; for the hard test it is 15 seconds.
Describe the Scores One More Time
The easy test has a mean of 10 seconds and a standard deviation of 1
second. The hard test has a mean of 100 seconds and a standard deviation
of 15 seconds. Your times were 8 and 70 seconds. With this extra
information, how would you describe your performances now? Why?
Hint
How many standard deviations below the mean is each time?
Answer
The easy-test time is 2 seconds below its mean: 2 ÷ 1 = 2 standard
deviations. The hard-test time is 30 seconds below its mean: 30 ÷ 15 = 2
standard deviations. So although the times look very different, both
performances are the same distance below their respective means in
standard-deviation units.
Did you spot the common theme? On each test, you were 2 standard deviations below the mean. So although 8 seconds and 70 seconds are very different times, and they fall a different number of seconds below their respective means, they occupy the same relative position in their respective distributions. The mean and standard deviation together provide context for the raw scores, giving us a common ruler with which to compare them.1
The \(z\)-score
This is the approach to comparing scores described in 1912 by the psychologist R. S. Woodworth of Columbia University. Woodworth didn’t claim to have invented the idea: “Such a method exists, and is certainly familiar to statisticians,” he wrote; “but it seems to be overlooked in many cases where it would prove of value” (Woodworth, 1912).
Woodworth noted that it was common practice to switch from quantitative data to qualitative statements: “to say, for example, that an individual who has done very well in the first test, well in the second, but rather poorly on the third, has on the whole, therefore, done rather well.” Maybe you’ve taken an approach like this if, say, you like most of a director’s films, except for one that you thought was just okay and another that you hated. You’d probably say that you generally like the director’s movies overall: a useful summary, but lacking precision. More numerical approaches, such as counting the tests on which someone performed above average or averaging their ranks, still threw away information. “What is needed,” Woodworth wrote, “is a method of combining results which shall preserve all the refinement of the original measurements.”
Woodworth didn’t give the resulting number a specific name or symbol, and didn’t formalize it as an equation. Here is his description:
Let the average in each case be counted as \(0\), i. e., let the individual’s standing be expressed as a deviation above or below the average; and further, let the measure of variability be taken as the unit deviation, and all deviations be expressed as fractions or multiples of this unit.2
Eventually the term \(z\)-score was attached to this approach. We can represent the population \(z\)-score mathematically: \[z = \frac{X - \mu}{\sigma} \tag{5.1}\]
\(X\) refers to a score, \(\mu\) to the population mean, and \(\sigma\) to the population standard deviation.
If we’re dealing with a sample rather than a population, the calculation is the same but we use the corresponding sample statistics:
\[z = \frac{X - M}{s} \tag{5.2}\]
Here \(M\) is the sample mean and \(s\) is the sample standard deviation.
The numerator in either equation should look familiar: it’s a deviation score, like we’ve encountered in the previous couple of chapters. Dividing a deviation score by the standard deviation expresses that deviation in standard-deviation units. The sign tells us which side of the mean the score falls on. A positive \(z\)-score is above the mean and a negative \(z\)-score is below it. (That doesn’t necessarily mean good or bad: in our example, a negative \(z\)-score means a faster completion time.) The absolute magnitude of the \(z\)-score, written \(|z|\), tells us how many standard deviations the score is from the mean. So if \(z = 0\), the score is exactly equal to the mean. As \(|z|\) increases, the score is farther from the mean.
Computing \(z\)-scores
Let’s look at this visually. Figure 5.1 shows a histogram of completion times for a sample of 30 people who took the easy math test. Remember, for this distribution \(M = 10\) seconds, and the standard deviation is \(s = 1\) second. The dashed line marks your time of 8 seconds, down near the lower end of the distribution.
Figure 5.1: Histogram of completion times on the easy math test for a sample of 30 people. The sample mean is 10 seconds and the sample standard deviation is 1 second; a vertical line marks a time of 8 seconds.
Histogram of completion times for a sample of 30 earlier test takers on the easy math test. The sample mean is 10 seconds and the sample standard deviation is 1 second; a vertical line marks a time of 8 seconds.
Putting those values into the sample equation gives us: \[z = \frac{8 - 10}{1} = \frac{-2}{1} = -2\]
For the hard test, the mean was 100 seconds, the standard deviation was 15 seconds, and your time was 70 seconds. The histogram below marks that original time.
Figure 5.2: Histogram of completion times on the hard math test for a sample of 30 people. The sample mean is 100 seconds and the sample standard deviation is 15 seconds; a vertical line marks a time of 70 seconds.
Histogram of completion times for a sample of 30 earlier test takers on the hard math test. The sample mean is 100 seconds and the sample standard deviation is 15 seconds; a vertical line marks a time of 70 seconds.
\[z = \frac{70 - 100}{15} = \frac{-30}{15} = -2\]
Calculating z-scores
Another person completes the hard test in 115 seconds. Using the
sample mean of 100 seconds and standard deviation of 15 seconds, what is
the \(z\)-score?
Answer
1
Explanation
\(z = (115 - 100) / 15 = 1\). This
time is one standard deviation above the mean, so it is slower than
average.
Converting back to raw scores
We can also work in the other direction, from a \(z\)-score back to a raw score, if we have the mean and standard deviation. For a sample, if \(z = \frac{X - M}{s}\), then
\[X = M + z \times s \tag{5.3}\]
Here’s what that looks like if we want to work backwards from the two \(z\)-scores to the original completion times.
\[
\begin{align}
\text{Easy test:} \quad X & = M + zs = 10 + (-2) \times 1 = 8 \\
\text{Hard test:} \quad X & = M + zs = 100 + (-2) \times 15 = 70
\end{align}
\]
Standardized distributions
So far, we’ve converted a couple of individual observations into \(z\)-scores. But what if we converted a whole distribution? The result would be a standardized distribution.
A useful way to picture this is as adding a new ruler underneath the distribution. The shape of the distribution doesn’t change; the scores keep the same order and relative positions, but the numbers attached to those positions change.
Here are the two math test distributions again, this time with \(z\)-score rulers underneath their original scales. The original rulers show completion times in seconds, while the new rulers show the corresponding standardized positions. This makes it even clearer how your two times of 8 seconds and 70 seconds occupy the same relative position in their respective distributions. We could compare any \(z\)-score position between the two distributions. Say we wanted to know how many seconds correspond to \(z = 2\)—two standard deviations above the mean—on each test, rather than \(z = -2\). The rulers make that easy to see: it would be 12 seconds on the easy test and 130 seconds on the hard test.
Figure 5.3: Both math test distributions, with a new ‘ruler’ beneath the original x-axis, showing the corresponding \(z\)-scores. The mean, corresponding to \(z = 0\), is marked with a dashed line, as are \(z = -1\) and \(z = 1\), the positions one standard deviation above and below the mean.
Histogram of easy-test completion times with aligned raw-score and z-score rulers; 8 seconds aligns with z = -2.
Histogram of hard-test completion times with aligned raw-score and z-score rulers; 70 seconds aligns with z = -2.
Other standard-score scales
A distribution of \(z\)-scores always has a mean of zero and standard deviation of one, because of how the \(z\)-score formula works. That is a useful standardization, but sometimes it’s not exactly what we want to deal with. So once we’ve converted scores to \(z\)-scores, we can then rescale them to use a more convenient mean and standard deviation, if we want to. These are usually called transformed standard scores.
You’ll already be familiar with some transformed scales like this. SAT section scores were designed to have a mean of 500 and a standard deviation of approximately 100 in a specified reference cohort. (That doesn’t mean every later group of test takers will have exactly those values; see College Board (2017)). Many IQ tests transform scores to use a mean of 100 and a standard deviation of 15. Those values are arbitrary; we could perfectly well just use \(z\)-scores for the job. But imagine telling someone you scored \(z = -0.32\) on the SATs or that your IQ is 1.03. It wouldn’t be wrong to give those measurements as \(z\)-scores, but since they usually entail decimals, negative values, and a relatively small numeric range, they are just a bit finicky and unintuitive. Transforming the means and standard deviations lets us deal mostly in positive whole numbers, which can be a bit easier to talk about.
The procedure to take a standardized distribution of \(z\)-scores and transform it to a new scale, with whatever mean and standard deviation your heart desires, is simple. After standardizing the original scores, so that the mean becomes \(0\) and the standard deviation becomes \(1\), we can then multiply those \(z\)-scores to set a new standard deviation, then add a constant to set a new mean. The tutorial below follows one score from its original scale to the \(z\) scale, then to simplified IQ- and SAT-style reporting scales.
Activity 5.1: From raw scores to IQ and SAT scales
A tutorial keeping one test-score distribution fixed while its aligned ruler changes from raw scores to z-scores, an IQ scale, and an SAT scale.
New Scores
A score has \(z = -1\). It is
transformed to a reporting scale with mean 100 and standard deviation
15. What is the transformed score?
Answer
85
Explanation
Multiply by the new standard deviation, then add the new mean: \(15 \times (-1) + 100 = 85\). The score is
still one standard deviation below the mean.
Using \(z\)-scores to identify unusual scores
Comparing scores from different scales is one use of \(z\)-scores. Another is asking whether a score seems unusual or extreme. As we learned in Chapter 4, the standard deviation quantifies a “typical” deviation from the mean. So in that general sense, a \(z\)-score of \(1\) or \(-1\) would be perfectly typical and expected. More extreme \(z\)-scores, like \(2\) or \(-2\) would be less typical (again, according to the standard deviation’s definition of typicality).
Consider again the fake movie ratings from Chapter 4. Figure 5.4 shows two of those sets of scores with \(z\)-score rulers added. (The third movie, for which all the ratings were exactly the same, had a standard deviation of zero, so it can’t be standardized: the \(z\) equation would have us divide by that SD of zero, so the answer would be undefined.)
Figure 5.4: Movie-rating distributions with raw-score and \(z\)-score rulers.
Block histogram of sample ratings for The Good, the Bad and the Okay with aligned raw-score and z-score rulers.
Block histogram of sample ratings for The Polarizing Express with aligned raw-score and z-score rulers.
Notice that The Good, the Bad and the Okay had most scores near the mean of 5, with two extreme scores out on either end. On the \(z\) scale, those extreme scores are beyond \(z = \pm 1\). According to our rough definition, whereby \(|z| = 1\) is considered typical, that seems right; those outlying scores are atypical for that distribution. But notice also that there aren’t actually any scores at \(z = \pm 1\). The z-score tells us the distance from the mean; how common scores are around that position still depends on the distribution’s shape. For The Polarizing Express, on the other hand, every score is within one standard deviation of the mean, simply because the standard deviation is so large. When every score is far from the mean, they are all typical in that sense.
Note the subtlety here: typical doesn’t necessarily have to imply close to the mean. But that was an unusual case. For reasons we will discuss over the next couple of chapters, we will more often find ourselves dealing with situations more like the ratings for The Good, the Bad and the Okay, where most scores are clustered around the mean and a few are out on the extremes. There, the standard deviation—and therefore, \(z\)-scores—generally tells us both how far a score is from the mean and how unusual it is.
Reaction times and radioactive spiders
Let’s think about this using a silly but hopefully memorable example. You know the gist of Spider-Man’s origin story, right? A high school kid gets bitten by a radioactive spider and develops superpowers, including Spidey Sense™️: the ability to sense and react to things unusually quickly. But how quick is unusually quick? Just how fast would a kid have to be for us to consider it evidence of superpowers? We can use \(z\)-scores to help answer that question.
Figure 5.5: My son’s toy Spider-Man action figure: he’s a big Spidey fan. That’s what inspired the goofy reaction time example.
First, I’d like you to have a go at testing your reaction times. Head over to humanbenchmark.com’s reaction time test. You’ll do a simple little test where you click as soon as the color on screen changes. You’ll do a few trials then note your average reaction time in milliseconds. When you’re done, make a note of your time (in milliseconds) by answering this quiz.
What was your average reaction time, in milliseconds?
Enter just the number, such as 284. There is no right or wrong answer
here; we’ll come back to your time once we have a distribution to
compare it with.
Now let’s imagine Peter Parker took the same test and averaged 189 milliseconds. That’s probably quicker than your time, but is it unusual enough to make us begin entertaining the possibility of superhuman reflexes?
Our interpretation depends on the reference distribution. The results page for the Human Benchmark reaction-time test reports a mean recorded click time of 284 milliseconds. The site doesn’t give a standard deviation, so for this example we’ll just pick a reasonable-sounding value of 50 milliseconds. We’ll also model the scores as normally distributed—with a single peak in the middle and tails extending symmetrically in both directions—even though in reality the data are actually somewhat positively skewed. (This is an example about radioactive spider bites, after all, so I don’t mind taking a little artistic license with the data.)
Figure 5.6: An illustrative normal model of reaction times, with a mean of 284 ms and an assumed standard deviation of 50 ms. Peter Parker’s reaction time of 189 ms is marked in the lower tail.
Normal curve of reaction times centered on 284 ms with standard deviation 50 ms; a vertical line marks Peter Parker's 189 ms well down in the lower tail.
With those assumptions in place, we can calculate Peter Parker’s \(z\)-score:
Under our assumed normal model, that location is pretty uncommon: it sits well down in the lower tail. Peter’s reaction time doesn’t look typical of this reference distribution. Maybe the spider bite really did change him. We’ll revisit and develop the idea of testing a hypothesis about unusual reaction times in Chapter 8.
What about you? You noted your own average a moment ago, and we now have the same two parameters to locate it against. Work out your own \(z\)-score the same way.
Now calculate your own reaction-time \(z\)-score
Peter Parker’s 189 milliseconds gave \(z =
-1.9\). Where does your own average fall on that same
distribution? Use the time you entered earlier, along with the mean and
standard deviation we’re assuming for the reference distribution:
A recap of the average reaction time you entered earlier, with the
assumed population mean of 284 milliseconds and standard deviation of 50
milliseconds.
Hint
Subtract the mean from your own time, then divide by the standard
deviation. A faster time than average gives a negative \(z\)-score.
Explanation
Substituting your own numbers into the population \(z\)-score equation, and marking that
position on the distribution:
Your own average reaction time substituted into the z-score equation,
and a normal curve of reaction times with aligned millisecond and
z-score rulers marking where your time falls.
The curve hasn’t moved and neither has your score. The upper ruler
labels that position in milliseconds and the lower ruler labels it in
standard deviations. A negative \(z\)-score means faster than the reference
average, while a positive one means slower. Peter Parker’s \(-1.9\) sits well down in the fast tail of
the normal model we’re using.
Clinical psychology and OCD treatment
We’ll have more fun with spider bites in Chapter 8, but for a more realistic example, let’s turn to clinical psychology and the treatment of mental health problems. This one is based on a published study that developed the 20-item Dimensional Obsessive-Compulsive Scale (DOCS) and used it to measure symptoms before and after exposure and response prevention, a form of cognitive behavioral therapy (Abramowitz et al., 2010).
Suppose the researchers found that prior to receiving treatment, people with diagnosed OCD scored 30.25 on that scale, with a standard deviation of 14.89. So the untreated population \(\mu = 30.25\) with \(\sigma = 14.89\).
Now suppose that you’re a clinical psychologist with a patient who has just completed a course of CBT and scored \(15.49\) on the DOCS. Where would that score place the patient within the untreated population? How far below average is it?
CBT \(z\)-score
A post-CBT symptom score is \(15.49\). The untreated population mean is
\(30.25\) with standard deviation \(14.89\). What is the \(z\)-score?
Answer
-0.99
Explanation
Under the chapter’s hypothetical population model, \(z = (15.49 - 30.25) / 14.89\), which is
about -0.99. The score is therefore almost one standard deviation below
the untreated mean.
Looking ahead
Those specific numbers I gave above came from the real study, but I deliberately misused them. In the actual research, \(15.49\) wasn’t one individual’s score, it was the average of a sample of 40 people who completed the DOCS scale after receiving treatment (and 30.25 was those same people’s average score before treatment, not a population parameter). I treated it as an individual score to give you another chance to calculate and interpret a lone observation converted to a \(z\)-score, but in real research we usually want to understand how typical or expected the sample as a whole is. One individual scoring lower on a measure of symptom severity might be suggestive that their treatment helped, but a sample of 40 people collectively scoring lower can provide more compelling evidence.
This brings us to the boundary between descriptive and inferential statistics. \(z\)-scores are an essential building block of the logic we’ll use to deal with samples, but we have more work to do before we can make real inferences about treatment effects. In Chapter 6, we’ll introduce more precise ways of thinking about probability and using \(z\)-scores to understand normally distributed scores, and then in Chapters 7 and 8 we’ll explore the tools that enable us to apply that understanding to samples.
Learning Checks
0 of 6
1
A \(z\)-score of 0 means the score
is exactly at the mean of its distribution.
Answer
True
Explanation
The numerator of the z-score equation is zero when X equals the
mean.
2
For a population with \(\mu = 50\)
and \(\sigma = 10\), what is the \(X\) value corresponding to \(z = 0.4\)?
Hint
Rearrange the \(z\)-score formula:
\(X = \mu + z\sigma\).
Answer
54
Explanation
\(X = \mu + z\sigma = 50 + 0.4 \times 10 =
54\).
3
Which score is farther from the mean of its distribution, measured in
standard-deviation units?
Answer
\(z=-2.5\)
Explanation
Distance from the mean in standard-deviation units is given by the
absolute magnitude of the \(z\)-score.
Because \(|-2.5| > |1.5|\), \(z=-2.5\) is farther from its mean. The
negative sign tells us the direction, not the distance.
4
What are the mean and standard deviation of a distribution of \(z\)-scores?
Answer
Mean 0, standard deviation 1
Explanation
Transforming scores to z-scores relabels the distribution so its mean
is 0 and its standard deviation is 1.
5
A positively skewed distribution is converted to \(z\)-scores. What shape does the
standardized distribution have?
Answer
It remains positively skewed
Explanation
Subtracting the mean and dividing by the standard deviation changes
the scale, but preserves the distribution’s shape. A mean of 0 and
standard deviation of 1 do not make a distribution normal.
6
Two distributions have different shapes. A score in each has \(z = -2\). What can we conclude about the
two scores from this information?
Answer
They are the same number of standard deviations below their
respective means
Explanation
Both scores are two standard deviations below their respective means.
Those standard deviations can represent different distances on the
original scales. The percentage of scores below each position depends on
the distribution’s shape, so equal \(z\)-scores alone do not establish equal
rarity.
Abramowitz, J. S., Deacon, B. J., Olatunji, B. O., Wheaton, M. G., Berman, N. C., Losardo, D., Timpano, K. R., McGrath, P. B., Riemann, B. C., Adams, T., Bjorgvinsson, T., Storch, E. A., & Hale, L. R. (2010). Assessment of obsessive-compulsive symptom dimensions: Development and evaluation of the dimensional obsessive-compulsive scale. Psychological Assessment, 22(1), 180–198. https://doi.org/10.1037/a0018260
Woodworth, R. S. (1912). Combining the results of several tests: A study in statistical method. Psychological Review, 19(2), 97–123.
Of course, we still need to be careful about what we can conclude from the comparison; a common ruler doesn’t necessarily mean the tests measure the same ability, or measure it equally well.↩︎
Note that Woodworth left the choice of variability measure open. The modern convention is to use the standard deviation, so that’s what we use here.↩︎
