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.
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:
The second test is more difficult:
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 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. How do you feel like you did? Good, bad, just okay?
Describe the Scores
You completed one test in 8 seconds and another in 70 seconds. Are the raw times enough to conclude which performance was better relative to other people?
No. Completion times from different tests are not directly comparable without knowing the distribution of times for each test.
A score only has meaning relative to the distribution it came from.
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 obviously 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. Now how do you feel about your performance?
Describe the Scores Again
Suppose the mean completion times are 10 seconds for the easy test and 100 seconds for the hard test. How do your times of 8 and 70 seconds compare with their respective means?
Both are faster than average. The easy-test time is 2 seconds below its mean, and the hard-test time is 30 seconds below its mean.
The raw deviations have different units of comparison until we know how much completion times vary on each test.
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. Now how do you feel about your performance?
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. You took 8 and 70 seconds, respectively. How do the times compare when each deviation from its mean is expressed in standard-deviation units?
Each time is 2 standard deviations below its respective mean.
The easy-test time is 2 seconds below its mean, and its standard deviation is 1 second. The hard-test time is 30 seconds below its mean, and its standard deviation is 15 seconds. Both deviations are therefore 2 standard deviations.
On each test, you were 2 standard deviations below the mean. So although 8 seconds and 70 seconds are very different raw times, they occupy the same relative position in their respective distributions. We now have a common ruler for comparing them. That doesn’t establish that the tests measure the same ability equally well, but it does let us compare the two performances more precisely.
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.” 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.
By the 1930s, the term \(z\)-score had been attached to this approach. (Note that the “unit deviation” was also deliberately general: Woodworth noted that standard deviation, variance, or even something like quartiles could be used. The modern convention is to use the standard deviation, so that’s what we use here.)
So now 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 far 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.
Let’s look at this visually. Here’s a histogram of completion times for a sample of 30 earlier test takers. Its sample mean, \(M\), is exactly 10 seconds, and its sample standard deviation, \(s\), is exactly 1 second. The vertical line marks your time of 8 seconds, down near the lower end of the distribution. Substituting those values into the sample equation gives us:
\[z = \frac{8 - 10}{1} = \frac{-2}{1} = -2\]
Now try the same calculation for the hard test. The mean was 100 seconds, the standard deviation was 15 seconds, and your time was 70 seconds.
Calculating z-scores
For a completion time of 70 seconds in a sample with mean 100 seconds and standard deviation 15 seconds, what is the z-score?
-2
z = (70 − 100)/15 = −2.
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 \(X = M + z \times s\). Here’s what that looks like if we want to work backwards from the two \(z\)-scores to the original completion times.
\[ \begin{align} \text{Easy test:} \quad X & = M + zs = 10 + (-2) \times 1 = 8 \\ \text{Hard test:} \quad X & = M + zs = 100 + (-2) \times 15 = 70 \end{align} \]
So far, we’ve converted a couple of individual observations into \(z\)-scores. But what if we converted every score in a distribution? The result would be a standardized distribution.
A distribution of \(z\)-scores has a mean of zero because its numerators are deviation scores, and the deviations sum to zero. It has a standard deviation of one because every deviation is divided by the original standard deviation. This works when we use matching conventions throughout: population \(\mu\) and \(\sigma\) for a population, or sample \(M\) and \(s\) for a sample.
A useful way to picture this is as changing the ruler underneath the distribution. The scores keep the same order and relative spacing, but the numbers attached to those positions change because we’ve changed the unit. A distance of two raw-score points, for example, might become a distance of one standard deviation.
Here are the two math test distributions again, this time with \(z\)-score rulers underneath their original scales. The original rulers show completion times, while the new rulers show the corresponding standardized positions.
Now let’s return to the fake movie reviews from the previous chapter. As before, we’ll treat the six observed ratings for each movie as a sample, using \(M\) and \(s\). The added rulers make the effect of their different standard deviations visible.
Notice where \(z=-1\) and \(z=1\) fall on the two rulers. The more variable ratings need a larger stretch of the original 0-to-10 scale to cover one standard deviation. I have left out Fine, Actually, the third movie from the earlier comparison, because all six of its ratings were 5. Its sample standard deviation is zero, so its ratings cannot be converted into \(z\)-scores.
Once we’ve converted scores to \(z\)-scores, we can rescale them to use a more convenient reporting mean and standard deviation. These are usually called transformed standard scores.
You’ll already be familiar with some reporting 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 (College Board, 2017). Many IQ tests use a mean of 100 and a standard deviation of 15 in their norming groups. One reason for using numbers like these is that \(z\)-scores, with their mean of zero and standard deviation of one, are just a bit finicky and unintuitive, since they usually entail decimals and negative values. (Imagine telling someone you scored 0.32 on the SATs.) Larger reporting means and standard deviations let us deal mostly in positive whole numbers, which people generally find a little bit easier to talk about.
The tutorial below follows one score from its original scale to the \(z\) scale, then to simplified IQ- and SAT-style reporting scales.
In each case the procedure is the same: standardize the original scores, multiply the \(z\)-scores to set a new standard deviation, then add a constant to set a new mean. More generally, we can write the transformation as \(Y=A+Bz\), where \(A\) is the desired mean and the positive value \(B\) is the desired standard deviation.
New Scores
A score has z = 2. It is transformed to a reporting scale with mean 100 and standard deviation 15. What is the transformed score?
130
Use Y = A + Bz = 100 + 15 × 2 = 130.
Comparing scores from different scales is one use of \(z\)-scores. Another is asking whether a score seems unusual. But a standardized distance from the mean is not quite the same thing as rarity.
We can calculate a \(z\)-score for a distribution of any shape. But a \(z\)-score close to zero is merely close to the mean; that doesn’t guarantee the score is common. Think back to a symmetrical bimodal distribution, whose mean can sit in the sparsely populated valley between two peaks. A score with \(z=0\) could then be relatively unusual, while scores farther from zero cluster around the modes. So the \(z\)-score gives us a location. Whether that location is common or unusual also depends on the shape of the distribution.
If scores are modeled as normally distributed, locations more than about two standard deviations from the mean are relatively uncommon. That isn’t a universal bright line for identifying unusual scores. In Chapter 6, we’ll use areas under the normal curve to say exactly how much of the modeled distribution lies beyond a particular \(z\)-score.
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.
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 159 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.1 The site doesn’t give a standard deviation, so for this example we’ll use 50 milliseconds. We’ll also model the scores as normally distributed even though reaction times are actually somewhat positively skewed. But this is an example about radioactive spider bites, so I don’t mind taking a little artistic license with the data.
With those assumptions in place, we can calculate Peter Parker’s \(z\)-score:
\[z = \frac{X - \mu}{\sigma} = \frac{159 - 284}{50} = -2.5\]
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 159 milliseconds gave z = −2.5. 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:
Subtract the mean from your own time, then divide by the standard deviation. A faster time than average gives a negative z-score.
Substituting your own numbers into the population z-score equation, and marking that position on the distribution:
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 −2.5 sits well down in the fast tail of the normal model we’re using.
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).
The numbers all come from that study, but I’m going to rearrange them slightly for teaching purposes. For the moment, suppose that the published pretreatment mean of 30.25 and standard deviation of 14.89 describe a hypothetical population of people who haven’t had treatment.
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?
-0.99
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.
Those numbers 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 posttreatment mean of a sample, and the researchers compared the same people’s symptoms before and after treatment. I treated it as an individual score to give you another chance to calculate and interpret a \(z\)-score, but when we scale up to a sample of scores, we can use another kind of \(z\)-score to better understand how typical or execpted the sample as a whole is.
This brings us to the boundary between descriptive and inferential statistics. In Chapter 6, we’ll use the normal curve to translate individual \(z\) locations into probabilities. In Chapter 7, we’ll build distributions of sample means and introduce the standard error. Then in Chapter 8, we’ll standardize a sample mean using that new distribution. The calculations will look related, but the score being located and the ruler used to locate it will have changed. z$-scores are an essential building block of the logic we’ll use, but we have more work to do before we can make real inferences about treatment effects.
Learning Checks
A z-score of 0 means the score is exactly at the mean of its distribution.
True
The numerator of the z-score equation is zero when X equals the mean.
For a population with μ = 50 and σ = 10, what is the X value corresponding to z = 0.4?
Rearrange the z-score formula: X = μ + zσ.
54
X = μ + zσ = 50 + 0.4 × 10 = 54.
Which score is farther from the mean of its distribution?
z = −2.5
Distance from the mean 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.
What are the mean and standard deviation of a distribution of z-scores?
Mean 0, standard deviation 1
Transforming scores to z-scores relabels the distribution so its mean is 0 and its standard deviation is 1.
That isn’t a universal parameter for human reaction time: the website’s users are self-selected, and the measurement includes delays from their computers, monitors, and input devices.↩︎