March 22, 2005
Statistics term of the Day: Standardized scores
The head's still pretty clogged, folks, so this one will be a quickie.
Now that we've learned some wonderful things about measures of spread and measures of central tendency, let's see a way we can use those to determine where someone's score lies within a distribution.
Let's say you have twin fifth-graders (bless your heart), enrolled in two different math classes. Bonnie comes home with a 75 on her math test, while Clyde has an 80. Clyde's score is higher, sure, but can you really compare the two? And how can you tell which of your kids is doing better within their class?
That's where z-scores (or standardized scores) come in. Z-scores tell you precisely where an observation lies within a distribution (they also tell you something about how representative a sample is of a population, but we'll get to that later).
Every score in a distribution has a corresponding z-score, or standardized score. All you need to calculate the z-score of a population is (a) the observed, or raw score, (b) the mean, and (c) the standard deviation. Let's consider Bonnie's class to be population #1, and Clyde's class to be population #2.
Bonnie's class has a mean of 71 and an SD of 2, while Clyde's class has a mean of 77 and an SD of 3. Hmm. We can tell already just from this information that both kids are above the mean in their class, so we know they're doing better than average. But how much better? To find the answer to this, we simply subtract the class mean from each score, and divide by the class's standard deviation.
Bonnie's z-score = (75-71)/2 = 4/2 = +2
Clyde's z-score = (80-77)/3 = 3/3 = +1
These scores of +2 and +1 are on the z-score metric, which has a mean of 0 and a standard deviation of 1. The positive sign tells us that Bonnie and Clyde's scores both fall above the mean, and their scores are in the standard deviation metric - Bonnie's score is 2 SDs above the class mean, while Clyde's is 1 SD above his class's mean. So, even though Clyde's raw score is higher, Bonnie is actually doing better in her class.
Can we really compare the two in this way? Yes, we can. This is one of things for which standardized scores are very useful. Without standardization, comparing Bonnie and Clyde's scores are like comparing apples to oranges.
One caveat to remember here. You can standardize any score, as long as you know the mean and SD. However, it's not as useful to standardize scores that come from a non-normal population, because the z-score transformation assumes that the underlying distribution is normal. For many psychological measures, it is, but it is important to be aware of any non-normality in the population. The assumption of normality doesn't matter so much for small comparisons like this; it matters a great deal when we get to probability and the use of z-scores to find areas underneath the curve.
One common error that people make is to misremember what standardization actually does. Standardization does not turn non-normal distributions into normal ones - a z-score distribution will always be the same as the underlying raw score distribution. The math2.org website gives this graphic to remind you of what the z-score distribution is assumed to be:
