Up till now, we've been analyzing one set of quantitative data values, discussing center, shape, spread, and their measures. Sometimes two different sets of quantitative data values have something to do with each other, or so we suspect. This post (and a few of those that follow), deal with how to analyze two sets of data that appear to be related to each other.
We all know now that smoking causes lung cancer. This fact took years and lots of statistical work to establish. The first step was to see if there was a connection between smoking and lung cancer. This was done by charting how many people smoked and how many people got lung cancer. This is where we'll start, only we'll do it with more current data for purposes of illustration. Below is a table showing the incidence rates of smoking and lung cancer per 100,000 people, as tracked by the Centers for Disease Control and Prevention, from 1999 - 2007.
Because smoking explains lung cancer incidence and not the other way around, we call the smoking rate variable the explanatory variable. The lung cancer rate variable is called the response variable.
Looking over this table, we can see that smoking rates -- and lung cancer rates -- have both decreased during these 9 years. To illustrate how we deal with two variables at a time, let's continue with this data. The first thing to do is to plot the data on an xy-grid, one point per year. For 1999, for example, we plot the point (23.3, 93.5). Continuing with all the points, we get a scatterplot:
Note that the explanatory variable is on the x-axis, while the response variable is on the y-axis.
When you view a scatterplot, you are looking for a straight-line, or linear, trend. What I do is sketch the narrowest possible oval around the dots. The narrower the oval, the stronger the linear relationship between the variables. As you can see below, the relationship between smoking and lung cancer is quite strong because the oval is skinny:
If our oval had looked more like a circle, we would conclude that there is no relationship. A fatter oval that has a discernable upward or downward direction would indicate a weak association.
This upward-reaching oval also tells us that the association is positive; that is, that as one variable increases, so does the other one. A downward-reaching oval would indicate a negative association, which means that as one variable increases, the other decreases.
When describing an association between two quantitative variables, we address form, strength, and direction. The form is linear, strength is strong, and direction is positive. So we would say, "The association between smoking and lung cancer between 1999 and 2007 is strong, positive, and linear."
Just so you know, there are other forms of association between two quantitative variables: quadratic (U-shaped), exponential, logarithmic, etc. But we limit ourselves for now to linear associations.
Another *really* important thing to remember is that just because there is a linear association between two quantitative variables, it doesn't necessarily mean that one variable causes the other. We know that in the case of smoking and lung cancer, there is a cause-effect relationship, but this was established by several controlled statistical experiments. Seeing the linear association was only the catalyst for further study...it wasn't the culmination!
In the next post, we'll continue with this same example to develop some of the finer points of analyzing associations between two quantitative variables. For now, here is a good term to know: Another way to say the end of that previous sentence is "...analyzing associations for bivariate data." Bivariate simply means "two variables."
For the purposes of this post, we will refer to the data values in a data set as "scores."
In the last post, we used an example N(14, 2) to illustrate the 68-95-99.7 Rule, which stands for the various percents of scores lying within 1, 2, and 3 standard deviations from the mean. We can generalize the diagram we used to represent N(0, 1), where 0 is the mean and 1 is the standard deviation. This makes the model easier to apply, because the units we're most accustomed to seeing -- -1, 0,1,2, and so on -- appear as standard deviation units. Take a look:
We call N(0,1) the Standard Normal model. We now can use the number line to locate points that are any number of standard deviations from the mean...even fractional numbers.
In any Normal model, we're going to want to see what how many standard deviations a particular score in the data set is from its mean. We can do this for any score, and it has to do with converting a "raw" score (a score from our data) to a "standardized" score (a score from the Standard Normal model. How do we do this? There's a formula, and it's really an easy one:
...where
X stands for the score you're trying to convert,
stands for the mean, and
stands for the standard deviation.
Suppose we're back in the Normal model N(14, 2) and we want to see how many standard deviations a score of 15 is from the mean. We would subtract our mean from 15, then divide by the standard deviation, 2. That is: (15 - 14) / 2 = 0.5. This mean that our score of 15 is 0.5 standard deviations from the mean. 0.5 is the score on the Standard Normal model that represents our score from N(14, 2). We call 0.5 our standardized score, also known as a z-score. Z-scores tell us how many standard deviations a given "raw" score is from the mean.
(Self-Test): In the Normal model N(50, 4), standardize a score of 55.
(Answer): Find the z-score using the formula:
z = (55 - 50) / 4 = 5 / 4 = 1.25. The score is 1.25 standard deviations above the mean.
(Self-Test): In the Normal model N(50, 4), find the z-score for 42.
(Answer): z-scores can be negative, too. in this problem, 42 is less than the mean. So it will lie to the left of 50, and its z-score will be negative. z = (42 - 50) / 4 = -8 / 4 = -2. The score is 2 standard deviations below the mean.
Why would we want to standardize our scores? There are actually two reasons.
1. It can help us see how unusual a score might be.
How? Well, the percents we have been talking about can also be thought of as probabilities. For example, the probability that a score is greater than the mean is 50%, the same as the probability that a score is less than the mean. In the last post, we marked off regions and computed percents. In statistics, we consider any z-score of 3 or more, or -3 or less, as unusual, because (as we saw in the last post) only 0.15% of scores are in each of those regions. In other words, the probability of seeing a score in one of these regions is at most 0.15%, less than even a quarter-percent. That's unusual.
2. It can allow us to compare apples to oranges.
How? Well, suppose you have just gotten back 2 tests you took: one in algebra and one in earth science. Suppose further that both score distributions follow a [different] Normal model. The algebra test's scores follow N(80, 5) and the earth science test's scores follow N(85, 8). Now imagine that you got a 90 on the algebra test and a 93 on the earth science test. You can easily see that percentage-wise, your score on the earth science test is higher than your algebra score. But relative to the distributions, on which test did you perform better?
To find out, figure out the z-score for each test score. For your algebra test, your z-score is (90 - 80) / 5 = 10 / 5 = 2 (2 standard deviations above the mean). For your earth science test, your z-score is (93 - 85) / 8 = 8 / 8 = 1 (1 standard deviation above the mean. Relatively speaking, your performance was better on the algebra test; that is, your score was more exceptional. Think of the probabilities. The probability of a score that's 2 standard deviations or more above the mean is 2.5%, whereas a score that's 1 standard deviation or more above the mean is 16%. Get the idea?
(Self-Test) Suppose Tom's algebra test score was 86 and his earth science test score was 86. In which test did Tom perform better, given that the test scores follow the Normal models we used above?
(Answer): Tom's z-score on the algebra test was z = (86 - 80) / 5 = 6/5 = 1.20. His z-score on the earth science test was z = (86 - 85) / 8 = 1/8 = 0.125. Because Tom's z-score on his algebra test (1.20) is higher than his earth science test z-score (0.125), his algebra performance was better than his earth science performance.
One more thing...let's use the z-score formula to go backwards: to convert a standardized score back to a raw score. Suppose, Nancy earned a score on the algebra test that was 1.6 standard deviations below the mean. (In other words, her z-score was 1.6.) What actual percentage score would that represent for her, assuming N(80, 5)? In this case, we would start with the z-score formula and fill in what we know. Then, using algebra (!) we would solve for the "raw" score...
Nancy's algebra test score was 72.
In the next post, we'll expand on the probability side of the Standard Normal model.
