Normality Check on TI-89
Copyright © 2012–2017 by Stan Brown
Copyright © 2012–2017 by Stan Brown
But just looking at a plot, you may not be sure whether it’s “close enough” to a straight line, especially with smaller data sets. Most of the time, you need to make some fairly gnarly computations to answer that question.
This note shows you how to make the plot and do the computations for an example, and then in an appendix it gives the theory behind all of this.
Consider these vehicle weights (in pounds):
Construct a plot to decide whether these vehicle weights seem to be normally distributed.
|Enter the data points.||
Cursor onto the label
Enter the data values.
|Sort the data, unless they’re already in ascending order.||
The test for normality is a normal probability plot. It’s a scatterplot, where x = the actual data and y = the z scores the data points would have if the data were perfectly normally distributed. The closer the plot is to a straight line, the closer the data set is to a normal distribution.
The calculator has a command to create a normal probability plot. But it’s really hard to look at a plot and tell whether it’s close enough to a straight line. You need a numerical test to help you in doubtful cases — and most cases are doubtful. So you end up having to duplicate some of the work the calculator would do.
The next calculations could all be combined into one very long command storing into list2, but it would be really easy to make mistakes in such a long formula. Instead, I’ve broken the calculation into two chunks.
|Get the normal probabilities into list2. (The normal probabilities are the probabilities of getting each data point or a lower one by random selection, if the data points are normally distributed. The appendix gives the formula (i−0.375)/(n+0.25), where i is the numbers 1, 2, 3, … and n is the sample size.||
Cursor to the column heading |
(You can see a part of that formula in the illustration at right.) The
Here’s some help with entering the formula.
|Find z scores that correspond to those probabilities. These are the z scores that your data would have if they were normally distributed, and they’ll also be used in the normal probability plot below.||Cursor to the column heading |
Here are some hints to help you:
Here’s the last part of the formula, and the resulting
the z scores in list3.
In this step, you disable any other plots and graphs that could overlay your normal probability plot.
|In this step, you make an xy scatterplot, where the x’s are the original data points and the y’s are the z scores you just computed.||Press [|
The plot setup dialog is shown at right. Select plot type “scatter” and set x to list1 and y to list3. You can’t use a command to paste the list names here, so use the [
If this plot is close to a straight line, then the data set is close to a normal distribution. But how close is close enough? If you’re very lucky, the plot will be an obvious straight line, or it will be very far from a straight line. Then you can declare the data normal or not normal, and stop. But usually the plot is iffy, and you could go either way just from looking at it.
What about the plot for this example? There are certainly some bumps, but is the plot too far from a straight line? It’s hard to say. This is one of those iffy cases that you usually get.
Fortunately, there’s a numerical test that helps you make a decision. The correlation coefficient answers the question “how close are these points to a straight line”, so let’s compute it.
|To find the correlation coefficient, the easy way is to do a linear regression.||
The graph took you out of Stats/List Editor, so press
Here’s the regression dialog. The X list is list1, and the Y
list is list3; use the [|
The correlation coefficient of .9599 (about .96) seems pretty good, but is it good enough? To answer this question, you have to compare it to a critical value. If r < CRIT, the data set is too far from normally distributed. If r > CRIT, the data set is close enough to normally distributed.
|The formula for CRIT is 1.0063−.6118/n+1.3505/n²−.1288/√n, where n is the sample size.||
Whew! r = 0.9599 and CRIT ≈ 0.9179. r > CRIT, and therefore you can say that the data set is close enough to a normal distribution.
The basic idea isn’t too bad. You make an xy scatterplot where the x’s are the data points, sorted in ascending order, and the y’s are the expected z scores for a normal distribution. (I’m going to abbreviate “normally distributed” or “normal distribution” as ND to save wear and tear on my keyboard and your eyes.)
Why would you expect that to be a straight line? Recall the formula for a z score: z = (x−x̅)/s. Breaking the one fraction into two, you have z = x/s−x̅/s. That’s just a linear equation, with slope 1/s and intercept x̅/s. So an xz plot of any theoretical ND, plotting each data point’s z score against the actual data value, would be a straight line.
Further, if your actual data points are ND, then their actual z scores will match their expected-for-a-normal-distribution z scores, and therefore a scatterplot of expected z scores against actual data values will also be a straight line.
Now, in real life no data set is ever exactly a ND, so you won’t ever see a perfectly straight line. Instead, you say that the closer the points are to a straight line, the closer the data set is to normal. If the data points are too far from a straight line — if their correlation coefficient r is lower than some critical value — then you reject the idea that the data set is ND.
Okay, so you have to plot the data points against what their z-scores should be if this is a ND, and specifically for a sample of n points from a ND, where n is your sample size. This must be built up in a sequence of steps:
1.0071−0.1371/√n−0.3682/n+0.7780/n² at α=0.10
0.9963−0.0211/√n−1.4106/n+3.1791/n² at α=0.01
The closer the points are to a straight line, the closer the data set is to fitting a normal model. In other words, a larger r indicates a ND, and a smaller r indicates a non-ND. You can draw one of two conclusions:
(If you haven’t studied hypothesis testing yet, another way to say it is that you’re pretty sure the data set doesn’t fit the normal model because there’s less than a 5% probability that it does.)
This doesn’t mean you are certain it does, merely that you can’t rule it out. Technically you don’t know either way, but practically it doesn’t matter. Remember (or you will learn later) that inferential statistics procedures like t tests are robust, meaning that they still work even if the data are moderately non-normal. But if your data were extremely non-normal, r would be less than the critical value. When r is greater than the critical value, you don’t know whether the data set comes from normal data or moderately non-normal data, but either way your inferential statistics procedures are okay.
So the bottom line is, if r > CRIT, treat the data as normal, and if r < CRIT, don’t.
The normal probability plot is just one of many possible ways to determine whether a data set fits the normal model. Another method, the D’Agostino-Pearson test, uses numerical measures of the shape of a data set called skewness and kurtosis to test for normality. For details, see Assessing Normality in Measures of Shape: Skewness and Kurtosis.
Updates and new info: https://BrownMath.com/ti83/