How to Perform an ANOVA Test on a TI-83 or TI-84
Hannah Schultheis and Julie Phillis
May 21, 2015
An ANOVA, or, Analysis Of Variance test, is used to determine if there is a significant
difference between the means of n number of groups where n >2. To begin, you must first create
a null and alternate hypothesis. The null hypothesis claims there is no difference between the
means of the data sets and they are all equal. The alternate, however, is the hypothesis we believe
to be true and which we are trying to find evidence in favor of, which is that there is a significant
difference between the means of each data set.
Example:
The National Transportation Safety Board (NTSB) wants to look at the safety of three different
car sizes. They collect data from three different sizes of cars. Using the data below, determine
whether the mean pressure applied to the driver's head during a crash is equal for each type of
car. Determine at α=.05 and α=.01
Compact Cars
Midsize Cars
Full-Size Cars
643
469
484
655
427
456
702
525
402
To perform this test on a TI-83 or a TI-84 calculator, you must first enter each set of data into it’s
own list. To do so, you click the “STAT” button and then select the “Edit...” option.
Once all of the numerical values have been entered into the list, press the “STAT” button and tab
over to the third tab titled, “TESTS.”
Go all the way to the bottom of this page to the entry titled, “ANOVA(“ and select this by
pushing the enter button. Then enter the list numbers by pushing the “2ND” button, then the 1
button, followed by a comma, followed by “2ND” and the 2 button, followed by a comma, and
lastly followed by “2ND” and the 3 button. This step must be altered if you enter more than three
sets of data into your calculator. In this case you must also select the additional data lists.
Click “ENTER” and the results of the ANOVA test will magically appear on the screen before
you.
The only two values you are concerned with are the F-Value and the p-Value; these are the first
two numbers on the results page. For this example the F-Value is 25.175 and the p-Value is
0.0012. Now that those two values are determined you have to compare the p-Value to the alpha
levels 0.05 and 0.01. Since the p-Value for the example is smaller than 0.01 it is estimated that
there is significant evidence to accept the alternate hypothesis that says there is a significant
difference in means. Had the p-Value been smaller than 0.05, but greater than 0.01 we would
accept the alternate hypothesis with strong evidence. When the ANOVA test gives you a p-Value
that is greater than both 0.05 and 0.01 the alternate hypothesis is to be rejected. Happy learning!