ANOVA Notes for GraphNCalc83
ANOVA: This type of analysis allows one to compare the means of multiple experimental groups. Basically, the question addressed is: “Do the means of the groups vary significantly?” The ANOVA program performs a comparison of the two kinds of variation seen in the data set, variation within each group and the variation between the means of the different groups. The basic concept is that a high ratio (greater than 1) of ‘between group variation: within group variation’ indicates that the variable, not chance, is the likely reason for the differing group means.
Interpretation: If the F value is larger than the F-critical, the null hypothesis can be rejected. This indicates that the variation seen in your data sets (i.e. your experimental treatments) appears to be real. The variation seems to be due to the treatment and not just to random chance such as sampling error (variation).
Conversely, if the F is equal to or lower than F-critical, the null hypothesis can be accepted. This suggests that the variation observed between your data sets is likely due to chance (sampling). You cannot say with confidence that the experimental treatments varied in effect.
Also, the p value gives a confidence level. Typically, a 0.05 number (95% confidence level) is used as a cutoff, though more stringent studies might wish to use a 0.01 cutoff. If your obtained p value is less than 0.05 (your cutoff), you are more than 95% confident that the null hypothesis can be rejected, meaning that you think the variation is statistically significant and most likely produced by the effect of the variable. Conversely, a p value greater than 0.05 suggests that you are not confident that the variation is statistically significant, so you are forced to accept the null hypothesis.
Reminder: the null hypothesis states that the experimental groups do not vary. That is, your treatment does not produce an effect outside of normal variation.
The example below shows how to compute aone way ANOVAusing aTI-83 Plus family, TI-84 Plus family andTI-Nspire handheld in TI-84 Plus mode.
Example: To compare the distance traveled by three different brands of golf balls when struck by a driver, a completely randomized design is used. A robotic golfer, using a driver, is set up to hit a random sample of 24 balls (8 of each brand) in a random sequence. The distance is recorded for each hit, and the results are shown in the table below, organized by brand.
Brand / Brand A / Brand B / Brand C
Distance / 264.3 / 262.9 / 241.9
258.6 / 259.9 / 238.6
266.4 / 264.7 / 244.9
256.5 / 254 / 236.2
182.7 / 191.2 / 167.3
181 / 189 / 165.9
177.6 / 185.5 / 162.4
187.3 / 192.1 / 172.5
Mean / 221.8 / 224.9 / 203.7
Standard Deviation / 42.8 / 38.08 / 39.4
N / 8 / 8 / 8
Follow the steps below:
1)Press [STAT] [1] to access the STATlist editor.
2)Input thedata for Brand A, Brand B and Brand C in L1, L2 and L3. Press [2nd] [MODE] to returnto the home screen.
3)Press [STAT], scroll to the left to highlight TESTS and press [ALPHA] [^] to select the H:ANOVA( option. This will paste the ANOVA( function onto the home screen.
4)Input L1,L2,L3. Press the [ ) ] key to complete the ANOVA function.
5)Press [ENTER] to calculate the ANOVA.
The p-value is 0.530657, which shows that there is not a significant difference between any of the mean distances traveled by the three brands of balls. In this case, the null hypothesis cannot be rejected.
There is no F critical value to utilize in this application.