In the prior discussion, we saw that there is a way of testing to see of all the means of several populations are not the same. Often, there are two factors involved and we want to see if the means are different within each factor. We will explain how this works with an example.
Suppose we want to look at students GPA's based on the type of their major (science, humanities, other) and their class status (freshmen, sophomore, junior, senior). We will use a level of significance of 0.05. We question find the GPA's for one randomly selected student for each of the 12 possible combinations. The results are shown in the table below.
Since there are two factors given, major and class status, we will have two separate pairs of hypotheses. H and H For the same reason we used the technique of ANOVA for a one-way table in the previous discussion, we will use ANOVA for this situation. In order to proceed, we need to make the following assumptions: - The measurements in each cell was
selected randomly from a normal distribution.
- The distributions from the cells all have
the same standard deviation.
- The values of each cell come from
independent samples.
- There are the same number of measurements in each cell (in the above example there was only one measurement taken per cell).
The calculation of the F statistic is not that enlightening to the elementary statistics student, so we will assume that a computer will be used for this calculation. The program (StatCrunch) that we have been using does not support 2-Way ANOVA; however there are free applets that do support 2-Way ANOVA. One such applet can be found at http://home.ubalt.edu/ntsbarsh/Business-stat/otherapplets/ANOVATwo.htm. We can think of the first factor as what block the student is in and the second factor a treatment that that student is given. This is the terminology that is given in the applet. The input and results using the above site are show below.
We can interpret the conclusions by saying that there is statistically sufficient evidence (P = 0.01224) to suggest that there is a difference between GPA's based on major, while there is not sufficient evidence (P = 0.10314) to suggest that the GPA's are different based on class status.
We have seen a method of testing for the difference between several means where three are two treatments. On the other hand, the example given showed only one value per cell. As we know, it is better to find a large sample, since the power of the test will be better. We can do this, but we must make sure that each cell is represented an equal number of times. This will only work if there is no interaction between the factors. Hence we will first test to see if there is an interaction. If there is an interaction, do not proceed. If there is not evidence for an interaction, then we can proceed as before. The next example illustrates this.
A researcher is interested in whether either people's hair color or their taste in music is a factor in cholesterol. Three people with each possible combination were tested for cholesterol. The results are shown below. Use a level of significance of 0 .05 to test researcher's hypothesis.
We will use a computer program found at http://home.ubalt.edu/ntsbarsh/Business-stat/otherapplets/ANOVA2Rep.htm The results are shown below.
We first notice that the P-Value for interactions is
0.89997 which is so large that we cannot reject the null hypothesis, that
is we do not have evidence to conclude that there is an interaction between the
two factors. Next, the P-Values for hair color is
0.63259, hence we do not have evidence to suggest that there any
particular hair color is more or less likely to predict a higher or lower
cholesterol. Finally, the music preference P-Value is
0.92041, hence there is also insufficient evidence to suggest that music
preference is a predictor of cholesterol.
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