Directions for using SigmaStat
One-Way Analysis of Variance

Besides learning how to do an ANOVA with SigmaStat, we are also going to examine how differences among the groups effect the mean square terms and ultimately our calculated F-Value.

Start SigmaStat. (by now you may have noticed that you can keep both your window for Netscape and a window for SigmaStat open and toggle back and forth between them).

Enter the following data in columns (one column for each group)
col 1     col 2      col 3      col 4      col 5
41          44         46        43          42
44          46         43        42          41
46          43         42        41          44
43          42         41        44          46
42          41         44        46          43
To run ANOVA: Choose Statistics, then Compare Many Groups, OneWay ANOVA. Then choose your data columns just as you did for a t-test, except that you should have choosen five columns. When you are done, click on Finish. Another window will come up asking you about a range test. For now, ignore this and click on Finish. The output is given below.

The output:
One Way Analysis of Variance	Wednesday, July 21, 1999, 09:13:14

Data source: Data 1 in Notebook

Normality Test:	Passed	(P = 0.216)

Equal Variance Test:	Passed	(P = 1.000)

Group	 N	Missing	
Col 1	 5	 0	
Col 2	 5	 0	
Col 3	 5	 0	
Col 4	 5	 0	
Col 5	 5	 0	

Group	 Mean	Std Dev	  SEM	
Col 1	43.200	1.924	 0.860	
Col 2	43.200	1.924	 0.860	
Col 3	43.200	1.924	 0.860	
Col 4	43.200	1.924	 0.860	
Col 5	43.200	1.924	 0.860	

Power of performed test with alpha = 0.050: 0.050

The power of the performed test (0.050) is below the desired power of 0.800.
You should interpret the negative findings cautiously.

Source of Variation	DF	 SS	 MS	  F	  P	
Between Treatments	 4      0.000   0.000   0.000   1.000	
Residual                20     74.000   3.700			
Total                   24     74.000				

The differences in the mean values among the treatment groups are not great
enough to exclude the possibility that the difference is due to random
sampling variability; there is not a statistically significant difference
(P = 1.000).

Explanation of the output:
Data Source:
Normality Test: Same test as that done for a two-sample t-test.
Equal Variance Test: Again a test for the equality of varainces.

Group: Label that you gave each column.
N: sample size for each group. In this case 5 for each column.
Missing: Number of blank cells. As we get into other models of ANOVA, missing values may preclude certain types of analysis.
Mean: Xbar for each group.
Std Dev: standard deviation for each group.
SEM: standard error of the mean for each group. In this case, note that Xbar, Std Dev, and SEM are all the same.

Power of ....: This is a claculation of power using the method presented in example 10.3 of Zar. Because all of our samples have the same observations, and thus there is no difference among the groups, our power is very low (equal to alpha). An interpretation of this value follows.

Source of Variation: This is the ANOVA table that is really the heart of the output. Between treatments is the same as the among groups source of variation. Residual is the within groups source of variation. This is often called the error source of variation. Total is the sum of the residual and between treatments source of variation. The column labeled DF is the degrees of freedom; between treatments is 4 (number of groups - 1 = 5 - 1 = 4), residual is 20 (number of observations minus the number of groups = 25 - 5 = 20). In the column labeled SS are the sum of squares. In the column labeled MS are the mean square for the sources of variation. These are obtained by dividing the sum of squares column by the degrees of freedom column. Note the total mean square is not used and is thus not calculated. The column labeled F is the calculated F-value, which is obtained by divided the between treatment mean square term by the residual mean square term. The column labeled P is the P-value.

Note that in this case the between treatments mean square term (among groups mean square) (0.000) is very small. Remember the mean square term is based on the variance of the Xbars around the grand mean. In this case, all the Xbars are equal to each other and to the grand mean, thus there is no variation among the Xbars, and thus between treatments terms are zero. As these terms are zero, the F-value is zero, and thus we fail to reject the null hypothesis that all of the means are equal.

Now start a new file, by clicking on File and then New and enter the follwoing set of data.
Column 1 is the same as above.
Column 2 is each observations in column 1 plus 1.
Column 3 is each observation in column 1 minus 1.
Column 4 is each observation in column 1 plus 10.
Coulmn 5 is each observation in column 1 minus 10.

You might take a moment and recall the effect of adding a constant to every observation on the smaple mean and sample variance.

col 1      col 2     col 3     col 4      col 5
 41         45         45       53         32
 44         47         42       52         31
 46         44         41       51         34
 43         43         40       54         36
 42         42         43       56         33
Rerun the analysis as above.

The output:
One Way Analysis of Variance	Wednesday, July 21, 1999, 09:16:08

Data source: Data 1 in Notebook

Normality Test:	Passed	(P = 0.216)

Equal Variance Test:	Passed	(P = 1.000)

Group	 N    Missing	
Col 1	 5	 0	
Col 2	 5	 0	
Col 3	 5	 0	
Col 4	 5	 0	
Col 5	 5	 0	

Group	 Mean  Std Dev    SEM	
Col 1	43.200	1.924	 0.860	
Col 2	44.200	1.924	 0.860	
Col 3	42.200	1.924	 0.860	
Col 4	53.200	1.924	 0.860	
Col 5	33.200	1.924	 0.860	

Power of performed test with alpha = 0.050: 1.000

Source of Variation     DF       SS        MS        F        P	
Between Treatments       4    1010.000	 252.500   68.243   <0.001	
Residual                20      74.000     3.700			
Total                   24    1084.000				

The differences in the mean values among the treatment groups are greater
than would be expected by chance; there is a statistically significant
difference  (P = <0.001).

All Pairwise Multiple Comparison Procedures (Tukey Test):

Comparisons for factor: 
Comparison	Diff of Means	 p	  q	P<0.05	
Col 4 vs. Col 5    20.000	 5	23.250	  Yes	
Col 4 vs. Col 3    11.000	 5	12.787	  Yes	
Col 4 vs. Col 1    10.000	 5	11.625	  Yes	
Col 4 vs. Col 2     9.000	 5	10.462	  Yes	
Col 2 vs. Col 5    11.000	 5	12.787	  Yes	
Col 2 vs. Col 3     2.000	 5	2.325	  No	
Col 2 vs. Col 1	    1.000	 5	1.162	  No	
Col 1 vs. Col 5    10.000	 5	11.625    Yes	
Col 1 vs. Col 3	    1.000	 5	1.162	  No	
Col 3 vs. Col 5     9.000	 5	10.462    Yes
Check that the sample means of each group have changed accoring to the constant that was added or subtracted from each group. Note the the Std Dev and the SEM did not change.

Compare the new ANOVA table to the one before. Note that the degrees of freedom did not change. Also note that the residual source of variation did not change (the sum of squares (74.00) and the mean square term (3.70) are the same as before). Note, however, that the between treatments (among groups) source of variation is very much larger. The resulting F-value (68.243) is much larger and the P-Value (P < 0.001) is much smaller. In fact, small enough that we can reject our null hypothesis.

The only thing that changed was a difference in the Xbars among the groups, thus you should be able to see how an analysis of variance becomes a test of the difference among the population means.

All Pairwise Multiple ....: Ignore this for now, we will come back to these types of tests later.

Return to Analysis of Variance .