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What is our verdict for the data? Do we reject the null? We are interested in the
statistics associated with the sources of variation entitled columns . Why? Because
in the original data used by Excel, the software product factor was located in the
columns of the table. Each treatment, Product a–d, contained a column of data for
the 5 block groups that were submitted to the experiment. According to the analysis
in Table 6.13, the F-Statistic, 31.182186 (cell E28) is much larger than the critical
F, 3.490295 (cell G28) and our p-value is 0.00000601728 (cell F28) which is much
smaller than the assumed
of 0.05. Given the results, we clearly must reject the
null hypothesis in favor of the alternative— at least one of the mean task comple-
tion times is significantly different from the others. If we reexamine the summary
statistics in D19:D22 of Table 6.13, we see that at least two of our averages, 29.2
(b) and 32 (d), are much larger than the others, 19.2 (a) and 17.6 (c). So, this casual
examination substantiates the results of the ANOVA.
6.7.2 Factorial Experimental Design Example
Now, let us return to our prisoner/non-prisoner and special training/ no special
training two factors example. Suppose we collect a new set of data for an exper-
imental study—24 observations of equal numbers of prisoners/non-prisoners and
standard trained/special trained. This implies a selection of two factors of interest:
prisoner status and training. The treatments for prisoner status are prisoner and non-
prisoner , while the treatments for training are trained and not-trained . The four cells
formed by the treatments each contain 6 replications (unique individual scores) and
lead to another type of ANOVA —ANOVA: Two-Factor with Replication .
Table 6.14 shows the 24 observations in the two-factor format and Table 6.15
shows the result of the ANOVA. The last section in Table 6.15, entitled ANOVA ,
provides the F-Statistics (E34:E36) and p-values (cells F34:F36) to reject the null
hypotheses related to the effect of both factors. In general, the null hypotheses
states that the various treatments of the factors do not lead to significantly different
averages for the scores.
Factor A (Training) and Factor B (Prisoner Status) are represented by the sources
of variation entitled Sample and Columns , respectively. Factor A has an F-Statistic
of 1.402199 (cell E34) and a critical value of 4.35125 (cell G34), thus we cannot
reject the null. The p-value, 0.250238 (cell F34), is much larger than the assumed
of 0.05.
Factor B has an F-Statistic of 4.582037 (cell E35) that is slightly larger than
the critical value of 4.35124 (cell G35). Also, the p-value, 0.044814 (cell F35), is
slightly smaller than 0.05. Therefore, for Factor B we can reject the null hypothesis,
but not with overwhelming conviction. Although the rule for rejection is quite clear,
a result similar to the one we have experienced with Factor B might suggest that fur-
ther experimentation is in order. Finally, the interaction of the factors does not lead
us to reject the null. The F-Statistic is rather small, 0.101639 (cell E36), compared
to the critical value, 4.35124 (cell G36).
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