Microsoft Office Tutorials and References
In Depth Information
If we reject the null hypothesis that there is no difference in the average scores,
then we are deciding in favor of the training indeed leading to a difference in scores.
As before, the decision will be made on the basis of a statistic that is calculated
from the sample data, in this case the z-Statistic , which is then compared to a
critical value. The critical value incorporates the decision maker’s willingness to
commit an error by possibly rejecting a true null hypothesis. Alternatively, we can
use the p-value of the test and compare it to the level of signiﬁcance which we have
adopted—frequently assumed to be 0.05. The steps in this procedure are quite sim-
ilar to the ones we performed in the chi-square analysis, with the exception of the
statistic that is calculated, z rather than chi-square.
6.5.2 Is There a Difference in Scores for SC Non-Prisoners and
EB Trained SC Prisoners?
The procedure for the analysis is shown in Exhibits 6.2 and 6.3. Exhibit 6.2 shows
the Data Analysis dialogue box in the Analysis group of the Data ribbon used to
perform the z-Test. We begin data entry for the z-Test in Exhibit 6.3 by identi-
fying the range inputs, including labels, for the two samples: 36 SC non-prisoner
standard trained scores (E1:E37) and 36 SC prisoners that receive special training
(G1:G37). Next, the dialog box requires a hypothesized mean difference. Since we
are assuming there is no difference in the null hypothesis, the input value is 0. This
Exhibit 6.2 Data analysis tool for z-test