656 / 762
656
Chapter 14: Hypothesis Tests – Small Samples
Conclusion
: Since - 66.1536 < -1.8596, reject the null hypothesis. That is,
there is sufficient sample evidence to indicate that the average for the
differences (after – before) is less than zero at the 5% level of significance.
This would indicate that the diet may have been effective in reducing body
weight. Also, this would indicate that the average for the weights after the
diet will be smaller than the average of the weights before the diet.
We can also use the
Small Sample Test for the Difference BetweenTwo
Dependent Means
workbook to help with the computations. This
workbookwill enable you to use summary information for the differences,
the actual differences, or the actual two sample data. On0e should be
cautious when entering both of the samples. For instance, if you needed the
test for (after – before), then you should enter “after” data as sample 1.
Figure 14-25
shows the output from the workbookwhen the actual data is
used in the computations.
Figure 14-25
: Display of the Small Sample Test for
for
Example 14-11
Since the
P
-value = 0.0000 for the left-tail test, then one will reject the null
hypothesis and conclude that the average weight after the diet is less than the
average weight before the diet. This would suggest that the diet was
effective in weight loss.