The rule is….if the absolute value of Calc-t >Critical-t then reject. From Excel (or from the tables on P101) we see that Critical-t9=1.833. As our calc-t = 1.73 the Null Hypothesis cannot be rejected. Therefore it seems that there is insufficient evidence to suggest that the revision course was effective. The Excel calculations are shown in Figure 2.28.
Figure 2.28: Excel spreadsheet showing hypothesis testing for a paired t-test
2.13 The t-test for independent samples In most cases, samples will be independent rather than paired, and if we want to test the significance of the difference between their means, we must use a different method to the one presented above. For example, the senior management of University A claim that on average their reward package is better than the reward package offered by their competitor University B. A random sample of 20 employees from University A and a random sample of 30 employees from University B were asked to score on several dimensions the reward scheme offered to them and the following scores were recorded as shown in Figure 2.29.
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