EXAMPLE 6.13

Average scholarship amount of borrowers and nonborrowers: The conclusion. In Example 6.12, we found that the P-value is

P = 2P(Z ≥ 1.20) = 2(0.1151) = 0.2302

There is an 23% chance of observing a difference as extreme as the $425 in our sample if the true population difference is zero. Because this P-value is larger than the α = 0.05 significance level, we conclude that our test result is not significant. We could report the result as “the data fail to provide evidence that would cause us to conclude that there is a difference in average scholarship amount between borrowers and nonborrowers (z = 1.20, P = 0.23).”