EXAMPLE 6.20

Average scholarship amount of borrowers and nonborrowers: Assessing significance. In Example 6.11 (page 365), we found the test statistic z = 1.20 for testing the null hypothesis that there was no difference in the mean scholarship amount between borrowers and nonborrowers. The alternative was two-sided. Under the null hypothesis, z has a standard Normal distribution, and from the last row in Table D, we can see that there is a 95% chance that z is between ±1.96. Therefore, we reject H₀ in favor of H_a whenever z is outside this range. Because our calculated value is 1.20, we are within the range and we do not reject the null hypothesis at the 5% level of significance.