Example 6.12

EXAMPLE 6.12

Average scholarship amount of borrowers and nonborrowers: The P-value. In Example 6.11, we found that the test statistic for testing

H₀: the true mean difference is 0

versus

H_a: there is a difference in the population means

If H₀ is true, then z is a single observation from the standard Normal, N(0, 1), distribution. Figure 6.9 illustrates this calculation. The P-value is the probability of observing a value of Z at least as extreme as the one that we observed, z = 1.20. From Table A, our table of standard Normal probabilities, we find

P(Z ≥ 1.20) = 1 − 0.8849 = 0.1151

The probability for being extreme in the negative direction is the same:

P(Z ≤ −1.20) = 0.1151

So the P-value is

P = 2P(Z ≥ 1.20) = 2(0.1151) = 0.2302

367

FIGURE 6.9 The P-value, Example 6.12. The P-value is the probability (when H₀ is true) that takes a value as extreme or more extreme than the actual observed value, z = 1.20. Because the alternative hypothesis is two-sided, we use both tails of the distribution.

This is the value that we reported on page 361. There is a 23% chance of observing a difference as extreme as the $425 in our sample if the true population difference is zero. This P-value tells us that our outcome is not particularly extreme. In other words, the data do not provide substantial evidence for us to doubt the validity of the null hypothesis.