Example 6.17

EXAMPLE 6.17 Bottle Fill Amount: The Test Statistic

CASE 6.1 For Example 6.13 (page 317), the null hypothesis is $H_{0} : μ = 473$ , and a sample gave $\bar{x} = 474.54$ . The test statistic for this problem is the standardized version of $\bar{x}$ :

$z = \frac{\bar{x} - μ}{σ / \sqrt{n}}$

This statistic is the distance between the sample mean and the hypothesized population mean in the standard scale of $z$ -scores. In this example,

$z = \frac{474.54 - 473}{2 / \sqrt{20}} = 3.44$

Reminder

68-95-99.7 rule, p. 43

Even without a formal probability calculation, by simply recalling the 68–95–99.7 rule for the Normal, we realize that a $z$ -score of 3.44 is an unusual value. This suggests incompatibility of the observed sample result with the null hypothesis.