TABLE 9-6
The quality control engineer for a furniture manufacturer is interested in the mean amount of force necessary to produce cracks in stressed oak furniture. She performs a two-tailed test of the null hypothesis that the mean for the stressed oak furniture is 650. The calculated value of the Z test statistic is a positive number that leads to a p-value of 0.080 for the test.
-Referring to Table 9-6, suppose the engineer had decided that the alternative hypothesis to test was that the mean was less than 650. Then if the test is performed with a level of significance of 0.10, the null hypothesis would be rejected.

Question

Quizplus · Accepted Answer

In Table 9-6, at a level of significance of 0.10 and for a one-tailed test (less than), the critical Z value is -1.28. Since the calculated Z value is positive, the null hypothesis would not be rejected. Therefore, the statement is false.

TABLE 9-6 The Quality Control Engineer for a Furniture Manufacturer Is Interested