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 greater than 650. What would be the p-value of this one-tailed test?
A) 0.840
B) 0.960
C) 0.040
D) 0.160

Question

Quizplus · Accepted Answer

For a one-tailed test where the alternative hypothesis is that the mean is greater than 650, the p-value is half of the two-tailed p-value. Therefore, 0.080 / 2 = 0.040.

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