Instruction 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 Instruction 9-6,suppose the engineer had decided that the alternative hypothesis to test was that the mean was greater 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

Since the p-value for the two-tailed test is 0.080, for a one-tailed test (mean greater than 650), the p-value would be half of that (0.040), which is less than the significance level of 0.10, leading to rejection of the null hypothesis.

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