The quality control engineer for a furniture manufacturer is interested in the m

Question

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-tail 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. Suppose the engineer had decided that the alternative hypothesis to test was that the mean was less than 650. What would be the p-value of this one-tail test? OA. 0.960 OB. 0.840 O c. 0.040 OD. 0.160

Explanation / Answer

The one-tail P value is half the two-tail P value.

Two tail p-value = 2*p(Z <z) = 0.08

One tail p-value= P(Z < z ) = 0.08/2= 0.04

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.