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

Question

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

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