To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 45 specimens are buried in soil for a 2-year period. The maximum penetration (in mils) for each specimen is then measured, yielding a sample average penetration of x52.6 and a sample standard deviation of s4.6. The conduits were manufactured with the specification that true average penetration be at most 50 mils. They will be used unless it can be demonstrated conclusively that the specification has not been met. What would you conclude? (Use a - 0.05.) State the appropriate null and alternative hypotheses. O HI> 50 H,I - 50 O Hi 50 H> 50 O M 50 H 50 H > 50 Calculate the test statistic and determine the Prvalue. (Round your test statistic to twe decimal places and your Prvalue to four decimal places.) Pvalue State the conclusion in the problem context. • Reject the null hypothesis. There is sufficient evidence to conclude that the true average penetration is more than 50 mils. O Do not reject the null hypothesis. There is sufficient evidence to conclude that the true average penetration is more than 30 mils. O Do not reject the null hypothesis. There is not sufficient evidence to conclude that the true average penetration is more than 50 mils. O Reject the null hypothesis. There is not sufficient evidence to conclude that the true average penetration is more than 50 mils.
To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 45 specimens are buried in soil for a 2-year period. The maximum penetration (in mils) for each specimen is then measured, yielding a sample average penetration of x52.6 and a sample standard deviation of s4.6. The conduits were manufactured with the specification that true average penetration be at most 50 mils. They will be used unless it can be demonstrated conclusively that the specification has not been met. What would you conclude? (Use a - 0.05.) State the appropriate null and alternative hypotheses. O HI> 50 H,I - 50 O Hi 50 H> 50 O M 50 H 50 H > 50 Calculate the test statistic and determine the Prvalue. (Round your test statistic to twe decimal places and your Prvalue to four decimal places.) Pvalue State the conclusion in the problem context. • Reject the null hypothesis. There is sufficient evidence to conclude that the true average penetration is more than 50 mils. O Do not reject the null hypothesis. There is sufficient evidence to conclude that the true average penetration is more than 30 mils. O Do not reject the null hypothesis. There is not sufficient evidence to conclude that the true average penetration is more than 50 mils. O Reject the null hypothesis. There is not sufficient evidence to conclude that the true average penetration is more than 50 mils.
A First Course in Probability (10th Edition)
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ISBN:9780134753119
Author:Sheldon Ross
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Chapter1: Combinatorial Analysis
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Transcribed Image Text:To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 45 specimens are buried in soil for a 2-year period. The maximum penetration (in mils) for each specimen is then measured, yielding a sample average penetration of x = 52.6 and a sample standard deviation of s = 4.6. The conduits were manufactured with the
specification that true average penetration be at most 50 mils. They will be used unless it can be demonstrated conclusively that the specification has not been met. What would you conclude? (Use a = 0.05.)
State the appropriate null and alternative hypotheses.
O H,i 4 > 50
H:H = 50
Ο Η, μέ 50
H:u > 50
O Hi = 50
Hoi H = 50
HH > 50
Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.)
P-value =
State the conclusion in the problem context.
O Reject the null hypothesis. There is sufficient evidence to conclude that the true average penetration is more than 50 mils.
O Do not reject the null hypothesis. There is sufficient evidence to conclude that the true average penetration
more than 50 mils.
O Do not reject the null hypothesis. There is not sufficient evidence to conclude that the true average penetration is more than 50 mils.
O Reject the null hypothesis. There is not sufficient evidence to conclude that the true average penetration is more than 50 mils.
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