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 = 52.2 and a sample standard deviation of s = 4.3. The conduits were manufactured with the specification that true average penetration be at most 50 mils. They willI 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 Ho: H = 50 H: H = 50 O Hoi H# 50 H: H > 50 O Ho: H > 50 H3:µ = 50 Ο H μ= 50 H:µ > 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 evidenc 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 50 mils. O Reject the null hypothesis. There is not sufficient evidence to conclude that the true average penetration is 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.
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 = 52.2 and a sample standard deviation of s = 4.3. The conduits were manufactured with the specification that true average penetration be at most 50 mils. They willI 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 Ho: H = 50 H: H = 50 O Hoi H# 50 H: H > 50 O Ho: H > 50 H3:µ = 50 Ο H μ= 50 H:µ > 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 evidenc 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 50 mils. O Reject the null hypothesis. There is not sufficient evidence to conclude that the true average penetration is 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.
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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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 = 52.2 and a sample standard deviation of s = 4.3. 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.
Ο H: μ 50
H: u # 50
O Ho: µ + 50
H3: H > 50
O Ho: H > 50
H: u = 50
O Ho: H = 50
H: u > 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.)
z =
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 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.
O Do not 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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