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 = 52.7 and a sample standard deviation of s = 4.9. 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 Họ: H+ 50 Hi > 50 • Hoi H - 50 Hiu > 50 O Hoi H > 50 H,il = 50 O Hạ: H = 50 HgiH # 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 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. • 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 not sufficient evidence to conclude that the true average penetration is more than 50 mils.

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.5: Comparing Sets Of Data
Problem 13PPS
icon
Related questions
Question
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 = 52.7 and a sample standard deviation of s = 4.9. 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.
Ho: u + 50
Haiu > 50
O Ho: µ = 50
H: µ > 50
Ho:H > 50
H: u = 50
Ho:u = 50
HaiH# 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.
Do not reject the null hypothesis. There is sufficient evidence to conclude that the true average penetration is more than 50 mils.
Reject the null hypothesis. There is not sufficient evidence to conclude that the true average penetration is more than 50 mils.
Reject the null hypothesis. There is sufficient evidence to conclude that the true average penetration is more than 50 mils.
Do not reject the null hypothesis. There is not sufficient evidence to conclude that the true average penetration is more than 50 mils.
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 = 52.7 and a sample standard deviation of s = 4.9. 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. Ho: u + 50 Haiu > 50 O Ho: µ = 50 H: µ > 50 Ho:H > 50 H: u = 50 Ho:u = 50 HaiH# 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. Do not reject the null hypothesis. There is sufficient evidence to conclude that the true average penetration is more than 50 mils. Reject the null hypothesis. There is not sufficient evidence to conclude that the true average penetration is more than 50 mils. Reject the null hypothesis. There is sufficient evidence to conclude that the true average penetration is more than 50 mils. Do not reject the null hypothesis. There is not sufficient evidence to conclude that the true average penetration is more than 50 mils.
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps with 4 images

Blurred answer
Knowledge Booster
Point Estimation, Limit Theorems, Approximations, and Bounds
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Glencoe Algebra 1, Student Edition, 9780079039897…
Glencoe Algebra 1, Student Edition, 9780079039897…
Algebra
ISBN:
9780079039897
Author:
Carter
Publisher:
McGraw Hill