To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 35 specimens are buried in soil for an extended period. The maximum penetration (in mils) is then measured for each specimen, yielding a sample mean penetration of 52.7 and a sample standard deviation of 4.8. The conduits were manufactured with the specification that true average penetration be at most 50 mils. Does the sample data indicate that specifications have not been met

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jtellezd jtellezd · Answer 1 · 2021-06-03T03:45:19+02:00

Answer:

t(s) > t(c) then t(s) is in the rejection region for H₀.

We reject H₀ we support the claim that the mean penetration is bigger than that of the specifications

Step-by-step explanation:

Sample Information:

sample size n = 35

sample mean x = 52.7

sample standard deviation s = 4.8

spec. true average = 50

Test Hypothesis:

Null Hypothesis H₀ x = 50

Alternative Hypothesis Hₐ x > 50

The standarddeviation of the population is unknown therefore even though we assume distribution is normal we have to use t-student distribution

The alternative hypothesis indicates that the test is a one-tail test, we will test at significance level α = 5 % CI = 95 %

for α = 0.05 t(c) is from t-student -table and with df = 35 - 1

df = 34 t(c) = 1.69

t(s) = ( x - μ ) / s/√n

t(s) = ( 52.7 - 50 ) / 4.8/√35

t(s) = 2.7 * 5.92 / 4.8

t(s) = 3.33

Comparing t(s) and t(c)

t(s) > t(c) then t(s) is in the rejection region for H₀.

We reject H₀ we support the claim that the mean penetration is bigger than that of the specifications