To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 25 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. According to the normal probability plot, we can observe a linear pattern using the sample we have. 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?
(a) What is the parameter of interest? Define it using any Greek letter you want.
(b) State hypotheses based on the question.
(c) If the population variance of the penetration is known as 36, what kind of the test statistics is the best to conduct the statistical test?
(d) Based on the (c), conduct the statistical test using the significance level 0.05 based on i. critical value
ii. p-value
(e) State the result of (d), formally, using necessary statistical terminology.
(f) Interpret the conclusion of (e) in the context of the question.
(g) Obtain 95% lower confidence bound for the parameter you defined in (a). Compare the calcu- lated confidence bound to the previous result of the statistical test.
(h) What is the power of the test when the true average penetration is 52? Interpret it.
(i) How many samples do we need to obtain the 80% power of the test when the significance level is 5%?
Part a)
parameter of interest is: Average penentration in mils it is denoted by µ
Part
b)
hypothesis is as follows:
Null hypothesis: µ ≤ 50 [ at the most 50 , not more than that]
Alternate hypothesis: µ > 50
[right tailed test, as M > 50 would leas to rejection of the specimen]
.
Part c)
Given population variance = σ^2 = 36
This implies population standard deviation σ = sqrt(36) = 6
Hence when population variance is provided one can easily use one sample Z test for the given scenario
.
Part d)
Z = ( x bar - µ) / (σ / sqrt(n))
Z = (52.7 -50) / (6/sqrt(25))
z = 2.7 / (6/5)
z = 2.7/1.2
Z = 2.25
P value = P(z > 2.25) = 1 - 0.9878 = 0.0122
.
Part e)
Interpretation: Since P value 0.0122 < alpha 0.05
we reject the null hypothesis
.
Part
f)
conclusion: Since we conclude to reject the null hypothesis, this
implies there is sufficient statistical evidence to support the
claim that the mean is not at the most 50, it is above 50
To obtain information on the corrosion-resistance properties of a certain type of steel conduit, ...
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 = 52.9 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...
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 = 52.8 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...
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To obtain format on o the comes on esistance proper es of a certain type of sted con t, 45 speamens are buried in soil or a 2-yaar penod. The ma mum penetration in m s or each specmen s then measured y na a sample average pen tration o x = 52.9 and a sample standard de ation at s 4 i The conduits were manifac red viltn the snec cation that true averaQA penetratiถn e at most sn...
Please use MATLAB. Part B (Based off week 11 workshop content) An engineering company has ordered steel to be used for a rail construction project. Before using the steel, they want to test it to ensure the material properties from the supplier are correct A random sample of the ordered steel is tested for its Yield Strength. The results are shown below. Yield Strength (MPa) 197203194196196197198205 1. Calculate the sample mean and sample variance by hand, then verify your answers...
Part B (Based off week 11 workshop content) An engineering company has ordered steel to be used for a rail construction project. Before using the steel, they want to test it to ensure the material properties from the supplier are correct. A random sample of the ordered steel is tested for its Yield Strength. The results are shown below. Yield Strength (MPa) 197 203 194 196 196 197 198 205 1. Calculate the sample mean and sample variance by hand,...