Question

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%?

Answer 1

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