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 = 4.2. 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 alpha = 0.05.) State the appropriate null and alternative hypotheses. H_0: mu = 50 H_a: mu notequalto 50 H_0: mu = 50 H_a: mu > 50 H_0: mu notequalto 50 H_a: mu > 50 H_0: mu > 50 H_a: mu = 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. 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. Reject the null hypothesis. There is not sufficient evidence to conclude that the true average penetration is more than 50 mils. Do not reject the null hypothesis. There is 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 a random sample of 45 specimens was taken and buried for two years.

The study variable is:

X: Max. penetration of a steel conduit.

The data of the sample

n= 45

sample mean X[bar]= 52.9

sample standard deviation S= 4.2

The conduits are manufactured to have a true average penetration of at most 50 mills, symbolically: μ ≤ 50

The hypothesis is:

H₀: μ ≤ 50

H₁: μ > 50

α: 0.05

To choose the corresponding statistic to use to study the population mean, the variable must have a normal distribution. There is no available information to check this, so I'll just assume that the variable has a normal distribution and, since the population variance is unknown and the sample is small, the statistic to use is a Student t.

Under the null hypothesis, the critical region and the p-value are one-tailed.

Critical value: [tex]t_{n- 1;1-\alpha} =t_{44;0.95} =1.68[/tex]

Rejection rule:

Reject the null hypothesis when t ≥ 1.68

t= [tex]\frac{52.9-50}{\frac{4.2}{\sqrt{45} } }[/tex]

t= 2.9-0.626 = 2.27

If the calculated value is greater than the critical value, the decision is to reject the null hypothesis.

p-value:

P(t ≥ 2.27) = 1 - P(t < 2.27) = < 0.00001.

The p-value is less than α so the decision is to reject the null hypothesis.

Since the null hypothesis was rejected, then the population average of the penetration of the conduits specimens is greater than 50 mils. It is not recommendable to use these conduits.

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