MATH SOLVE

3 months ago

Q:
# 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 = 53.4 and a sample standard deviation of s = 4.4. 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 α = 0.05.) State the appropriate null and alternative hypotheses.

Accepted Solution

A:

Answer:There is not enough evidence to support the claim that the true average penetration is at most 50 mils.Step-by-step explanation:We are given the following in the question: Population mean, μ = 50Sample mean, [tex]\bar{x}[/tex] = 52.8Sample size, n = 45Alpha, α = 0.05 Sample standard deviation, s = 4.5First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu \leq 50\text{ mils}\\H_A: \mu > 50\text{ mils}[/tex] We use one-tailed t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex] Putting all the values, we have [tex]t_{stat} = \displaystyle\frac{52.8 - 50}{\frac{4.5}{\sqrt{45}} } = 4.1739[/tex] Now, [tex]t_{critical} \text{ at 0.05 level of significance, 44 degree of freedom } = 1.6802[/tex] Since,
The calculated test statistic is greater than the critical value, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.Conclusion:There is not enough evidence to support the claim that the true average penetration is at most 50 mils.