Question

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 = 4.5. 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.)

Answer 1

We conclude that specification has not been met as the true average penetration comes out to be greater than 50 mils.

Explanation:

We are given 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 = 4.5.

We have to test if the true average penetration would be at most 50 mils.

Let, NULL HYPOTHESIS,
`H_0` :
`\mu \leq` 50 mils {means that the true average penetration is at most 50 mils}

ALTERNATE HYPOTHESIS,
`H_a` :
`\mu` > 50 mils {means that the true average penetration is greater than 50 mils}

The test statistics that will be used here is One-sample t-test;

T.S. =
`(\bar X - \mu)/((s)/(√(n) ) )` ~
`t_n_-_1`

where,
`\bar X` = sample average penetration = 52.8

s = sample standard deviation = 4.5

n = sample of specimens = 45

So, test statistics =
`(52.8-50)/((4.5)/(√(45) ) )` ~
`t_4_4`

= 4.174

Now, at 0.05 significance level, t table gives a critical value of 1.6808 at 44 degree of freedom. Since our test statistics is way higher than the critical value of t so we have sufficient evidence to reject null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the true average penetration is greater than 50 mils.

Answer 2

There is not enough evidence to support the claim that the true average penetration is at most 50 mils.

Explanation:

We are given the following in the question:

Population mean, μ = 50

Sample mean,
`\bar{x}` = 52.8

Sample size, n = 45

Alpha, α = 0.05

Sample standard deviation, s = 4.5

First, we design the null and the alternate hypothesis

`H_(0): \mu \leq 50\text{ mils}\\H_A: \mu > 50\text{ mils}`

We use one-tailed t test to perform this hypothesis.

Formula:

`t_(stat) = \displaystyle\frac{\bar{x} - \mu}{(s)/(√(n)) }`

Putting all the values, we have

`t_(stat) = \displaystyle(52.8 - 50)/((4.5)/(√(45)) ) = 4.1739`

Now,
`t_(critical) \text{ at 0.05 level of significance, 44 degree of freedom } = 1.6802`

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.