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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 \(\overline x = 52.7\) and a sample standard deviation of s = 4.8. 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?

Short Answer

Expert verified

Reject null hypothesis.

Step by step solution

01

Step 1:Null hypothesis.

The null hypothesis, denoted by H0, is the claim that is initially assumed to be true (the 鈥減rior belief鈥 claim). The alternative hypothesis, denoted by Ha, is the assertion that is contradictory to H0.

The null hypothesis will be rejected in favour of the alternative hypothesis only if sample evidence suggests that H0 is false. If the sample does not strongly contradict H0, we will continue to believe in the plausibility of the null hypothesis. The two possible conclusions from a hypothesis-testing analysis are then reject H0 or fail to reject H0.

02

To find z-value.

The hypothesis of interest are \({H_0}:\mu = 50\) versus \({H_a}:\mu < 50\)

Assumption that the sample is from normal population distribution with known standard deviation allows using test statistic:

\(z = \frac{{\overline X - {\mu _0}}}{{\sigma /\sqrt n }}\)

The \(n\) is big enough to assume this. This z statistic value is

\(\begin{array}{l}z = \frac{{52.7 - 50}}{{4.8/\sqrt {45} }}\\z = 3.77\end{array}\)

03

Hypothesis is reject or not.

The critical value \({z_\alpha }\), for \(\alpha = 0.05\)(other level of significance could have been chosen), because of one sided testing is,

\(\begin{array}{l}{z_\alpha } = {z_{0.05}}\\{z_\alpha } = 1.645\end{array}\)

The fact ,

\(\begin{array}{l}z = 3.77 > 1.65\\z = {z_\alpha }\end{array}\)

Implies that the null hypothesis is rejected at \(0.05\) level of significance. The conclusion is that the true average exceed \(50\) miles.

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