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To determine whether the pipe welds in a nuclear power plant meet specifications, a random sample of welds is selected, and tests are conducted on each weld in the sample. Weld strength is measured as the force required to break the weld. Suppose the specifications state that mean strength of welds should exceed \(100 \mathrm{Ib} / \mathrm{in}^{2}\), the inspection team decides to test \(H_{0}: \mu=100\) versus \(H_{a}: \mu>100\). Explain why it might be preferable to use this \(H_{\mathrm{a}}\) rather than \(\mu<100\).

Short Answer

Expert verified
Using \(H_a: \mu > 100\) ensures weld strength meets safety specifications, focusing on compliance rather than weakness.

Step by step solution

01

Understanding Hypotheses

In hypothesis testing, we set up a null hypothesis \(H_0\) and an alternative hypothesis \(H_a\). The null hypothesis \(H_0: \mu = 100\) represents the status quo, where the mean strength of welds is 100 lb/in². The alternative hypothesis \(H_a: \mu > 100\) is what we are trying to find evidence for - whether the mean strength is greater than 100 lb/in².
02

Choosing the Right Alternative Hypothesis

The choice between \(H_a: \mu > 100\) and \(H_a: \mu < 100\) depends on the context and the objective of the test. In this case, the specification requires that the mean strength should exceed 100 lb/in², which means we want to ensure the strength is not just less than or equal to 100. Hence, \(H_a: \mu > 100\) aligns with the objective to verify if the welds are strong enough to exceed the minimum standard.
03

Implications of Each Hypothesis

Using \(H_a: \mu > 100\) focuses on proving that the weld strength meets or exceeds the safety specifications, which is crucial for operational safety. On the other hand, using \(H_a: \mu < 100\) would only test if the welds are weaker than required, which is not directly assessing if the specifications are met. Therefore, \(H_a: \mu > 100\) is more relevant and aligned with safety and compliance goals.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Null Hypothesis
In hypothesis testing, the *null hypothesis* is a fundamental concept. It is the statement we assume to be true until we have evidence proving otherwise. In this exercise, the null hypothesis, denoted as \(H_0: \mu = 100\), suggests that the average weld strength of the pipes is 100 lb/in².

The null hypothesis acts as a baseline or default position. Essentially, it represents the current accepted norm. We use it to determine if sufficient evidence exists to favor a change, represented by the alternative hypothesis. By adhering to the null hypothesis initially, scientists can objectively test their assumptions and seek clarity when analyzing their data.

It is important for the null hypothesis to be precise and testable. When performing tests, scientists use statistical methods to decide whether to "reject" or "fail to reject" the null hypothesis based on the evidence collected. Thus, it serves as a cornerstone for conducting fair and unbiased statistical analyses.
Alternative Hypothesis
Once the null hypothesis is set, we consider the *alternative hypothesis*. This is the hypothesis researchers are actually interested in proving. In this case, the alternative hypothesis \(H_a: \mu > 100\) puts forward that the mean strength of the welds is greater than 100 lb/in².

Choosing the appropriate alternative hypothesis is crucial as it directs the analysis towards what researchers are trying to establish or provide evidence for. The decision to pose \(H_a: \mu > 100\) instead of \(\mu < 100\) or \(\mu eq 100\) stems from the need to ensure that welds have adequate strength to meet safety specifications.

The alternative hypothesis challenges the status quo and prompts an investigation. It aligns more with the objective of testing – here to prove that the weld strength is substantial enough for safety standards. Emphasizing a greater strength directly supports nuclear safety compliance goals.
Statistical Significance
In hypothesis testing, determining *statistical significance* is a key step. It indicates whether the results of a test are strong enough to reject the null hypothesis. Statistical significance is typically determined using p-values or confidence intervals.

A p-value is the probability of obtaining test results at least as extreme as the observed data, assuming the null hypothesis is true. In our scenario, if the p-value is less than a predetermined significance level (often 0.05), we have enough evidence to reject \(H_0\) and accept \(H_a: \mu > 100\).

Achieving statistical significance provides confidence that the results are not due to random chance, but instead reflect true differences in weld strength. It reassures stakeholders that the test methods are valid and that the findings can be used to make informed decisions, thereby upholding the integrity of nuclear safety standards.
Nuclear Safety Testing
When dealing with critical infrastructure like nuclear power plants, *nuclear safety testing* cannot be overemphasized. It is crucial for ensuring that all components, such as pipe welds, meet stringent safety standards to prevent catastrophic failures.

Testing procedures often follow a rigorous protocol that involves hypothesis testing as just one of many assurance steps. Here, the goal is to affirm that the weld strength greatly exceeds the minimum set by specifications, thus confirming safety and reliability.

By establishing thorough testing protocols, nuclear facilities can continuously monitor and maintain their systems. This vigilance ensures the operational integrity of the plant and protects the environment and human lives. The choice of hypotheses in these tests is directly linked to these safety objectives, as only results that confirm meeting or exceeding safety criteria are acceptable.

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Most popular questions from this chapter

Let \(\mu\) denote the true average reaction time to a certain stimulus. For a \(z\) test of \(H_{0}: \mu=5\) versus \(H_{a}: \mu>5\), determine the \(P\)-value for each of the following values of the \(z\) test statistic. \(\begin{array}{lllllll}\text { a. } 1.42 & \text { b. } .90 & \text { c. } 1.96 & \text { d. } 2.48 & \text { e. }-.11\end{array}\)

The melting point of each of 16 samples of a certain brand of hydrogenated vegetable oil was determined, resulting in \(\bar{x}=94.32\). Assume that the distribution of the melting point is normal with \(\sigma=1.20\). a. Test \(H_{0}: \mu=95\) versus \(H_{a}: \boldsymbol{\mu} \neq 95\) using a twotailed level .01 test. b. If a level .01 test is used, what is \(\beta(94)\), the probability of a type II error when \(\mu=94\) ? c. What value of \(n\) is necessary to ensure that \(\beta(94)=.1\) when \(\alpha=.01\) ?

Before agreeing to purchase a large order of polyethylene sheaths for a particular type of high-pressure oilfilled submarine power cable, a company wants to see conclusive evidence that the true standard deviation of sheath thickness is less than \(.05 \mathrm{~mm}\). What hypotheses should be tested, and why? In this context, what are the type I and type II errors?

The paint used to make lines on roads must reflect enough light to be clearly visible at night. Let \(\mu\) denote the true average reflectometer reading for a new type of paint under consideration. A test of \(H_{0}: \mu=20\) versus \(H_{a}: \mu>20\) will be based on a random sample of size \(n\) from a normal population distribution. What conclusion is appropriate in each of the following situations? a. \(n=15, t=3.2, \alpha=.05\) b. \(n=9, t=1.8, \alpha=.01\) c. \(n=24, t=-.2\)

Reconsider the paint-drying problem discussed in Example 8.5. The hypotheses were \(H_{0}: \mu=75\) versus \(H_{\mathrm{a}}: \mu<75\), with \(\sigma\) assumed to have value 9.0. Consider the alternative value \(\mu=74\), which in the context of the problem would presumably not be a practically significant departure from \(H_{0}\) - a. For a level \(.01\) test, compute \(\beta\) at this alternative for sample sizes \(n=100,900\), and 2500 . b. If the observed value of \(\bar{X}\) is \(\bar{x}=74\), what can you say about the resulting \(P\)-value when \(n=2500\) ? Is the data statistically significant at any of the standard values of \(\alpha\) ? c. Would you really want to use a sample size of 2500 along with a level .01 test (disregarding the cost of such an experiment)? Explain.

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