Question

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 that the specifications state that the mean strength of welds should exceed 100 \(\mathrm{lb} / \mathrm{in} .^{2}\). The inspection team decides to test \(H_{0}: \mu=100\) versus \(H_{a}: \mu>100 .\) Explain why this alternative hypothesis was chosen rather than \(\mu<100\).

Short answer

The alternative hypothesis \(H_{a}: \mu>100\) was chosen because the specifications state that the mean strength of the welds should exceed 100 lb/in.^{2}, indicating that the team wants to verify if the strength is more than the specified limit. On the other hand, \(\mu<100\) would suggest looking for deficiencies, i.e., if the weld strength is below the required specifications, which is not the goal in this case.

Step-by-step solution

Understand the Hypothesis Testing Concept

A hypothesis test in statistics is a method where an assumption about a parameter in a statistical population is tested. The null hypothesis \(H_{0}\) represents a theory that has been put forward, either because it is believed to be true or because it is to be used as a basis for argument, but has not been proved. The alternative hypothesis, \(H_{a}\), represents a theory that will challenge the null hypothesis. In our case, \(H_{0}: \mu=100\) is the assumption that the mean strength of welds is exactly 100 lb/in.^{2}, and the alternative hypothesis is designed to test if the welds exceed that strength.

Interpret the Given Context

In the given context, the specification states that the mean strength of welds should exceed 100 lb/in.^{2}, not less than or equal to. So, the alternative hypothesis should be based on exceeding the mean strength of 100 lb/in.^{2}, rather than finding a deficiency in strength. Hence, \(H_{a}: \mu>100\) was chosen instead of \(\mu<100\).

Explaining Why \(\mu

An alternative hypothesis of \(\mu<100\) would suggest that the inspection team is looking for deficiencies in strength, i.e., they suspect the weld strength to be below the required specifications. However, in this case, they wish to confirm that the weld strength meets or exceeds the specifications. Hence, the choice of \(H_{a}: \mu>100\) is appropriate and aligns with the team's goal

Key concepts

Null Hypothesis (H0)

The null hypothesis, symbolized as \(H_0\), is a fundamental aspect of hypothesis testing in statistics. It is a statement used as a default assumption that there is no effect or no difference; it posits that any observed variation is due to chance or random occurrences. Essentially, when we talk about \(H_0\), we're referring to the status quo or a specific value for a statistical population parameter that we aim to test.

For instance, in weld strength analysis, when inspectors posit that \(H_0: \mu=100\) lb/in\(^2\), they are saying, based on their current understanding or the established standards, weld strength is expected to average 100 lb/in\(^2\). If evidence collected through testing shows otherwise, they may reject \(H_0\) in favor of the alternative hypothesis. In creating \(H_0\), it is crucial to be precise, as it determines the structure of the hypothesis test and what the evidence must address.

Alternative Hypothesis (Ha)

In contrast to the null hypothesis, the alternative hypothesis, denoted \(H_a\) or \(H_1\), represents a statement that directly contradicts \(H_0\). It is what a researcher seeks to demonstrate to be a plausible explanation when \(H_0\) is rejected. This hypothesis suggests that there is a statistically significant effect or difference, and it's what's tested against the null hypothesis.

In the context of weld strength, \(H_a: \mu>100\) lb/in\(^2\) implies that the inspectors have a suspicion or want to prove that the actual average strength of the welds is greater than the specification. The choice of this specific alternative hypothesis over \(\mu<100\) indicates a directionality in their testing: they want evidence to support the claim that the weld quality exceeds, rather than falls short of, the standard. This directional nature of \(H_a\) guides the statistical test deployed and influences how the statistical evidence is interpreted.

Statistical Population Parameter

A statistical population parameter is a value that represents a particular characteristic of an entire population. Parameters such as the mean (average), standard deviation, proportion, and variance are used to summarize and describe the prominent features of a large set of data.

In hypothesis testing, we often focus on estimating or testing these parameters to make inferences about the population. For example, specifying \(\mu=100\) lb/in\(^2\) as the mean weld strength sets the benchmark for a nuclear plant's entire population of welds. Inspectors use samples to estimate this parameter and determine if the population parameter likely differs from the specified value. It's through this process they arrive at decisions concerning the population as a whole based on findings from sample data.

Weld Strength Analysis

Weld strength analysis involves examining the integrity and robustness of weldments. It is essential in fields like construction and manufacturing where safety is paramount. By testing the force required to break a weld, inspectors can assess whether it meets established specifications. This information is critical to ensure structures are durable and reliable.

In statistical terms, weld strength forms the dataset of a particular characteristic (force required to break) from a sample of welds. By applying hypothesis testing to this dataset, inspectors can conclude whether the average weld strength in the overall population surpasses the minimum threshold. The procedures involve collecting data, making calculations, such as the mean and standard deviation, and running the appropriate statistical tests (like t-tests or z-tests) to evaluate the null hypothesis against the alternative.