A company that manufactures steel wires guarantees that the mean breaking strength (in kilonewtons) of the wires is greater than . They measure the strengths for a sample of wires and test Hv0 : u = 50 versus Hv1 : u >50.

Question

If a Type I error is made, what conclusion will be drawn regarding the mean breaking strength?

use ti-84 calculator

Answer 1

In hypothesis testing, a Type I error occurs when we reject the null hypothesis \( H_0 \) when it is actually true. In your scenario, the null hypothesis is:

• \( H_0: \mu = 50 \) (the mean breaking strength is equal to 50 kN)

and the alternative hypothesis is:

• \( H_1: \mu > 50 \) (the mean breaking strength is greater than 50 kN).

If a Type I error is made in this test, it means that you concluded that the mean breaking strength of the wires is greater than 50 kN (i.e., you rejected \( H_0 \) in favor of \( H_1 \)), even though the true mean breaking strength is actually equal to 50 kN.

In practical terms, this means that you would falsely conclude that the steel wires are stronger than they actually are, which could lead to decisions like increasing production or marketing the wires as having higher strength than they genuinely possess.

Regardless of using a TI-84 calculator, the explanation remains the same. If you would like to conduct the hypothesis test using your TI-84 calculator, it's useful for calculating p-values or critical values based on sample data, but the nature of the Type I error will be consistent as described.

Answer 2

If a Type II error is made, what conclusion will be drawn regarding the mean breaking strength?

Answer 3

In hypothesis testing, a Type II error occurs when we fail to reject the null hypothesis \( H_0 \) when it is actually false. In your scenario, the hypotheses are as follows:

• Null hypothesis: \( H_0: \mu = 50 \) (the mean breaking strength is equal to 50 kN)
• Alternative hypothesis: \( H_1: \mu > 50 \) (the mean breaking strength is greater than 50 kN)

If a Type II error is made, it means that you did not reject the null hypothesis (you concluded that there is not enough evidence to support the claim that the mean breaking strength is greater than 50 kN) when, in fact, the true mean breaking strength is greater than 50 kN.

In practical terms, this means that you would falsely conclude that the steel wires do not have a mean breaking strength greater than 50 kN, even though they actually do. As a result, you might miss an opportunity to enhance production, improve marketing strategies, or invest in further development based on the stronger properties of the wires.

Thus, in summary, if a Type II error is made, it leads to the incorrect conclusion that the mean breaking strength is 50 kN or less, despite the reality that it is actually greater than 50 kN.

Answer 4

This test uses a one-tailed alternative hypothesis. Explain why a one-tailed hypothesis is more appropriate than a two-tailed hypothesis in this situation.

Answer 5

In the context of testing the mean breaking strength of steel wires, the choice between a one-tailed and a two-tailed hypothesis depends on the specific research question or the alternative hypothesis you are interested in investigating.

One-Tailed Hypothesis

In your case, the test is set up to determine whether the mean breaking strength \( \mu \) is greater than 50 kN. The null hypothesis and alternative hypothesis are defined as follows:

• \( H_0: \mu = 50 \)
• \( H_1: \mu > 50 \)

This setup reflects a one-tailed hypothesis test because you are testing for a difference in one specific direction (greater than 50 kN). You are interested only in whether the breaking strength exceeds 50 kN, not whether it is simply different from 50 kN.

Reasons for Choosing a One-Tailed Hypothesis

1. Directionality of the Research Question: A one-tailed test is appropriate when you have a specific direction in mind for your alternative hypothesis. In your case, the company asserts that the wires' breaking strength is greater than 50 kN. You are not just interested in whether the strength is different from 50 kN; rather, you are specifically concerned with whether it is greater.

2. Increased Power: A one-tailed test can provide more statistical power to detect an effect in the specified direction. Since all of the significance level (e.g., \(\alpha = 0.05\)) is allocated to one tail of the distribution, it reduces the critical value needed for rejecting the null hypothesis. This means that if the mean breaking strength is truly above 50 kN, you have a better chance of detecting that difference with a one-tailed test compared to a two-tailed test.

3. Context of the Industry: In manufacturing and quality assurance, companies often set minimum acceptable standards. If the main goal is to demonstrate that a product exceeds a certain threshold (in this case, breaking strength), a one-tailed test aligns with the operational needs of ensuring that safety and performance standards are met.

Conclusion

In summary, a one-tailed hypothesis test is more appropriate in this situation because you have a specific interest in determining whether the mean breaking strength of the steel wires is greater than 50 kN, not just whether it is different. This focus improves the power of the test and aligns with the company's goals for quality control and safety, making a one-tailed test the most suitable choice for this hypothesis testing scenario.