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Suppose a new production method will be implemented if a hypothesis test supports the conclusion that the new method reduces the mean operating cost per hour. a. State the appropriate null and alternative hypotheses if the mean cost for the current production method is \(\$ 220\) per hour. b. What is the Type I error in this situation? What are the consequences of making this error? c. What is the Type II error in this situation? What are the consequences of making this error?

Short Answer

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a. - Null: \( H_0: \mu \geq 220 \) - Alternative: \( H_a: \mu < 220 \) b. Type I error: Concluding the new method reduces costs when it does not. c. Type II error: Missing out on cost savings by not adopting the better method.

Step by step solution

01

Stating the Hypotheses

To determine the null and alternative hypotheses, we need to compare the mean operating cost of the new method against the current cost, which is \(\\(220\) per hour. We set:- **Null Hypothesis (\(H_0\))**: The mean cost under the new method is greater than or equal to \(\\)220\) per hour, written as \(\mu \geq 220\).- **Alternative Hypothesis (\(H_a\))**: The mean cost under the new method is less than \(\$220\) per hour, written as \(\mu < 220\).
02

Understanding Type I Error

A Type I error occurs when we reject the null hypothesis when it is actually true. In this scenario, it means concluding that the new method reduces costs when, in fact, it does not. The consequence of making this error is implementing a new method that does not offer any cost savings over the current method, potentially wasting resources.
03

Understanding Type II Error

A Type II error occurs when we fail to reject the null hypothesis when the alternative hypothesis is actually true. Here, it means not adopting the new method when it actually does reduce operating costs. The consequence of making this error is missing out on the potential cost savings and efficiency gains that the new method could have provided.

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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 an initial statement that we attempt to test for possible rejection under statistical scrutiny. It is usually a statement of no effect or no difference. For this exercise, the null hypothesis, denoted as \(H_0\), assumes that the new production method does not reduce the mean operating cost per hour.

Mathematically, we express this as \(\mu \geq 220\). This means the average cost under the new method is either higher than or equal to the current method at $220 per hour.

This forms the basis or the default position we start with before conducting any testing. If the evidence from the data is strong enough to reject the null hypothesis, then an alternative scenario, the alternative hypothesis, is suggested.
Alternative Hypothesis
Contrary to the null hypothesis, the alternative hypothesis represents what we expect to be true if the null hypothesis is false. It is the hypothesis that suggests there is an effect or a difference. In this exercise, it posits that the new production method does lead to a reduction in the mean operating cost per hour.

The alternative hypothesis, denoted as \(H_a\), is expressed as \(\mu < 220\). This indicates that the average cost using the new method is less than \$220 per hour, suggesting the new method is more economical than the current one.

The alternative hypothesis is what the test aims to support if the null hypothesis does not hold. Determining the correct alternative hypothesis is crucial, because it affects the direction and type of statistical test being carried out.
Type I Error
In hypothesis testing, a Type I error occurs when the null hypothesis is rejected even though it is true. It's like a false positive, where we incorrectly determine a difference or effect exists.

For the scenario regarding production methods, a Type I error would occur if we conclude that the new method reduces cost while, in reality, it does not.

The main consequence of making a Type I error in this context could be the unnecessary adoption of a new method that does not offer any cost benefits over the existing process. This could lead to wastage of resources and possibly introduce inefficiencies.
Type II Error
A Type II error happens when we fail to reject the null hypothesis despite the alternative hypothesis actually being true. It's kind of like a false negative, where an effect or difference goes undetected.

In the production method scenario, a Type II error means not recognizing that the new method does indeed reduce costs, therefore continuing with the more expensive current method.

The consequence of making a Type II error is missing out on potential cost savings and improvements that the new method might bring. This could result in lost opportunities for gaining a competitive advantage or increasing operational efficiency.

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Most popular questions from this chapter

Wall Street securities firms paid out record year-end bonuses of \(\$ 125,500\) per employee for 2005 (Fortune, February 6,2006 ). Suppose we would like to take a sample of employees at the Jones \(\&\) Ryan securities firm to see whether the mean year-end bonus is different from the reported mean of \(\$ 125,500\) for the population. a. State the null and alternative hypotheses you would use to test whether the year-end bonuses paid by Jones \& Ryan were different from the population mean. b. Suppose a sample of 40 Jones \(\&\) Ryan employees showed a sample mean year- end bonus of \(\$ 118,000\), Assume a population standard deviation of \(\sigma=\$ 30,000\) and compute the \(p\) -value. c. With \(a=.05\) as the level of significance, what is your conclusion? d. Repeat the preceding hypothesis test using the critical value approach.

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Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu=22 \\ H_{\mathrm{a}}: \mu \neq 22 \end{array} \\] A sample of 75 is used and the population standard deviation is \(10 .\) Compute the \(p\) -value and state your conclusion for each of the following sample results. Use \(\alpha=.01\) a. \(\bar{x}=23\) b. \(\bar{x}=25.1\) c. \(\quad \bar{x}=20\)

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