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Many older homes have electrical systems that use fuses rather than circuit breakers. A manufacturer of 40 -amp fuses wants to make sure that the mean amperage at which its fuses burn out is in fact 40 . If the mean amperage is lower than 40 , customers will complain because the fuses require replacement too often. If the mean amperage is higher than 40 , the manufacturer might be liable for damage to an electrical system as a result of fuse malfunction. To verify the mean amperage of the fuses, a sample of fuses is selected and tested. If a hypothesis test is performed using the resulting data, what null and alternative hypotheses would be of interest to the manufacturer?

Short Answer

Expert verified
The null hypothesis \(H_0\) for the manufacturer would be: The mean amperage at which the fuses burn out is 40 (\(H_0: \mu = 40\)). The alternative hypothesis \(H_a\) would be: The mean amperage at which the fuses burn out is not 40 (\(H_a: \mu \neq 40\)).

Step by step solution

01

Formulate the Null Hypothesis

The null hypothesis, denoted by \(H_0\), is a statement of no difference. In this context, it would state that the mean amperage at which the fuses burn out is 40, i.e. \(H_0: \mu = 40\). This represents the status quo or the manufacturer's requirement.
02

Formulate the Alternative Hypothesis

The alternative hypothesis, denoted by \(H_a\) or \(H_1\), is a statement that suggests a change. For this manufacturer, the alternative hypothesis would be that the mean amperage at which the fuses burn out is not 40, i.e. \(H_a: \mu \neq 40\). This would mean that there is a problem with the amperage at which the fuses burn out, causing either customer complaints or potential legal issues.

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Key Concepts

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

Null Hypothesis
When we talk about the null hypothesis, denoted by H0, we're referring to a basic presumption that there is no effect or no difference in the context of a statistical test. Think of it as the 'status quo' or a baseline to compare against.

In the case of our fuse manufacturer, the null hypothesis asserts that the mean amperage at which the fuses blow is exactly 40 amperes (40A). Mathematically, this would be written as H0: μ = 40. Here, μ symbolizes the population mean for the amperage. The null hypothesis serves as a starting point for hypothesis testing. If data gathered from testing fuses suggest that the null hypothesis is unlikely to be true, the hypothesis may be rejected, leading to further investigation.
Alternative Hypothesis
In contrast to the null hypothesis, the alternative hypothesis, labeled as Ha or H1, represents a statement that indicates some statistical effect or difference that we want to test for. It's the hypothesis that researchers typically hope to provide evidence for.

For our example, the alternative hypothesis posits that the mean amperage isn't 40A - it could be lower or higher, thus it's a two-tailed test. Expressed in formula, it's Ha: μ ≠ 40. If the mean is indeed not 40A, then the fuses are not performing as intended. This could result in either customer dissatisfaction or a risk of product liability, both outcomes the manufacturer aims to avoid. Validating the alternative hypothesis would guide the manufacturer to make necessary changes to the production process.
Mean Amperage
The term 'mean amperage' describes the average current at which a batch of fuses is expected to burn out. This average is crucial for manufacturers, as it's a benchmark for product consistency and safety. The importance of this cannot be overstated; a fuse with too low of a mean amperage rating might blow frequently, causing an annoyance to consumers. Conversely, a fuse with a too high mean amperage rating creates a risk of electrical damage or fire.

In our context, the sample of fuses that the manufacturer tests will provide an empirical mean amperage. This observed mean can then be compared to the hypothesized mean (40A) to see if it falls within an acceptable range of variation, or if it significantly differs, implying potential issues with the fuses' design or manufacture.
Statistical Significance
The notion of 'statistical significance' is used to determine if the observed effect in a study is unlikely to have occurred by random chance. In other words, it helps us decide whether to accept or reject the null hypothesis. Statistical significance is generally determined using a p-value, which is calculated from hypothesis testing. A low p-value (<0.05 is a common threshold) indicates that the observed data is unlikely under the assumption that the null hypothesis is true, thus leading to the rejection of the null hypothesis.

In the example of the fuse manufacturer, if the testing of the sample fuses results in a mean amperage significantly different from 40A, and this difference is statistically significant, it compels the manufacturer to review their processes and product specifications. Conversely, if the result is not statistically significant, it suggests that the sampled fuses are burning out at the expected mean amperage, supporting the null hypothesis.

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Most popular questions from this chapter

Water samples are taken from water used for cooling as it is being discharged from a power plant into a river. It has been determined that as long as the mean temperature of the discharged water is at most \(150^{\circ} \mathrm{F}\), there will be no negative effects on the river's ecosystem. To investigate whether the plant is in compliance with regulations that prohibit a mean discharge water temperature above \(150^{\circ} \mathrm{F}\), researchers will take 50 water samples at randomly selected times and record the temperature of each sample. The resulting data will be used to test the hypotheses \(H_{0}: \mu=150^{\circ} \mathrm{F}\) versus \(H_{a}: \mu>150^{\circ} \mathrm{F}\). In the context of this example, describe Type I and Type II errors. Which type of error would you consider more serious? Explain.

A hot tub manufacturer advertises that with its heating equipment, a temperature of \(100^{\circ} \mathrm{F}\) can be achieved in at most \(15 \mathrm{~min}\). A random sample of 25 tubs is selected, and the time necessary to achieve a \(100^{\circ} \mathrm{F}\) temperature is determined for each tub. The sample average time and sample standard deviation are \(17.5\) min and \(2.2\) min, respectively. Does this information cast doubt on the company's claim? Carry out a test of hypotheses using significance level \(.05 .\)

A credit bureau analysis of undergraduate students credit records found that the average number of credit cards in an undergraduate's wallet was \(4.09\) ("Undergraduate Students and Credit Cards in 2004," Nellie Mae, May 2005). It was also reported that in a random sample of 132 undergraduates, the sample mean number of credit cards carried was 2.6. The sample standard deviation was not reported, but for purposes of this exercise, suppose that it was \(1.2\). Is there convincing evidence that the mean number of credit cards that undergraduates report carrying is less than the credit bureau's figure of \(4.09\) ?

In a study of computer use, 1000 randomly selected Canadian Internet users were asked how much time they spend using the Internet in a typical week (Ipsos Reid, August 9,2005 ). The mean of the 1000 resulting observations was \(12.7\) hours. a. The sample standard deviation was not reported, but suppose that it was 5 hours. Carry out a hypothesis test with a significance level of \(.05\) to decide if there is convincing evidence that the mean time spent using the Internet by Canadians is greater than \(12.5\) hours. b. Now suppose that the sample standard deviation was 2 hours. Carry out a hypothesis test with a significance level of . 05 to decide if there is convincing evidence that the mean time spent using the Internet by Canadians is greater than \(12.5\) hours.

When a published article reports the results of many hypothesis tests, the \(P\) -values are not usually given. Instead, the following type of coding scheme is frequently used: \({ }^{*} p=.05,{ }^{* *} p=.01,{ }^{* * *} p=.001,{ }^{* * * *} p=.0001\). Which of the symbols would be used to code for each of the following \(P\) -values? a. \(.037\) c. 072 b. \(.0026\) d. \(.0003\)

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