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Chapter 10: Problem 20

Suppose that you are an inspector for the Fish and Game Department and that you are given the task of determining whether to prohibit fishing along part of the Oregon coast. You will close an area to fishing if it is determined that fish in that region have an unacceptably high mercury content. a. Assuming that a mercury concentration of \(5 \mathrm{ppm}\) is considered the maximum safe concentration, which of the following pairs of hypotheses would you test: \(H_{0}: \mu=5\) versus \(H_{a}: \mu>5\) or \(H_{0}: \mu=5\) versus \(H_{a}: \mu<5\) Give the reasons for your choice. b. Would you prefer a significance level of \(.1\) or \(.01\) for your test? Explain.

Short Answer

Expert verified
a) The hypotheses to be tested should be \(H_{0}: \mu = 5\) versus \(H_{a}: \mu > 5\) since the concern is potentially unsafe (high) mercury levels. b) A significance level of .01 is preferred because it reduces the risk of falsely determining that the region is safe for fishing when mercury levels may actually be unsafe.

Step by step solution

01

Choosing the Appropriate Hypotheses

For this task, the primary concern is too high a mercury concentration, which could lead to fishing restriction in the region. Hence, the hypotheses to be tested should be: \(H_{0}: \mu = 5\) versus \(H_{a}: \mu > 5\). The null hypothesis represents the safe maximum concentration, while the alternative hypothesis signifies an unsafe high mercury level.
02

Determining the Significance Level

The chosen significance level exhibits the tolerance for erroneously rejecting the null hypothesis (\( \mu = 5 ppm \)). If a significance level of .1 is chosen, it would mean that one in ten times, the null hypothesis might incorrectly be rejected when it's true. If a level of .01 is chosen, the chance of erroneous rejection is reduced to one in a hundred. Given the severe health implications linked to high mercury levels, it's preferable to adopt a lower risk of incorrect rejection and therefore choose a more stringent significance level of .01.

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Key Concepts

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

Null and Alternative Hypothesis
Understanding the null and alternative hypotheses is crucial to hypothesis testing in statistics. These two hypotheses are competing assertions about the true nature of a population parameter, such as the average mercury content in fish.

The null hypothesis, denoted as H0, represents a statement of no effect or no difference. It is the default assumption that there is nothing unusual happening. In the mercury content example, the null hypothesis is H0: 渭 = 5, suggesting that the fish have a safe mercury level at the threshold of 5 parts per million (ppm).

The alternative hypothesis, denoted as Ha or H1, posits that there is an effect or a difference. It challenges the status quo of the null hypothesis. In this case, the alternative hypothesis is Ha: 渭 > 5, meaning the inspector is concerned about mercury levels exceeding the safe limit. Selection of the alternative hypothesis is directly related to the consequences of interest; in this scenario, the inspector would want to ensure public health, so the focus is on detecting higher mercury levels.
Significance Level
The significance level of a statistical hypothesis test is a critical decision point that affects the reliability of test results. It is denoted by alpha (伪) and represents the threshold at which we decide whether to reject the null hypothesis on the basis of the probability of observing the test statistic.

Choosing a significance level is like setting a strictness level for our statistical test. For instance, a significance level of 0.01 is more stringent than 0.1. This means that with 伪=0.01, there is only a 1% chance we'd reject the null hypothesis when it is actually true (Type I error), which in our case could mean wrongly concluding that the area has unsafe mercury levels when they are indeed safe.

A lower significance level is preferred when the consequences of a Type I error are severe. In the fish mercury content example, a higher significance level (e.g., 0.1) implies a greater chance of a Type I error, which isn't desirable given the potential health risks. Hence, a more conservative level of 0.01 is advised to minimize this risk.
Mercury Content Hypothesis
). This sets the framework for statistical testing, where evidence from sample data will either strengthen the case against the null hypothesis or fail to do so, leading to a decision on whether action, such as prohibiting fishing, is warranted.

In practice, hypothesis testing uses sample data to make inferences about the population. If the sample evidence strongly suggests that the population mean exceeds the safe mercury threshold, the null hypothesis will be rejected in favor of the alternative. Choosing the correct hypotheses and significance level is paramount to protect public health without unnecessarily affecting the fishing industry.

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Most popular questions from this chapter

Medical personnel are required to report suspected cases of child abuse. Because some diseases have symptoms that mimic those of child abuse, doctors who see a child with these symptoms must decide between two competing hypotheses: \(H_{0}:\) symptoms are due to child abuse \(H_{a}:\) symptoms are due to disease (Although these are not hypotheses about a population characteristic, this exercise illustrates the definitions of Type I and Type II errors.) The article "Blurred Line Between Illness, Abuse Creates Problem for Authorities" (Macon Telegraph, February 28, 2000) included the following quote from a doctor in Atlanta regarding the consequences of making an incorrect decision: "If it's disease, the worst you have is an angry family. If it is abuse, the other kids (in the family) are in deadly danger." a. For the given hypotheses, describe Type I and Type II errors. b. Based on the quote regarding consequences of the two kinds of error, which type of error does the doctor quoted consider more serious? Explain.

A certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of a writing machine) is at least 10 hours. A random sample of 18 pens is selected, the writing lifetime of each is determined, and a normal probability plot of the resulting data supports the use of a one-sample \(t\) test. The relevant hypotheses are \(H_{0}: \mu=10\) versus \(H_{a}: \mu<10 .\) a. If \(t=-2.3\) and \(\alpha=.05\) is selected, what conclusion is appropriate?

Occasionally, warning flares of the type contained in most automobile emergency kits fail to ignite. A consumer advocacy group wants to investigate a claim against a manufacturer of flares brought by a person who claims that the proportion of defective flares is much higher than the value of 1 claimed by the manufacturer. A large number of flares will be tested, and the results will be used to decide between \(H_{0}: p=.1\) and \(H_{a}: p>.1\), where \(p\) represents the proportion of defective flares made by this manufacturer. If \(H_{0}\) is rejected, charges of false advertising will be filed against the manufacturer. a. Explain why the alternative hypothesis was chosen to be \(H_{a}: p>.1\). b. In this context, describe Type I and Type II errors, and discuss the consequences of each.

The mean length of long-distance telephone calls placed with a particular phone company was known to be \(7.3\) minutes under an old rate structure. In an attempt to be more competitive with other long-distance carriers, the phone company lowered long-distance rates, thinking that its customers would be encouraged to make longer calls and thus that there would not be a big loss in revenue. Let \(\mu\) denote the mean length of long-distance calls after the rate reduction. What hypotheses should the phone company test to determine whether the mean length of long-distance calls increased with the lower rates?

Pairs of \(P\) -values and significance levels, \(\alpha\), are given. For each pair, state whether the observed \(P\) -value leads to rejection of \(H_{0}\) at the given significance level. a. \(\quad P\) -value \(=.084, \alpha=.05\) b. \(\quad P\) -value \(=.003, \alpha=.001\) c. \(\quad P\) -value \(=.498, \alpha=.05\) d. \(\quad P\) -value \(=.084, \alpha=.10\) e. \(P\) -value \(=.039, \alpha=.01\) f. \(\quad P\) -value \(=.218, \alpha=.10\)
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