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Chapter 9: Problem 4

Let \(\mu\) denote the true average radioactivity level (picocuries per liter). The value \(5 \mathrm{pCi} / \mathrm{L}\) is considered the dividing line between safe and unsafe water. Would you recommend testing \(H_{0}: \mu=5\) versus \(H_{\mathrm{a}}: \mu>5\) or \(H_{0}: \mu=5\) versus \(H_{\mathrm{a}}: \mu<5\) ? Explain your reasoning. [Hint: Think about the consequences of a type I and type II error for each possibility.]

Short Answer

Expert verified
Test \(H_0: \mu = 5\) versus \(H_a: \mu > 5\) to prioritize public safety.

Step by step solution

01

Understanding the Hypotheses

For this problem, we have two different sets of hypotheses to consider. The null hypothesis \(H_0\) for both scenarios states that the true average radioactivity level, \(\mu\), is equal to 5 \(\text{pCi/L}\). The alternative hypotheses are different: \(H_\text{a}: \mu > 5\) suggests the water is unsafe due to higher radioactivity, while \(H_\text{a}: \mu < 5\) suggests the water is safe due to lower radioactivity.
02

Considering Type I and Type II Errors for \(H_0: \mu = 5\) versus \(H_\text{a}: \mu > 5\)

A Type I error in this scenario would mean rejecting \(H_0\) (concluding \(\mu > 5\)) when, in fact, \(\mu = 5\). The consequence is declaring the water unsafe when it is exactly at the safety threshold. A Type II error would mean failing to reject \(H_0\) when \(\mu > 5\), resulting in not identifying water as unsafe when it is indeed unsafe.
03

Considering Type I and Type II Errors for \(H_0: \mu = 5\) versus \(H_\text{a}: \mu < 5\)

Here, a Type I error would mean rejecting \(H_0\) (concluding \(\mu < 5\)) when \(\mu = 5\), thus declaring the water safe when it is at the threshold. A Type II error would mean failing to reject \(H_0\) when \(\mu < 5\), meaning we don't confirm the water is safe even though it is below the threshold.
04

Evaluating the Consequences

Given the importance of public safety, the consequences of failing to identify unsafe water (Type II error for \(H_a: \mu > 5\)) are more severe than unnecessarily expensive remediation measures (Type I error for \(H_a: \mu < 5\)). Thus, it is more critical to accurately detect when water becomes unsafe.
05

Making the Recommendation

Based on the analysis, it is recommended to test \(H_0: \mu = 5\) versus \(H_a: \mu > 5\) to prioritize identifying any conditions of water being unsafe and ensure public safety.

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Key Concepts

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

type I error
In hypothesis testing, a Type I error is made when we incorrectly reject a true null hypothesis. This means we conclude that there is an effect or difference when in reality, there isn't one. It's like getting a false alarm - we think something significant is present when it's not. In the context of the exercise, if we test the hypothesis that the average radioactivity level is exactly at the safety threshold and accidentally reject this hypothesis, we might wrongly assume the water is unsafe.
  • This error can lead to unnecessary measures or panic.
  • The level of significance, denoted by \( \alpha \), is the probability of making a Type I error.
  • Commonly, \( \alpha \) is set at 0.05, meaning there is a 5% chance of making this error.
type II error
A Type II error occurs when we fail to reject a false null hypothesis. Essentially, we miss detecting an effect or difference that's actually there. This is the opposite of a Type I error, like not noticing a fire alarm when there's smoke. In the case of our exercise, failing to reject the null hypothesis when the water is indeed unsafe (radioactivity levels above the safety threshold) would be a Type II error.
  • This mistake might delay necessary action, exposing people to unsafe water
  • The probability of making a Type II error is denoted by \( \beta \).
  • Reducing \( \beta \) requires increasing the power of the test, which measures the test's ability to detect an effect.
null hypothesis
The null hypothesis is the foundation of statistical testing. It asserts that there is no effect or difference. It's our starting assumption which we aim to test against. In simpler words, it's like assuming the status quo until proven otherwise. In our problem regarding water safety, the null hypothesis could be that the radioactivity level is at the critical threshold \( \mu = 5 \).
  • We test this hypothesis to check if any action or change is necessary.
  • It's crucial because it provides a baseline for the statistical test.
  • Rejecting the null hypothesis requires sufficient evidence because it implies accepting an alternative scenario.
alternative hypothesis
The alternative hypothesis provides a statement that reflects our research question or suspicion. It's what we hope to demonstrate through our hypothesis testing. In statistical terms, it's the rival against the null hypothesis. For the exercise in question, the alternative could be that the radioactivity is above the critical level, \( \mu > 5 \), indicating unsafe water conditions.
  • The alternative hypothesis is considered likely if the null hypothesis is rejected.
  • It represents the potential change we are investigating.
  • Choosing the right alternative hypothesis helps frame the research correctly.

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Most popular questions from this chapter

The calibration of a scale is to be checked by weighing a 10-kg test specimen 25 times. Suppose that the results of different weighings are independent of one another and that the weight on each trial is normally distributed with \(\sigma=.200 \mathrm{~kg}\). Let \(\mu\) denote the true average weight reading on the scale. a. What hypotheses should be tested? b. Suppose the scale is to be recalibrated if either \(\bar{x} \geq 10.1032\) or \(\bar{x} \leq 9.8968\). What is the probability that recalibration is carried out when it is actually unnecessary? c. What is the probability that recalibration is judged unnecessary when in fact \(\mu=10.1\) ? When \(\mu=9.8\) ? d. Let \(z=(\bar{x}-10) /(\sigma / \sqrt{n})\). For what value \(c\) is the rejection region of part (b) equivalent to the "two-tailed" region either \(z \geq\) cor \(z \leq-c\) ? e. If the sample size were only 10 rather than 25 , how should the procedure of part (d) be altered so that \(\alpha=.05\) ? f. Using the test of part (e), what would you conclude from the following sample data? \(\begin{array}{lllll}9.981 & 10.006 & 9.857 & 10.107 & 9.888 \\ 9.728 & 10.439 & 10.214 & 10.190 & 9.793\end{array}\) g. Re-express the test procedure of part (b) in terms of the standardized test statistic \(Z=(\bar{X}-10) /(\sigma / \sqrt{n)} .\)

The article "Statistical Evidence of Discrimination" (J. Amer. Statist. Assoc., 1982: 773-783) discusses the court case Swain v. Alabama (1965), in which it was alleged that there was discrimination against blacks in grand jury selection. Census data suggested that \(25 \%\) of those eligible for grand jury service were black, yet a random sample of 1050 people called to appear for possible duty yielded only 177 blacks. Using a level \(.01\) test, does this data argue strongly for a conclusion of discrimination?

Give as much information as you can about the \(P\)-value of a \(t\) test in each of the following situations: a. Upper-tailed test, \(\mathrm{df}=8, t=2.0\) b. Lower-tailed test, \(\mathrm{df}=11, t=-2.4\) c. Two-tailed test, \(\mathrm{df}=15, t=-1.6\) d. Upper-tailed test, df \(=19, t=-.4\) e. Upper-tailed test, df \(=5, t=5.0\) f. Two-tailed test, df \(=40, t=-4.8\)

The accompanying observations on residual flame time (sec) for strips of treated children's nightwear were given in the article "An Introduction to Some Precision and Accuracy of Measurement Problems" (J. Test. Eval., 1982: 132-140). Suppose a true average flame time of at most \(9.75\) had been mandated. Does the data suggest that this condition has not been met? Carry out an appropriate test after first investigating the plausibility of assumptions that underlie your method of inference. \(\begin{array}{lllllll}9.85 & 9.93 & 9.75 & 9.77 & 9.67 & 9.87 & 9.67 \\ 9.94 & 9.85 & 9.75 & 9.83 & 9.92 & 9.74 & 9.99 \\ 9.88 & 9.95 & 9.95 & 9.93 & 9.92 & 9.89 & \end{array}\)

Chapter 8 presented a CI for the variance \(\sigma^{2}\) of a normal population distribution. The key result there was that the rv \(\chi^{2}=(n-1) S^{2} / \sigma^{2}\) has a chi-squared distribution with \(n-1\) df. Consider the null hypothesis \(H_{0}: \sigma^{2}=\sigma_{0}^{2}\) (equivalently, \(\sigma=\sigma_{0}\) ). Then when \(H_{0}\) is true, the test statistic \(\chi^{2}=(n-1) S^{2} / \sigma_{0}^{2}\) has a chi-squared distribution with \(n-1\) df. If the relevant alternative is \(H_{\mathrm{a}}: \sigma^{2}>\sigma_{0}^{2}\), rejecting \(H_{0}\) if \((n-1) S^{2} / \sigma_{0}^{2} \geq\) \(\chi_{\alpha, n-1}^{2}\) gives a test with significance level \(\alpha\). To ensure reasonably uniform characteristics for a particular application, it is desired that the true standard deviation of the softening point of a certain type of petroleum pitch be at most \(.50^{\circ} \mathrm{C}\). The softening points of ten different specimens were determined, yielding a sample standard deviation of \(.58^{\circ} \mathrm{C}\). Does this strongly contradict the uniformity specification? Test the appropriate hypotheses using \(\alpha=.01\).
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