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Chapter 8: Problem 27

Show that for any \(\Delta>0\), when the population distribution is normal and \(\sigma\) is known, the two-tailed test satisfies \(\beta\left(\mu_{0}-\Delta\right)=\beta\left(\mu_{0}+\Delta\right)\), so that \(\beta\left(\mu^{\prime}\right)\) is symmetric about \(\mu_{0}\).

Short Answer

Expert verified
The power function \(\beta(\mu')\) is symmetric about \(\mu_0\).

Step by step solution

01

Understand the Two-Tailed Test

A two-tailed test is used when we are interested in deviations on either side of the null hypothesis mean, \(\mu_0\). This involves rejecting the null hypothesis when the test statistic falls into the critical regions on both tails.
02

Define the Rejection Regions

For a given significance level \(\alpha\), the critical values are found in the tails of the normal distribution. These are typically \(z_{\alpha/2}\) and \(-z_{\alpha/2}\), as the normal distribution is symmetric around the mean. The rejection regions are \(Z > z_{\alpha/2}\) or \(Z < -z_{\alpha/2}\), where \(Z\) is the standard normal variable.
03

Calculate the Power Function

The power function, \(\beta(\mu')\), of a test is defined as the probability of rejecting the null hypothesis when the true mean is \(\mu'\). Since \(Z = \frac{\bar{X} - \mu_0}{\sigma/\sqrt{n}}\), the power function becomes \(\beta(\mu') = P(Z > z_{\alpha/2}) + P(Z < -z_{\alpha/2})\) when \(\mu' eq \mu_0\).
04

Evaluate Symmetry of the Power Function

For any \(\Delta > 0\), evaluate \(\beta(\mu_0 - \Delta)\) and \(\beta(\mu_0 + \Delta)\). Compute \(Z = \frac{\bar{X} - \mu_0 + \Delta}{\sigma/\sqrt{n}}\) for \(\mu' = \mu_0 - \Delta\) and similarly for \(\mu_0 + \Delta\). These calculations reflect symmetry: \(P(Z > z_{\alpha/2})\) and \(P(Z < -z_{\alpha/2})\) are equal for both \(\mu_0 - \Delta\) and \(\mu_0 + \Delta\), therefore \(\beta(\mu_0 - \Delta) = \beta(\mu_0 + \Delta)\).

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Key Concepts

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

Power Function
The power function is a critical aspect of hypothesis testing, especially in the context of a two-tailed test with a normally distributed population where the standard deviation (\( \sigma \) is known. It represents the probability of correctly rejecting the null hypothesis when the alternative hypothesis is true. In technical terms, it鈥檚 labeled as \( \beta(\mu') \) and calculated as the probability of obtaining a test statistic in the critical region, either greater than or less than the critical values.

Key aspects of the power function:
  • It measures the effectiveness of a test; a higher power means a higher chance of detecting an effect if there is one.
  • The computation involves determining the probability that the test statistic, \( Z \), falls in the rejection region鈥擻( Z > z_{\alpha/2} \) or \( Z < -z_{\alpha/2} \).
  • For any given \( \Delta > 0 \), the power function maintains symmetry concerning deviations around the true mean \( \mu_0 \), which means \( \beta(\mu_0 - \Delta) = \beta(\mu_0 + \Delta) \).
In essence, the power function allows us to evaluate the performance of a test across various potential true means, turning it into a very informative tool in statistical testing scenarios.
Symmetric Distribution
The symmetric distribution plays a pivotal role in the calculation of critical values and understanding the behavior of the power function. A distribution is symmetric if it looks the same on both sides from a central point, typically the mean.

Key attributes of symmetric distributions:
  • In the context of the normal distribution, symmetry is consistent about the mean, allowing easy determination of critical values and rejection regions.
  • This symmetry is vital as it assures that probabilities on either side of the mean are equal, enhancing the fairness of hypothesis testing involving two-tailed tests.
  • The symmetry ensures that when computing \( \beta(\mu') \), if the power function at \( \mu' = \mu_0 + \Delta \) matches \( \mu_0 - \Delta \), the test's power balances effectively across the tails.
Thus, symmetric distribution attributes ensure that statistical tests correctly and effectively identify real effects by keeping calculations manageable and ensuring equal weight in the tails.
Critical Values
Critical values are thresholds that determine the boundary inside which the null hypothesis is not rejected, and outside which it is. These values are vital for conducting hypothesis tests, particularly in two-tailed tests.

Defining characteristics of critical values:
  • They are derived based on a specified significance level \( \alpha \) which dictates the probability of incorrectly rejecting the null hypothesis when it is true鈥攃ommonly set at levels like 0.05 or 0.01.
  • In a symmetric normal distribution, critical values are denoted as \( z_{\alpha/2} \) and \( -z_{\alpha/2} \), located in the two tails of the distribution.
  • These values mark the edges of the rejection regions: one where the test statistic might be exceedingly high and another where it is exceedingly low relative to the mean.
Understanding and identifying the accurate critical values is crucial as they dictate the outcome of hypothesis tests, ensuring tests are conducted with precision and reliability.

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Most popular questions from this chapter

Reconsider the paint-drying problem discussed in Example 8.5. The hypotheses were \(H_{0}: \mu=75\) versus \(H_{\mathrm{a}}: \mu<75\), with \(\sigma\) assumed to have value 9.0. Consider the alternative value \(\mu=74\), which in the context of the problem would presumably not be a practically significant departure from \(H_{0}\) - a. For a level \(.01\) test, compute \(\beta\) at this alternative for sample sizes \(n=100,900\), and 2500 . b. If the observed value of \(\bar{X}\) is \(\bar{x}=74\), what can you say about the resulting \(P\)-value when \(n=2500\) ? Is the data statistically significant at any of the standard values of \(\alpha\) ? c. Would you really want to use a sample size of 2500 along with a level .01 test (disregarding the cost of such an experiment)? Explain.

Let \(\mu\) denote the true average reaction time to a certain stimulus. For a \(z\) test of \(H_{0}: \mu=5\) versus \(H_{a}: \mu>5\), determine the \(P\)-value for each of the following values of the \(z\) test statistic. \(\begin{array}{lllllll}\text { a. } 1.42 & \text { b. } .90 & \text { c. } 1.96 & \text { d. } 2.48 & \text { e. }-.11\end{array}\)

The article "Uncertainty Estimation in Railway Track Life-Cycle Cost" (J. of Rail and Rapid Transit, 2009) presented the following data on time to repair (min) a rail break in the high rail on a curved track of a certain railway line. distribution of repair time is at least approximately normal. The sample mean and standard deviation are \(249.7\) and \(145.1\), respectively. a. Is there compelling evidence for concluding that true average repair time exceeds 200 min? Carry out a test of hypotheses using a significance level of .05. b. Using \(\sigma=150\), what is the type II error probability of the test used in (a) when true average repair time is actually \(300 \mathrm{~min}\) ? That is, what is \(\beta(300)\) ?

The calibration of a scale is to be checked by weighing a \(10-\mathrm{kg}\) 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. With the sample mean itself as the test statistic, what is the \(P\)-value when \(\bar{x}=9.85\), and what would you conclude at significance level .01? c. For a test with \(\alpha=.01\), what is the probability that recalibration is judged unnecessary when in fact \(\mu=\) 10.1? When \(\mu=9.8\) ?

A regular type of laminate is currently being used by a manufacturer of circuit boards. A special laminate has been developed to reduce warpage. The regular laminate will be used on one sample of specimens and the special laminate on another sample, and the amount of warpage will then be determined for each specimen. The manufacturer will then switch to the special laminate only if it can be demonstrated that the true average amount of warpage for that laminate is less than for the regular laminate. State the relevant hypotheses, and describe the type I and type II errors in the context of this situation.
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