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Measurement error in a particular situation is normally distributed with mean value \(\mu\) and standard deviation 4. Consider testing \(H_{0}: \mu=0\) versus \(H_{\mathrm{a}}: \mu \neq 0\) based on a sample of \(n=16\) measurements. a. Verify that the usual test with significance level \(.05\) rejects \(H_{0}\) if either \(\bar{x} \geq 1.96\) or \(\bar{x} \leq-1.96 .\) [Note: That this test is unbiased follows from the fact that the way to capture the largest area under the \(z\) curve above an interval having width \(3.92\) is to center that interval at 0 (so it extends from \(-1.96\) to \(1.96\) ).] b. Consider the test that rejects \(H_{0}\) if either \(\bar{x} \geq 2.17\) or \(\bar{x} \leq-1.81\). What is \(\alpha\), that is, \(\pi(0)\) ? c. What is the power of the test proposed in (b) when \(\mu=.1\) and when \(\mu=-.1\) ? (Note that . 1 and \(-.1\) are very close to the null value, so one would not expect large power for such values). Is the test unbiased? d. Calculate the power of the usual test when \(\mu=.1\) and when \(\mu=-.1\). Is the usual test a most powerful test? [Hint: Refer to your calculations in (c).] [Note: It can be shown that the usual test is most powerful among all unbiased tests.]

Short Answer

Expert verified
The usual test with critical values \(-1.96\) and \(1.96\) is unbiased and powerful, whereas the alternative test does not provide a better power performance.

Step by step solution

01

Identify the Standardized Test Statistic

For the given normally distributed errors, we use the sample mean \(\bar{x}\) to test the hypothesis. The test statistic for the sample mean is given by the formula \(z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\), where \(\sigma = 4\) and \(n = 16\).
02

Calculate the Standard Error

Compute the standard error \(SE = \frac{\sigma}{\sqrt{n}} = \frac{4}{4} = 1\). The standard deviation of the sample mean \(\bar{x}\) is 1.
03

Usual Test Rejection Region at 0.05 Significance Level

The usual two-tailed test at a 0.05 significance level uses the critical values of the standard normal distribution, which are \(-1.96\) and \(1.96\). Thus, we reject \(H_0\) if \(\bar{x} \leq -1.96\) or \(\bar{x} \geq 1.96\).
04

Verify the Usual Test

Given the critical values and standard error from previous steps, verify that the computed \(z\)-scores for \(\bar{x} = -1.96\) and \(\bar{x} = 1.96\) correspond to a cumulative probability of \(0.025\) in each tail, summing to \(0.05\) to match the significance level.
05

Calculate Alpha for New Test

For the test that rejects when \(\bar{x} \geq 2.17\) or \(\bar{x} \leq -1.81\), calculate \(\alpha\) as \(P(\bar{x} \geq 2.17) + P(\bar{x} \leq -1.81)\). Use the standard normal distribution to find \(P(z \geq 2.17)\) and \(P(z \leq -1.81)\). Look up these probabilities in a standard normal table.
06

Calculate Power for New Test at \(\mu = 0.1\) and \(\mu = -0.1\)

Power is the probability of rejecting \(H_0\) when \(\mu = 0.1\) or \(\mu = -0.1\). Calculate new \(z\)-values like \(z = \frac{2.17 - 0.1}{1}\) or \(z = \frac{-1.81 + 0.1}{1}\), and use these values to find probabilities from the normal distribution table.
07

Hypothesis Test Bias Analysis

Analyze the symmetry or coverage of \(H_0\) rejection across plausible errors to determine biased behavior. Check probabilities to see if they symmetrically cover conditions around \(\mu = 0.1\) and \(\mu = -0.1\).
08

Calculate Power for Usual Test at \(\mu = 0.1\) and \(\mu = -0.1\)

Follow a process similar to Step 6 but substitute \(1.96\) and \(-1.96\) for \(2.17\) and \(-1.81\) to compute power values. Compare these with those from Step 6 to evaluate performance.
09

Evaluate Test Power Maximum

Consider calculated powers and determine if the usual test (Steps 3 and 8) presents equal or higher power for similar or greater \(|\mu|\) influence over alternatives, confirming maximum unbiased power potential.

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Key Concepts

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

Measurement Error
Measurement errors are a crucial aspect of statistical analysis that can occur when estimating or measuring a specific variable. In this particular exercise, we are dealing with errors that follow a normal distribution, which is common in real-world scenarios. A normal distribution of errors suggests that the errors have a bell-shaped and symmetric spread around a central value, which in this case, is the mean \( \mu \). The standard deviation, given as 4, indicates the typical amount by which the measurement deviates from the mean. A small measurement error implies high precision, whereas a large error indicates greater variability in the recorded measurements.

Addressing measurement errors correctly ensures the validity and reliability of hypothesis testing. When performing tests, researchers aim to minimize these errors to reach accurate conclusions about the data. Understanding the nature and impact of measurement error assists in effectively establishing the context within which hypothesis testing proceeds.
Significance Level
In hypothesis testing, the significance level, commonly denoted as \( \alpha \), is a critical concept. It represents the probability of rejecting the null hypothesis \( H_0 \) when it is actually true, thus making a Type I error. The significance level is often set at 0.05, although other levels like 0.01 or 0.10 can also be used based on the context and importance of avoiding Type I errors.

To illustrate, in this exercise, we use a significance level of 0.05. This implies that there is a 5% risk of concluding that the mean measurement error \( \mu eq 0 \), even if the true mean is zero. To apply this in a test, we determine the rejection region for our test statistic (often a \( z \)-score), based on the chosen \( \alpha \). This region is the set of values that, if the test statistic falls within, we will reject \( H_0 \). For our example, critical values \(-1.96\) and \(1.96\) are derived from the normal distribution given a two-tailed test.

Choosing the level of significance requires balancing between being conservative (a lower \( \alpha \)) and being more permissive (a higher \( \alpha \)), depending on the consequences of making a Type I error.
Power of a Test
The power of a test is an essential concept in hypothesis testing, reflecting the test's ability to correctly reject a false null hypothesis. It is defined as \( 1 - \beta \), where \( \beta \) is the probability of making a Type II error (failing to reject \( H_0 \) when the alternative hypothesis \( H_a \) is true). Higher power implies a greater likelihood of detecting true effects if they exist.

In our exercise, we calculate the power of tests when the alternative means \( \mu \) are slightly different from zero (such as 0.1 or -0.1). To compute power, we determine the probability that the test statistic would fall into the rejection region under these alternative hypotheses. Calculating these probabilities requires evaluating the \( z \)-scores using the shifted mean values and comparing against critical values.

A test is deemed powerful when it effectively detects small true differences. Power analysis also assists in decision-making about the sample size, where larger samples yield more power, holding all else constant. Ensuring that a test has adequate power is key to gaining reliable results from statistical testing.
Normal Distribution
Normal distribution is foundational to many statistical methods, including hypothesis testing. This distribution is characterized by its bell-shaped curve, symmetric around the mean. It is defined by two parameters: the mean \( \mu \) and the standard deviation \( \sigma \).

In hypothesis testing, the assumption of normality allows us to use the standard normal distribution table to find probabilities associated with our test statistics. In our exercise, having measurement errors that are normally distributed enables us to apply \( z \)-tests for hypothesis evaluation.

This symmetry simplifies calculations of probabilities and test statistics like the \( z \)-score, which is computed as \( z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \). Understanding normal distribution is crucial because many statistical tests and confidence intervals assume data follow this pattern. It ensures that established critical values reflect the desired significance level correctly, allowing precise decision-making during tests.

The normal distribution serves as the backbone of inferential statistics, providing a standardized framework for evaluating the likelihood of observing certain outcomes under an assumed population parameter.

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Most popular questions from this chapter

The error \(X\) in a measurement has a normal distribution with mean value 0 and variance \(\sigma^{2}\). Consider testing \(H_{0}: \sigma^{2}=2\) versus \(H_{\mathrm{a}}: \sigma^{2}=3\) based on a random sample \(X_{1}, \ldots, X_{n}\) of errors. a. Show that a most powerful test rejects \(H_{0}\) when \(\sum x_{i}^{2} \geq c\). b. For \(n=10\), find the value of \(c\) for the test in (a) that results in \(\alpha=.05\). c. Is the test of (a) UMP for \(H_{0}: \sigma^{2}=2\) versus \(H_{\mathrm{a}}: \sigma^{2}>2 ?\) Justify your assertion.

One method for straightening wire before coiling it to make a spring is called "roller straightening." The article "The Effect of Roller and Spinner Wire Straightening on Coiling Performance and Wire Properties" (Springs, 1987: 27-28) reports on the tensile properties of wire. Suppose a sample of 16 wires is selected and each is tested to determine tensile strength \(\left(\mathrm{N} / \mathrm{mm}^{2}\right)\). The resulting sample mean and standard deviation are 2160 and 30 , respectively. a. The mean tensile strength for springs made using spinner straightening is \(2150 \mathrm{~N} / \mathrm{mm}^{2}\). What hypotheses should be tested to determine whether the mean tensile strength for the roller method exceeds 2150 ? b. Assuming that the tensile strength distribution is approximately normal, what test statistic would you use to test the hypotheses in part (a)? c. What is the value of the test statistic for this data? d. What is the \(P\)-value for the value of the test statistic computed in part (c)?

The times of first sprinkler activation for a series of tests with fire prevention sprinkler systems using an aqueous film-forming foam were (in sec) \(\begin{array}{lllllllllll}27 & 41 & 22 & 27 & 23 & 35 & 30 & 33 & 24 & 27 & 28 & 22 & 24\end{array}\) (see "Use of AFFF in Sprinkler Systems," Fire Tech., 1976: 5). The system has been designed so that true average activation time is at most \(25 \mathrm{~s}\) under such conditions. Does the data strongly contradict the validity of this design specification? Test the relevant hypotheses at significance level \(.05\) using the \(P\)-value approach.

The error in a measurement is normally distributed with mean \(\mu\) and standard deviation 1 . Consider a random sample of \(n\) errors, and show that the likelihood ratio test for \(H_{0}: \mu=0\) versus \(H_{\mathrm{a}}: \mu \neq 0\) rejects the null hypothesis when either \(\bar{x} \geq c\) or \(\bar{x} \leq-c\). What is \(c\) for a test with \(\alpha=.05\) ? How does the test change if the standard deviation of an error is \(\sigma_{0}\) (known) and the relevant hypotheses are \(H_{0}: \mu=0\) versus \(H_{\mathrm{a}}: \mu \neq \mu_{0}\) ?

To test the ability of auto mechanics to identify simple engine problems, an automobile with a single such problem was taken in turn to 72 different car repair facilities. Only 42 of the 72 mechanics who worked on the car correctly identified the problem. Does this strongly indicate that the true proportion of mechanics who could identify this problem is less than \(.75\) ? Compute the \(P\)-value and reach a conclusion accordingly.

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