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Let \(\mu\) denote the true average diameter for bearings of a certain type. A test of \(H_{0}: \mu=0.5\) versus \(H_{a}: \mu \neq 0.5\) will be based on a sample of \(n\) bearings. The diameter distribution is believed to be normal. Determine the value of \(\beta\) in each of the following cases: a. \(n=15, \alpha=.05, \sigma=0.02, \mu=0.52\) b. \(n=15, \alpha=.05, \sigma=0.02, \mu=0.48\) c. \(n=15, \alpha=.01, \sigma=0.02, \mu=0.52\) d. \(n=15, \alpha=.05, \sigma=0.02, \mu=0.54\) e. \(n=15, \alpha=.05, \sigma=0.04, \mu=0.54\) f. \(n=20, \alpha=.05, \sigma=0.04, \mu=0.54\) \(\mathrm{g}\). Is the way in which \(\beta\) changes as \(n, \alpha, \sigma\), and \(\mu\) vary consistent with your intuition? Explain.

Short Answer

Expert verified
The exact numerical value for \(\beta\) would depend on the values obtained after computating the z-score for each case. Trends that can be inferred generally include: as the sample size increases, the value of \(\beta\) decreases (the power of the test increases). Furthermore, \(\beta\) decreases when the true mean \(\mu\) moves further from \(\mu_0\) and when the standard deviation decreases. Also, \(\beta\) decreases when the significance level \(\alpha\) increases.

Step by step solution

01

Computing Z-Score for each case

Calculate the z-score for each case using the formula: \[ Z= \frac{\mu - \mu_0}{\sigma/\sqrt{n}} \] Where; \(\mu\) is the average diameter for each case, \(\mu_0\) is the average diameter under the null hypothesis (0.5 in this case), \(\sigma\) is the standard deviation, and n is the sample size.
02

Calculating \(\beta\) for each case

The type II error \(\beta\) is the probability of accepting the null hypothesis when the alternative hypothesis is true. It can be found using the standard normal distribution table (z-table) after the z-score is calculated. \(\beta\) can be expressed as \(\beta = P(Z \geq Z_{\alpha})\) or \(\beta = P(Z \leq -Z_{\alpha})\) depending on the case where \(Z_{\alpha}\) is the Z value at the given significance level (\(\alpha\)).
03

Evaluating variations of \(\beta\)

Discuss the trends of \(\beta\) as \(n\), \(\alpha\), \(\sigma\), and \(\mu\) vary, by comparing the calculated \(\beta\) values for each case.
04

Explanation of observed trends

Analyze the trends observed in the calculation. Relate the changes of \(\beta\) values with the variations in \(n\), \(\alpha\), \(\sigma\), and \(\mu\) for each case. Is the trend in agreement with intuition?

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Key Concepts

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

Type II Error
Understanding a Type II error is crucial when dealing with hypothesis testing in statistics. This type of error, often denoted by \( \beta \), occurs when a test fails to reject a false null hypothesis; in other words, it's a false negative. Imagine you are a quality control specialist and your job is to detect defective products. A Type II error signifies that a defective product was deemed acceptable and missed by your quality control process.

Its impact is significant because it means that the research or quality check might conclude that there is no effect or difference when in fact there is one. The probability of making a Type II error is influenced by factors such as the sample size, the significance level, and the true effect size. Therefore, interpreting and minimizing \( \beta \) is an integral part of reliable statistical analysis and is directly connected to the test's power, which is 1 - \( \beta \).
Standard Normal Distribution
The standard normal distribution is a pivotal concept in the realm of statistics, and it serves as the foundation for hypothesis testing. It is essentially a normal distribution with a mean of zero and a standard deviation of one. Known also as the Z-distribution, it allows us to convert any normal distribution to a standard form, enabling easy calculation of probabilities and critical values using the standard normal table, colloquially known as the z-table.

When a statistic, such as the sample mean, is transformed into a Z-score using the formula \( Z= \frac{\mu - \mu_0}{\sigma/\sqrt{n}} \), it tells us how many standard deviations the sample mean is from the hypothesized population mean under the null hypothesis. In hypothesis testing, we use the standard normal distribution to determine critical values that, in turn, help us decide if we should reject the null hypothesis.
Significance Level
The significance level, denoted as \( \alpha \), is a threshold that determines the probability of rejecting a true null hypothesis, known as a Type I error. In hypothesis testing, the significance level is a critical value that researchers choose before conducting the test. It shows how much risk we are willing to take to mistakenly reject the null hypothesis. Common \( \alpha \) values include 0.01, 0.05, and 0.10.

A lower \( \alpha \) means a more stringent test where we require stronger evidence before we reject the null hypothesis. This conservative approach reduces the likelihood of a Type I error but can increase the chances of a Type II error. Therefore, the choice of \( \alpha \) involves a trade-off between the risks of making Type I and Type II errors.
Sample Size
Sample size, denoted by \( n \), is a term that stands out in statistical analysis, particularly in hypothesis testing. The sample size is the number of observations or data points that are collected and used to infer the properties of a larger population. In hypothesis testing, the sample size has a profound effect on the ability to detect a true effect.

A larger sample size typically leads to more accurate estimates of the population parameters, as it reduces the standard error of the mean. This increase in accuracy makes it easier to detect whether a significant difference or effect exists. Therefore, increasing the sample size is an effective way to reduce a Type II error, improve the power of the test, and provide more trustworthy results. Practically, there's always a constraint on how large a sample size can be due to time, cost, or logistical considerations, but understanding its influence helps in designing effective experiments and studies.

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Most popular questions from this chapter

The true average diameter of ball bearings of a certain type is supposed to be \(0.5\) in. What conclusion is appropriate when testing \(H_{0}: \mu=0.5\) versus \(H_{a}: \mu \neq 0.5 \mathrm{in}\) each of the following situations: a. \(n=13, t=1.6, \alpha=.05\) b. \(n=13, t=-1.6, \alpha=.05\) c. \(n=25, t=-2.6, \alpha=.01\) d. \(n=25, t=-3.6\)

The amount of shaft wear after a fixed mileage was determined for each of 7 randomly selected internal combustion engines, resulting in a mean of \(0.0372\) in. and a standard deviation of \(0.0125 \mathrm{in}\). a. Assuming that the distribution of shaft wear is normal, test at level \(.05\) the hypotheses \(H_{0}: \mu=.035\) versus \(H_{a}:\) \(\mu>.035\) b. Using \(\sigma=0.0125, \alpha=.05\), and Appendix Table 5 . what is the approximate value of \(\beta\), the probability of a Type II error, when \(\mu=.04\) ? c. What is the approximate power of the test when \(\mu=\) \(.04\) and \(\alpha=.05\) ?

Teenagers (age 15 to 20 ) make up \(7 \%\) of the driving population. The article "More States Demand Teens Pass Rigorous Driving Tests" (San Luis Obispo Tribune, January 27,2000 ) described a study of auto accidents conducted by the Insurance Institute for Highway Safety. The Institute found that \(14 \%\) of the accidents studied involved teenage drivers. Suppose that this percentage was based on examining records from 500 randomly selected accidents. Does the study provide convincing evidence that the proportion of accidents involving teenage drivers differs from \(.07\), the proportion of teens in the driving population?

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: \(^{*} p<.05,^{* *} p<.01,^{*} *^{*} p<.001, *\) Which of the symbols would be used to code for each of the following \(P\) -values? a. 037 c. \(.072\) b. \(.0026\) d. \(.0003\)

To determine whether the pipe welds in a nuclear power plant meet specifications, a random sample of welds is selected and tests are conducted on each weld in the sample. Weld strength is measured as the force required to break the weld. Suppose that the specifications state that the mean strength of welds should exceed 100 \(\mathrm{lb} / \mathrm{in} .^{2} .\) The inspection team decides to test \(H_{0}: \mu=100\) versus \(H_{a}: \mu>100 .\) Explain why this alternative hypothesis was chosen rather than \(\mu<100\).

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