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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 \(\alpha=.05 ?\)

Short Answer

Expert verified
In conclusion, the test results do not provide enough evidence to reject the null hypothesis at the 0.05 significance level. The approximation of the Type II error probability when the true mean wear is 0.04 is about 0.40. The test power under this condition is 0.60.

Step by step solution

01

Test the Hypotheses

Firstly, with the given sample mean, standard deviation, and sample size, compute the test statistic (Z-score). The formula used is \(Z = \frac{\bar{x} - \mu_{0}}{S/ \sqrt{n}}\), where \(\bar{x}\) is sample mean, \(\mu_{0}\) is hypothesized mean, S is the standard deviation of the sample, and n is the sample size. So Z-score computation evaluates to \( Z = \frac{0.0372 - 0.035}{0.0125/ \sqrt{7}} = 0.549\). Refer table for a single-tail test with \(\alpha=0.05\), the critical Z-score is 1.645. Here, calculated Z-score is less than the critical value, we cannot reject the null hypothesis.
02

Calculate the Probability of a Type II error

Type II error (beta) occurs when we fail to reject a null hypothesis that is false. Let's calculate it for \(\mu=.04\). We need to find Z for this new \(\mu\) using formula \(Z = \frac{\mu - \mu_{0}}{S / \sqrt{n}}\). This gives \(Z = \frac{0.04 - 0.035}{0.0125 / \sqrt{7}} = 1.393\). Next, we need to compute the difference between this Z value and the critical Z value i.e., \(Z_{\beta} = Z_{\alpha} - Z\) = \(1.645 - 1.393\) = 0.252. Lookup the value corresponding to this Z in the Z-table gives \(\beta\) approx.= 0.40.
03

Calculate the Test Power

The power of the test is the probability of correctly rejecting a false null hypothesis, which can be found by subtracting the probability of a Type II error from 1. Hence, Power = 1 - \(\beta\) = 1 - 0.40 = 0.60. This means that, with \(\mu = 0.04\) and \(\alpha = 0.05\), there is a 60% probability of correctly rejecting null hypothesis whenever it's false.

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Key Concepts

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

Type II Error
Understanding Type II error is crucial in hypothesis testing, as it pertains to a scenario where we do not reject a null hypothesis (\(H_0\)) that is actually false. This error occurs because the sample does not provide enough evidence against the null hypothesis. In the context of shaft wear in internal combustion engines, if the actual mean wear is more than the hypothesized value of 0.035 inches, but our test statistic leads us not to reject the null hypothesis, a Type II error has been committed.

A practical aspect of Type II errors is their association with the chosen significance level (\ralpha). In the exercise, \ralpha is set at 0.05, indicating a 5% risk of mistakenly rejecting e true null hypothesis (Type I error), which inversely impacts the risk of committing a Type II error; a larger \ralpha usually decreases the risk of a Type II error, but increases the risk of a Type I error.
Normal Distribution
The normal distribution, often referred to as the bell curve, is a foundational concept in statistics that describes how the values of a variable are distributed. It's symmetrical, with most of the observations clustering around the central peak and the probabilities for values farther away from the mean tapering off equally in both directions.

In hypothesis testing, the assumption of normality allows us to use the Z-distribution to calculate probabilities and critical values. In our engine shaft wear exercise, it is assumed that the shaft wear follows a normal distribution, which justifies the use of the Z-score formula to compute test statistics and ultimately make decisions about the null hypothesis. In applications, the normality assumption can often be validated using graphical methods like Q-Q plots or statistical tests such as the Shapiro-Wilk test.
Test Statistic Calculation
The test statistic is a standardized value that is calculated from sample data during a hypothesis test. It's used to decide whether to support or reject the null hypothesis. In the case of the shaft wear problem, the test statistic is the Z-score.\rThe Z-score represents how many standard deviations an element is from the mean. The formula for its calculation is \rZ = (\(\bar{x}\) - \(\mu_0\))/(S/\(\sqrt{n}\)), where \(\bar{x}\) is the sample mean, \(\mu_0\) is the hypothesized population mean, S is the sample standard deviation, and n is the sample size.

By computing the Z-score, we can compare it against critical values from the standard normal distribution to determine the likelihood of observing such a mean if the null hypothesis were true. This process is integral to the hypothesis testing procedure.
Statistical Power Analysis
Statistical power analysis is used to determine the probability that a test will correctly reject a false null hypothesis. This is the test's power, and it is important because it reflects the sensitivity of the test to detect effects of a certain size. Power is calculated as 1 minus the probability of a Type II error (beta).

In the given exercise, the power of the test when \(\mu = 0.04\) and \(\alpha = 0.05\) is approximately 60%. This means that if the true mean shaft wear is 0.04 inches, there's a 60% chance the test will correctly identify that the mean wear is greater than 0.035 inches. Higher power is better, as it means there's a greater chance of detecting an effect when there is one. Power can be increased by raising the significance level, increasing the sample size, or improving the precision of measurements.

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Most popular questions from this chapter

A certain university has decided to introduce the use of plus and minus with letter grades, as long as there is evidence that more than \(60 \%\) of the faculty favor the change. A random sample of faculty will be selected, and the resulting data will be used to test the relevant hypothe ses. If \(\pi\) represents the true proportion of all faculty that favor a change to plus- minus grading, which of the following pair of hypotheses should the administration test: $$ H_{0}: \pi=.6 \text { versus } H_{a}: \pi<.6 $$ or $$ H_{0}: \pi=.6 \text { versus } H_{a}: \pi>.6 $$ Explain your choice.

Let \(\mu\) denote the true average lifetime for a certain type of pen under controlled laboratory conditions. A test of \(H_{0}: \mu=10\) versus \(H_{a}: \mu<10\) will be based on a sample of size 36. Suppose that \(\sigma\) is known to be \(0.6\), from which \(\sigma_{x}=0.1\). The appropriate test statistic is then $$ z=\frac{\bar{x}-10}{0.1} $$ a. What is \(\alpha\) for the test procedure that rejects \(H_{0}\) if \(z \leq\) \(-1.28 ?\) b. If the test procedure of Part (a) is used, calculate \(\beta\) when \(\mu=9.8\), and interpret this error probability. c. Without doing any calculation, explain how \(\beta\) when \(\mu=9.5\) compares to \(\beta\) when \(\mu=9.8\). Then check your assertion by computing \(\beta\) when \(\mu=9.5\). d. What is the power of the test when \(\mu=9.8 ?\) when \(\mu=9.5 ?\)

An article titled "Teen Boys Forget Whatever It Was" appeared in the Australian newspaper The Mercury (April 21, 1997). It described a study of academic performance and attention span and reported that the mean time to distraction for teenage boys working on an independent task was 4 min. Although the sample size was not given in the article, suppose that this mean was based on a random sample of 50 teenage Australian boys and that the sample standard deviation was \(1.4\) min. Is there convincing evidence that the average attention span for teenage boys is less than 5 min? Test the relevant hypotheses using \(\alpha=.01\).

The success of the U.S. census depends on people filling out and returning census forms. Despite extensive advertising, many Americans are skeptical about claims that the Census Bureau will guard the information it collects from other government agencies. In a USA Today poll (March 13,2000 ), only 432 of 1004 adults surveyed said that they believe the Census Bureau when it says the information you give about yourself is kept confidential. Is there convincing evidence that, despite the advertising campaign, fewer than half of U.S. adults believe the Census Bureau will keep information confidential? Use a significance level of \(.01\).

Students at the Akademia Podlaka conducted an experiment to determine whether the Belgium-minted Euro coin was equally likely to land heads up or tails up. Coins were spun on a smooth surface, and in 250 spins, 140 landed with the heads side up (New Scientist, January 4 , 2002). Should the students interpret this result as convincing evidence that the proportion of the time the coin would land heads up is not .5? Test the relevant hypotheses using \(\alpha=.01\). Would your conclusion be different if a significance level of \(.05\) had been used? Explain.

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