/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 The article "Uncertainty Estimat... [FREE SOLUTION] | 91Ó°ÊÓ

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The article "Uncertainty Estimation in Railway Track LifeCycle 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. A normal probability plot of the data shows a reasonably linear pattern, so it is plausible that the population 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 \(200 \mathrm{~min}\) ? 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 \(\beta(300)\) ?

Short Answer

Expert verified
Test if mean > 200 using t-test at 0.05 level; calculate Type II error for mean=300.

Step by step solution

01

State the Hypotheses

We are testing if the true average repair time exceeds 200 minutes. Therefore, the null and alternative hypotheses are:- Null Hypothesis (\(H_0\)): \(\mu = 200\)- Alternative Hypothesis (\(H_a\)): \(\mu > 200\)This is a one-sided test.
02

Identify Significance Level

The significance level (\(\alpha\)) for the test is given as 0.05. This means we will reject the null hypothesis if the probability of observing a test statistic as extreme as the one we calculate is less than 0.05 if the null hypothesis is true.
03

Calculate the Test Statistic

Use the formula for the test statistic for a normal distribution when the population standard deviation is unknown:\[\text{Test Statistic } (t) = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}\]Where:- \(\bar{x} = 249.7\)- \(\mu_0 = 200\)- \(s = 145.1\)The sample size \(n\) is not directly given, so it cannot be calculated without it. Let's assume this is a missing data point that would come from the context in which this problem is provided (or use some estimation method to find out \(n\)). Nevertheless, proceed to use generalized values (assuming \(n\)) to practice.
04

Compare Test Statistic to Critical Value

Determine the critical value from the t-distribution table corresponding to \(\alpha = 0.05\) for a one-tailed test (with estimated \(n-1\) degrees of freedom). If the calculated test statistic is greater than the critical value, reject \(H_0\).
05

Calculate Type II Error Probability for Part b

To compute the Type II error probability (\(\beta\)), we need to calculate the probability that we do not reject \(H_0\) when \(\mu = 300\). Assume the same standard deviation \(\sigma = 150\) (given specifically for this part). The new test statistic under true mean 300 is\[t' = \frac{300 - 200}{150/\sqrt{n}}\]Again, compare with critical values for deciding acceptance regions under the specific \(H_a\).Calculate \(\beta(300)\) using integration or software if necessary.

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Key Concepts

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

Normal Distribution
Imagine a bell curve—you know, the symmetrical curve that rises in the middle and tapers on both ends. This is essentially what a normal distribution looks like, and it's a powerful way to represent data when it follows certain natural patterns.
Normal distribution, often referred to as Gaussian distribution, is paramount in statistics because it appears frequently in nature, such as in heights, test scores, and yes, even repair times for railway tracks!
  • **Shape:** A famous bell curve, symmetric about the center.
  • **Mean, Median, Mode:** All are equal and located at the center of the distribution.
  • **Standard Deviation:** It determines the width or the spread of the distribution. For larger standard deviations, the curve is wider.
In the context of the exercise, since the given data of repair times achieves a linear pattern in a normal probability plot, it's fair to say the data likely follows a normal distribution. Recognizing this shape allows statisticians to apply particular rules and calculations, like hypothesis testing using a t-test, which we discuss next.
Type II Error
In hypothesis testing, understanding errors is crucial. A Type II error, denoted as \( \beta \), occurs when you fail to reject a false null hypothesis. To put it simply, it’s when we might mistakenly say there's not enough evidence for something that’s actually true.
  • **Type II Error:** Incorrectly retains the null hypothesis when it is false.
  • **Consequences:** Can lead to missed opportunities or wrong assumptions in practical scenarios.
In the exercise step, calculating the Type II error probability was required when the true mean repair time was actually \(300 \text{ min}\), not \(200 \text{ min}\). Knowing this error helps to understand the robustness of the test, especially in deciding whether the test is sensitive enough to detect the true situation or not.Suppose we compute \( \beta(300) \) with given standard deviation \(\sigma = 150\). It shows us what proportion of the time a true repair time of \(300 \text{ min}\) might be inaccurately judged as \(\leq 200 \text{ min}\). By assessing \( \beta \), we can better understand the risk of such an oversight. Increasing sample size or choosing a more appropriate significance level can help reduce these errors.
Significance Level
The significance level, denoted as \( \alpha \), is an essential part of hypothesis testing. It's the threshold we set for deciding when to reject the null hypothesis. In simple terms, it’s your willingness to take a risk of making a Type I error, which is rejecting a true null hypothesis.
  • **Popular Choices:** Typically set at \(0.05\) or \(0.01\).
  • **Meaning:** At \(\alpha = 0.05\), you're accepting a 5% risk of being wrong when you reject the null hypothesis.
  • **Role:** Determines critical values against which test statistics are compared.
In the exercise, a significance level of \(0.05\) was used to test whether the average repair time was greater than \(200\) minutes. This means if the probability of observing our test statistic under the null hypothesis is less than \(0.05\), we'll reject the null hypothesis. This threshold aids in balancing the trade-off between Type I and Type II errors.
Test Statistic
The test statistic is a core element in hypothesis testing. It's an objective measure that tells us how far our sample is from the expected outcome under the null hypothesis. Using this, we can determine whether the sample data supports rejecting the null hypothesis based on how extreme the test statistic is.
  • **Common Form:** For a t-test, the test statistic \( t \) can be calculated as: \[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \]
  • **Roles:** Helps compare sample data to what we’d expect if the null hypothesis were true.
In the problem, our calculated test statistic was based on a sample mean \( \bar{x} = 249.7 \), null hypothesis mean \( \mu_0 = 200 \), and standard deviation \( s = 145.1 \). It’s used to check how convincingly the data argues against \( \mu = 200 \).
Once you have a test statistic, it’s placed against critical values from the t-distribution table based on sample size and significance level. If the statistic exceeds the critical value in a one-tailed test, we have statistically significant evidence to reject the null hypothesis.

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Most popular questions from this chapter

Have you ever been frustrated because you could not get a container of some sort to release the last bit of its contents? The article "Shake, Rattle, and Squeeze: How Much Is Left in That Container?" (Consumer Reports, May 2009: 8) reported on an investigation of this issue for various consumer products. Suppose five \(6.0 \mathrm{oz}\) tubes of toothpaste of a particular brand are randomly selected and squeezed until no more toothpaste will come out. Then each tube is cut open and the amount remaining is weighed, resulting in the following data (consistent with what the cited article reported): \(.53, .65, .46, .50, .37\). Does it appear that the true average amount left is less than \(10 \%\) of the advertised net contents? a. Check the validity of any assumptions necessary for testing the appropriate hypotheses. b. Carry out a test of the appropriate hypotheses using a significance level of \(.05\). Would your conclusion change if a significance level of .01 had been used? c. Describe in context type I and II errors, and say which error might have been made in reaching a conclusion.

The melting point of each of 16 samples of a certain brand of hydrogenated vegetable oil was determined, resulting in \(\bar{x}=94.32\). Assume that the distribution of the melting point is normal with \(\sigma=1.20\). a. Test \(H_{0}: \mu=95\) versus \(H_{\mathrm{a}}: \mu \neq 95\) using a two- tailed level \(.01\) test. b. If a level \(.01\) test is used, what is \(\beta(94)\), the probability of a type II error when \(\mu=94\) ? c. What value of \(n\) is necessary to ensure that \(\beta(94)=.1\) when \(\alpha=.01\) ?

Let \(X_{1}, \ldots, X_{n}\) denote a random sample from a normal population distribution with a known value of \(\sigma\). a. For testing the hypotheses \(H_{0}: \mu=\mu_{0}\) versus \(H_{\mathrm{a}}: \mu>\mu_{0}\) (where \(\mu_{0}\) is a fixed number), show that the test with test statistic \(\bar{X}\) and rejection region \(\bar{x} \geq \mu_{0}+2.33 \sigma / \sqrt{n}\) has significance level \(.01\). b. Suppose the procedure of part (a) is used to test \(H_{0}: \mu \leq \mu_{0}\) versus \(H_{\mathrm{a}}: \mu>\mu_{0}\). If \(\mu_{0}=100, n=25\), and \(\sigma=5\), what is the probability of committing a type I error when \(\mu=99\) ? When \(\mu=98\) ? In general, what can be said about the probability of a type I error when the actual value of \(\mu\) is less than \(\mu_{0}\) ? Verify your assertion.

The article "Caffeine Knowledge, Attitudes, and Consumption in Adult Women" (J. of Nutrition Educ., 1992: 179-184) reports the following summary data on daily caffeine consumption for a sample of adult women: \(n=47\), \(\bar{x}=215 \mathrm{mg}, s=235 \mathrm{mg}\), and range \(=5-1176\). a. Does it appear plausible that the population distribution of daily caffeine consumption is normal? Is it necessary to assume a normal population distribution to test hypotheses about the value of the population mean consumption? Explain your reasoning. b. Suppose it had previously been believed that mean consumption was at most \(200 \mathrm{mg}\). Does the given data contradict this prior belief? Test the appropriate hypotheses at significance level . 10 and include a \(P\)-value in your analysis.

Water samples are taken from water used for cooling as it is being discharged from a power plant into a river. It has been determined that as long as the mean temperature of the discharged water is at most \(150^{\circ} \mathrm{F}\), there will be no negative effects on the river's ecosystem. To investigate whether the plant is in compliance with regulations that prohibit a mean discharge water temperature above \(150^{\circ}, 50\) water samples will be taken at randomly selected times and the temperature of each sample recorded. The resulting data will be used to test the hypotheses \(H_{0}: \mu=150^{\circ}\) versus \(H_{\mathrm{a}}: \mu>150^{\circ}\). In the context of this situation, describe type I and type II errors. Which type of error would you consider more serious? Explain.

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