/*! 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 47 Supercavitation is a propulsion ... [FREE SOLUTION] | 91Ó°ÊÓ

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Supercavitation is a propulsion technology for undersea vehicles that can greatly increase their speed. It occurs above approximately 50 meters per second when pressure drops sufficiently to allow the water to dissociate into water vapor, forming a gas bubble behind the vehicle. When the gas bubble completely encloses the vehicle, supercavitation is said to occur. Eight tests were conducted on a scale model of an undersea vehicle in a towing basin with the average observed speed \(\bar{x}=102.2\) meters per second. Assume that speed is normally distributed with known standard deviation \(\sigma\) \(=4\) meters per second. (a) Test the hypothesis \(H_{0}: \mu=100\) versus \(H_{1}: \mu<100\) using \(\alpha=0.05\) (b) What is the \(P\) -value for the test in part (a)? (c) Compute the power of the test if the true mean speed is as low as 95 meters per second. (d) What sample size would be required to detect a true mean speed as low as 95 meters per second if you wanted the power of the test to be at least \(0.85 ?\) (e) Explain how the question in part (a) could be answered by constructing a one-sided confidence bound on the mean speed.

Short Answer

Expert verified
(a) Fail to reject \(H_0\); (b) \(P\) = 0.059; (c) Power ≈ 1; (d) Sample size = 35; (e) One-sided bound supports \(H_0\).

Step by step solution

01

State the Hypotheses

For part (a), we need to test the hypothesis. The null hypothesis is \( H_0: \mu = 100 \) and the alternative hypothesis is \( H_1: \mu < 100 \).
02

Determine the Test Statistic

The test statistic for the hypothesis test is calculated as follows:\[Z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}}\]where \( \bar{x} = 102.2 \), \( \mu_0 = 100 \), \( \sigma = 4 \), and \( n = 8 \). Substituting these values, we have:\[Z = \frac{102.2 - 100}{4/\sqrt{8}} = \frac{2.2}{1.414} \approx 1.556\]
03

Find the Critical Value

For \( \alpha = 0.05 \) in a one-tailed test, the critical Z-value from the standard normal distribution is approximately \(-1.645\).
04

Compare Test Statistic and Critical Value

Since the calculated Z-value \(1.556\) is greater than the critical value \(-1.645\), we fail to reject the null hypothesis \(H_0\).
05

Calculate the P-value

The P-value for the test is the probability that a Z-value is at least 1.556 standard deviations below the mean. This can be found using the standard normal distribution table or calculator. \( \text{P-value} \approx 0.059 \).
06

Compute the Power of the Test

To compute the power of the test for \( \mu = 95 \), we first calculate the Z-prime value where the alternate mean is 95:\[Z' = \frac{95 - 100}{4/\sqrt{8}} = \frac{-5}{1.414} \approx -3.536\]The critical Z-value threshold remains \(-1.645\). The power of the test is \( P(Z < -3.536) + P(Z > 1.645) = 1.000 \).
07

Determine Required Sample Size for Power

To find the required sample size for a power of 0.85:Calculate the noncentrality parameter:\[n = \left( \frac{(Z_{\alpha} + Z_{\beta}) \cdot \sigma}{\mu - \mu_0} \right)^2 = \left( \frac{(1.645 + 1.036) \cdot 4}{95 - 100} \right)^2 \approx 34.56\]Thus, we need a sample size of at least 35 for the desired power.
08

Construct a One-Sided Confidence Bound

To construct a one-sided confidence interval for the mean, use:\[\bar{x} - Z_{\alpha} \cdot \frac{\sigma}{\sqrt{n}}\]\[102.2 - 1.645 \cdot 1.414 \approx 99.87\]This bound implies the mean is likely not less than 100, failing to reject \(H_0\).

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Key Concepts

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

Supercavitation
Supercavitation is a fascinating technology that allows undersea vehicles to travel at extremely high speeds. This process involves the formation of a gas bubble that completely encloses the vehicle, minimizing the friction against water. It occurs when the speed of the vehicle exceeds 50 meters per second, causing a significant drop in pressure and allowing the water to turn into vapor.
For students, understanding supercavitation is crucial because it highlights a practical application of physics and fluid dynamics. It demonstrates how pressure and speed interact and can result in innovative technologies capable of enhancing performance and efficiency.
  • Supercavitation enables undersea vehicles to move faster compared to traditional methods.
  • The technology is highly reliant on achieving specific velocity to reach the necessary pressure drop.
  • This phenomenon reduces contact with water, creating a more efficient travel method undersea.
Normal Distribution
The normal distribution is a fundamental concept in statistics, often referred to as a bell curve due to its shape. It describes how data points are likely to be distributed around the mean value. In many natural phenomena, including the speed of objects like vehicles in the supercavitation example, the normal distribution is an excellent approximation.
Understanding the normal distribution makes it easier to predict probabilities and conduct various types of analysis. In hypothesis testing, it assists in determining whether deviations from the average are significant.
  • It is characterized by its mean (\(mu\)) and standard deviation (\(sigma\)).
  • The curve is symmetrical around the mean, displaying how data is spread.
  • Most of the data points lie close to the mean, within one standard deviation.
Confidence Intervals
Confidence intervals provide a range of values that is likely to contain a population parameter, such as a mean, with a certain degree of confidence. They add depth to our understanding of data by showing where the true value might lie based on sample data.
In the context of hypothesis testing, constructing a one-sided confidence interval can help validate or refute a hypothesis. For example, if the average speed of supercavitation is known, calculating a confidence interval gives insight into potential fluctuations.
  • Confidence intervals are used to estimate parameters with a specified confidence level, typically 95% or 99%.
  • They aid in making informed decisions by quantifying the uncertainty of an estimate.
  • A one-sided interval provides an upper or lower bound, depending on the nature of the test.
Sample Size Determination
Determining the sample size is an essential part of designing a study, especially when testing hypotheses and ensuring adequate power for the test. It involves calculating how many samples are needed to detect an effect of a given size with a certain probability.
Sample size affects the reliability of results. Larger samples provide more accurate estimates of the population parameters, while smaller samples can lead to greater variability and less reliable conclusions. In our exercise, calculating the right sample size ensures that the power of the test is sufficient to detect true differences in speed due to supercavitation.
  • Sample size determination helps in designing a study that is efficient and cost-effective.
  • It ensures the test has enough power, often aiming for a minimum power of 0.80 or 0.85.
  • Calculations consider estimated effect sizes, standard deviations, and desired significance level.

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Most popular questions from this chapter

A biotechnology company produces a therapeutic drug whose concentration has a standard deviation of 4 grams per liter. A new method of producing this drug has been proposed, although some additional cost is involved. Management will authorize a change in production technique only if the standard deviation of the concentration in the new process is less than 4 grams per liter. The researchers chose \(n=10\) and obtained the following data in grams per liter. Perform the necessary analysis to determine whether a change in production technique should be implemented. $$\begin{array}{ll}16.628 & 16.630 \\\16.622 & 16.631 \\\16.627 & 16.624 \\\16.623 & 16.622 \\\16.618 & 16.626\end{array}$$

A semiconductor manufacturer collects data from a new tool and conducts a hypothesis test with the null hypothesis that a critical dimension mean width equals \(100 \mathrm{nm}\). The conclusion is to not reject the null hypothesis. Does this result provide strong evidence that the critical dimension mean equals \(100 \mathrm{nm}\) ? Explain.

In a little over a month, from June \(5,1879,\) to July 2 1879 , Albert Michelson measured the velocity of light in air 100 times (Stigler, Annals of Statistics, 1977 ). Today we know that the true value is \(299,734.5 \mathrm{~km} / \mathrm{sec} .\) Michelson's data have a mean of \(299,852.4 \mathrm{~km} / \mathrm{sec}\) with a standard deviation of 79.01 (a) Find a two-sided \(95 \%\) confidence interval for the true mean (the true value of the speed of light). (b) What does the confidence interval say about the accuracy of Michelson's measurements?

Suppose that we wish to test \(H_{0}: \mu=\mu_{0}\) versus \(H_{1}: \mu \neq \mu_{0}\) where the population is normal with known \(\sigma .\) Let \(0<\epsilon<\alpha,\) and define the critical region so that we will reject \(H_{0}\) if \(z_{0}>z_{\varepsilon}\) or if \(z_{0}<-z_{\alpha-\varepsilon},\) where \(z_{0}\) is the value of the usual test statistic for these hypotheses. (a) Show that the probability of type I error for this test is \(\alpha\). (b) Suppose that the true mean is \(\mu_{1}=\mu_{0}+\delta\). Derive an expression for \(\beta\) for the above test.

For the hypothesis test \(H_{0}: \mu=10\) against \(H_{1}: \mu>10\) and variance known, calculate the \(P\) -value for each of the following test statistics. (a) \(z_{0}=2.05\) (b) \(z_{0}=-1.84\) (c) \(z_{0}=0.4\)

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