/*! 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 13 A mechanical engineer wishes to ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A mechanical engineer wishes to compare strength properties of steel beams with similar beams made with a particular alloy. The same number of beams, \(n\), of each type will be tested. Each beam will be set in a horizontal position with a support on each end, a force of \(2500 \mathrm{lb}\) will be applied at the center, and the deflection will be measured. From past experience with such beams, the engineer is willing to assume that the true standard deviation of deflection for both types of beam is .05 in. Because the alloy is more expensive, the engineer wishes to test at level .01 whether it has smaller average deflection than the steel beam. What value of \(n\) is appropriate if the desired type II error probability is \(.05\) when the difference in true average deflection favors the alloy by \(.04\) in.?

Short Answer

Expert verified
The appropriate sample size is 13 beams of each type.

Step by step solution

01

Understanding the Problem

The engineer wants to test if the alloy beams have a smaller average deflection than steel beams under a force of 2500 lb, with an assumed standard deviation of 0.05 in. for both beam types. We need to find an appropriate sample size, \( n \), to ensure a type II error probability of 0.05, given that the average deflection for alloy beams is 0.04 in. less than for steel beams, at a significance level of 0.01.
02

Formulating the Hypotheses

Define the null hypothesis \( H_0: \mu_{steel} - \mu_{alloy} = 0 \) against the alternative hypothesis \( H_1: \mu_{steel} - \mu_{alloy} > 0.04 \). This is a one-tailed test because we are testing if the alloy beams have smaller deflection than steel beams.
03

Calculating the Test Statistic

The test statistic for the difference between two means with known standard deviations is given by \( Z = \frac{(\mu_{steel} - \mu_{alloy}) - \Delta}{\sigma \sqrt{\frac{2}{n}}} \), where \( \Delta = 0.04 \) is the hypothesized mean difference and \( \sigma = 0.05 \) is the standard deviation. We use this to determine \( n \) that satisfies the type II error probability of 0.05.
04

Determining Critical Z-values

For a significance level of \( \alpha = 0.01 \), the critical Z-value for a one-tailed test is approximately \( Z_{\alpha} = 2.33 \). For a type II error probability of \( \beta = 0.05 \), the corresponding Z-value is approximately \( Z_{\beta} = -1.645 \).
05

Solving for the Sample Size \( n \)

Using the formula for the test statistic:\[ 2.33 = \frac{(0 + 0.04) - 0.04}{0.05 \sqrt{\frac{2}{n}}} - 1.645 \]Rearranging the equation to solve for \( n \):\[ 2.33 + 1.645 = \frac{0.04}{0.05} \sqrt{\frac{2}{n}} \]\[ 3.975 = \frac{0.04 \sqrt{2n}}{0.05} \]\[ 3.975 \times 0.05 = 0.04 \sqrt{2n} \]\[ 0.19875 = 0.04 \sqrt{2n} \]\[ \frac{0.19875}{0.04} = \sqrt{2n} \]\[ 4.96875 = \sqrt{2n} \]Squaring both sides gives:\[ n = \frac{24.687\ldots}{2} \approx 12.34 \]Since \( n \) must be an integer, round up to \( n = 13 \).
06

Conclusion

An appropriate sample size is \( n = 13 \) beams of each type to achieve the desired power of the test with a significance level of 0.01 and type II error of 0.05.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Hypothesis Testing
Hypothesis testing is a vital statistical method used to determine whether there is enough evidence in a sample to conclude that a condition holds true for the entire population. In this scenario, the hypothesis testing revolves around comparing the deflection properties of steel beams with those made from a specific alloy. The engineer's goal is to verify whether the alloy beams have a lower average deflection than steel beams.

This is set up using two placeholders: the null hypothesis and the alternative hypothesis. The null hypothesis, denoted as \( H_0 \), posits that there is no difference in deflection (\( \mu_{steel} - \mu_{alloy} = 0 \)) between the two types of beams. Conversely, the alternative hypothesis \( H_1 \) states that the alloy beams have a smaller deflection by a specified amount (e.g., \( \mu_{steel} - \mu_{alloy} > 0.04 \)). This represents a one-tailed test since the direction of the test (i.e., smaller deflection for the alloy) matters here.
Statistical Power
Statistical power is an essential concept in designing experiments and analyzing data in hypothesis testing. It quantifies the test's ability to correctly reject the null hypothesis when the alternative hypothesis is true. In other words, it measures the likelihood that the study will detect an effect when an effect actually exists.

In the given problem, the engineer aims to achieve a statistical power that minimizes making a type II error, set at a probability of 0.05. High power is desirable because it means there is a high chance of detecting a real effect (in this case, the alloy having a smaller deflection) if it exists. The statistical power depends on several factors like sample size, the variability of the data, and the significance level (type I error rate). Higher sample sizes usually increase power, enabling more reliable conclusions.
Type I and Type II Errors
Type I and Type II errors are fundamental terms when discussing errors in hypothesis testing. A Type I error occurs when the null hypothesis is incorrectly rejected, essentially seeing an effect that isn't truly there. The probability of making a Type I error is denoted by \( \alpha \), and, in this problem, it is set at 0.01, meaning a 1% risk of incorrectly rejecting the null hypothesis.

Conversely, a Type II error occurs when the null hypothesis is wrongly accepted, overlooking a real effect. This probability is represented by \( \beta \). In this study, the engineer wishes to keep \( \beta \) at 0.05, showing a 5% chance of missing a true smaller deflection effect of the alloy beam. Understanding and managing these errors is crucial in statistical decision-making, where balancing between unneeded alarm and missing significant findings is the key.
Standard Deviation
Standard deviation is a vital statistical measure that reveals the variability or spread in a data set. In this particular exercise, the engineer assumes a standard deviation of 0.05 inches for both steel and alloy beams, representing the usual fluctuation in deflection values.

In hypothesis testing and sample size determination, standard deviation is crucial because it influences the test's accuracy and precision. The smaller the standard deviation, the more tightly clustered the data points are around the mean, making the test results more reliable. Standard deviation also impacts the calculation of the test statistic, which is used to determine whether the null hypothesis can be rejected. Hence, knowing the deflection variance helps in accurately ascertaining the required sample size for the desired power of the test. It provides guidance on the consistency of the measurements, a foundational aspect in hypothesis testing.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The level of monoamine oxidase (MAO) activity in blood platelets ( \(\mathrm{nm} / \mathrm{mg}\) protein/h) was determined for each individual in a sample of 43 chronic schizophrenics, resulting in \(\bar{x}=2.69\) and \(s_{1}=2.30\), as well as for 45 normal subjects, resulting in \(\bar{y}=6.35\) and \(s_{2}=4.03\). Does this data strongly suggest that true average MAO activity for normal subjects is more than twice the activity level for schizophrenics? Derive a test procedure and carry out the test using \(\alpha=.01\). [Hint: \(H_{0}\) and \(H_{2}\) here have a different form from the three standard cases. Let \(\mu_{1}\) and \(\mu_{2}\) refer to true average MAO activity for schizophrenics and normal subjects, respectively, and consider the parameter \(\theta=2 \mu_{1}-\mu_{2}\). Write \(H_{0}\) and \(H_{a}\) in terms of \(\theta\), estimate \(\theta\), and derive \(\hat{\sigma}_{\dot{\theta}}\) ("Reduced Monoamine Oxidase Activity in Blood Platelets from Schizophrenic Patients," Nature, July 28, 1972: 225-226).]

An experimenter wishes to obtain a CI for the difference between true average breaking strength for cables manufactured by company I and by company II. Suppose breaking strength is normally distributed for both types of cable with \(\sigma_{1}=30 \mathrm{psi}\) and \(\sigma_{2}=20\) psi. a. If costs dictate that the sample size for the type I cable should be three times the sample size for the type II cable, how many observations are required if the \(99 \%\) CI is to be no wider than 20 psi? b. Suppose a total of 400 observations is to be made. How many of the observations should be made on type I cable samples if the width of the resulting interval is to be a minimum?

The article "Enhancement of Compressive Properties of Failed Concrete Cylinders with Polymer Impregnation" \((J\). of Testing and Evaluation, 1977: 333-337) reports the following data on impregnated compressive modulus \(\left(\mathrm{psi} \times 10^{6}\right)\) when two different polymers were used to repair cracks in failed concrete. \(\begin{array}{lllll}\text { Epoxy } & 1.75 & 2.12 & 2.05 & 1.97 \\ \text { MMA prepolymer } & 1.77 & 1.59 & 1.70 & 1.69\end{array}\) Obtain a \(90 \%\) CI for the ratio of variances by first using the method suggested in the text to obtain a general confidence interval formula.

The article "Flexure of Concrete Beams Reinforced with Advanced Composite Orthogrids" (J. of Aerospace Engr. 1997: 7-15) gave the accompanying data on ultimate load (kN) for two different types of beams. \begin{tabular}{lccc} Type & Sample Size & Sample Mean & Sample SD \\ \hline Fiberglass grid & 26 & \(33.4\) & \(2.2\) \\ Commercial carbon grid & 26 & \(42.8\) & \(4.3\) \\ \hline \end{tabular} a. Assuming that the underlying distributions are normal, calculate and interpret a \(99 \%\) CI for the difference between true average load for the fiberglass beams and that for the carbon beams. b. Does the upper limit of the interval you calculated in part (a) give a \(99 \%\) upper confidence bound for the difference between the two \(\mu\) 's? If not, calculate such a bound. Does it strongly suggest that true average load for the carbon beams is more than that for the fiberglass beams? Explain.

An experiment was performed to compare the fracture toughness of high-purity \(18 \mathrm{Ni}\) maraging steel with commercial-purity steel of the same type (Corrosion Science, 1971: 723-736). For \(m=32\) specimens, the sample average toughness was \(\bar{x}=65.6\) for the highpurity steel, whereas for \(n=38\) specimens of commercial steel \(\bar{y}=59.8\). Because the high-purity steel is more expensive, its use for a certain application can be justified only if its fracture toughness exceeds that of commercialpurity steel by more than 5 . Suppose that both toughness distributions are normal. a. Assuming that \(\sigma_{1}=1.2\) and \(\sigma_{2}=1.1\), test the relevant hypotheses using \(\alpha=.001\). b. Compute \(\beta\) for the test conducted in part (a) when \(\mu_{1}-\mu_{2}=6\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.