/*! 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 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 value of \(n\) is 8.

Step by step solution

01

Understanding the Problem

We need to determine the sample size \(n\) given the conditions of the hypothesis test: to test if the alloy has a smaller average deflection than steel. The null hypothesis is that the mean deflection \(\mu_A\) for the alloy is equal or greater than that of the steel \(\mu_S\), while the alternative hypothesis is that \(\mu_A < \mu_S\). The known parameters are the significance level \(\alpha = 0.01\), desired power \(1-\beta = 0.95\), and the true standard deviation of \(0.05\)% for both materials.
02

Define the Hypotheses and Significance Level

Let the null hypothesis be \(H_0: \mu_A - \mu_S = 0\) and the alternative hypothesis be \(H_1: \mu_A - \mu_S = -0.04\). The engineer wants to test this at a significance level of \(\alpha = 0.01\).
03

Identify the Critical Value for \(\alpha = 0.01\)

The critical value \(z_{\alpha}\) is needed to determine the rejection region under the standard normal distribution. We use the significance level of \(0.01\) to find \(z_{0.01}\), which typically corresponds to \(-2.33\) since we are checking for a one-tailed test (testing smaller).
04

Determine the Critical Value for \(\beta = 0.05\)

For a desired power of \(0.95\), we find the critical value \(z_{\beta} = z_{0.05}\), which corresponds typically to \(1.645\) in a one-tailed standard normal distribution.
05

Use the Sample Size Formula for Two Means

For two population means with known equal standard deviations, the formula for finding \(n\) is:\[ n = \left(\frac{(z_{\alpha} + z_{\beta}) \cdot \sigma}{\Delta}\right)^2 \]where \(\Delta = 0.04\) is the true mean difference, and \(\sigma = 0.05\).
06

Substitute Known Values into Formula

Substituting the known values, we calculate:\[ n = \left(\frac{(-2.33 + 1.645) \times 0.05}{0.04}\right)^2 \] Simplifying this gives:\[ n = \left(\frac{-0.685 \times 0.05}{0.04}\right)^2 = \left(\frac{-0.03425}{0.04}\right)^2 \approx 0.7344^2 \approx 0.539 \approx 1.0 \] rounds mathematically to 2.

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.

Type I Error
In hypothesis testing, a Type I error occurs when we mistakenly reject a true null hypothesis. Essentially, it's a false positive. In the context of the mechanical engineering exercise regarding beam strength, a Type I error would happen if we conclude that the alloy beam has a smaller average deflection when, in fact, it does not. This error is connected to the significance level, represented by \(\alpha\), which is set at 0.01 in the exercise. This value indicates the engineer's tolerance for incorrectly rejecting the null hypothesis.

Think of the significance level \(\alpha = 0.01\) as a 1% chance of making a Type I error. A lower \(\alpha\) reduces the likelihood of a Type I error, making the test more conservative but also possibly more challenging to detect actual effects. This is crucial in scenarios where the cost of a false positive is high, such as using a more expensive material without real benefit.
Type II Error
A Type II error in hypothesis testing happens when we fail to reject a false null hypothesis. This is akin to a false negative. For the beam strength test, a Type II error would be accepting that there is no significant difference between the alloy and steel beams when in reality, the alloy beam indeed exhibits a smaller average deflection.

The probability of a Type II error is denoted by \(\beta\), and its complement, \(1 - \beta\), represents the power of the test. In this exercise, the engineer desires a power of 0.95, implying an emphasis on minimizing Type II errors to 5% (\(\beta = 0.05\)). Therefore, achieving a high test power ensures that if the alloy truly has a smaller deflection, it will be properly detected.
Significance Level
The significance level, denoted by \(\alpha\), is a threshold set by researchers to determine when to reject the null hypothesis. Essentially, it measures the probability of committing a Type I error. For the beam strength problem, a significance level of 0.01 is chosen.

Choosing a significance level involves balancing sensitivity and reliability. A lower significance level (like 0.01) means stricter criteria for detecting differences, thus reducing the chance of false positives. In engineering contexts, where even small errors can have significant repercussions, selecting a stringent significance level is often prudent. This careful choice allows the engineer to be more certain of discovering a real difference in beam deflection.
Sample Size Calculation
Sample size calculation is a fundamental part of hypothesis testing, ensuring that the study has enough power (\(1 - \beta\)) to detect a meaningful effect if it exists. Determining an adequate sample size is essential, as too small a sample may miss true effects, while one too large may waste resources.

For the mechanical engineering exercise, the formula for calculating the necessary sample size for comparing two means when the standard deviation is known is given by:
  • Standard deviation of the population: \(\sigma = 0.05\)
  • Mean difference to detect: \(\Delta = 0.04\)
  • Critical values for \(\alpha = 0.01\) and \(\beta = 0.05\)
Using these values, the required sample size \(n\) was calculated to ensure that if there's a 0.04 inch improvement in deflection using the alloy, it will be reliably detected with high confidence. Correct sample size calculation not only validates the findings but also optimizes resource usage, which is vital in engineering projects where costs for materials and testing are substantial.

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 article "Fatigue Testing of Condoms" cited in Exercise \(7.32\) reported that for a sample of 20 natural latex condoms of a certain type, the sample mean and sample standard deviation of the number of cycles to break were 4358 and 2218 , respectively, whereas a sample of 20 polyisoprene condoms gave a sample mean and sample standard deviation of 5805 and 3990 , respectively. Is there strong evidence for concluding that true average number of cycles to break for the polyisoprene condom exceeds that for the natural latex condom by more than 1000 cycles? Carry out a test using a significance level of .01. [Note: The cited paper reported \(P\)-values of \(t\) tests for comparing means of the various types considered.]

The slant shear test is widely accepted for evaluating the bond of resinous repair materials to concrete; it utilizes cylinder specimens made of two identical halves bonded at \(30^{\circ}\). The article "Testing the Bond Between Repair Materials and Concrete Substrate" (ACI Materials J., 1996: 553-558) reported that for 12 specimens prepared using wire-brushing, the sample mean shear strength \(\left(\mathrm{N} / \mathrm{mm}^{2}\right)\) and sample standard deviation were \(19.20\) and \(1.58\), respectively, whereas for 12 hand-chiseled specimens, the corresponding values were \(23.13\) and \(4.01\). Does the true average strength appear to be different for the two methods of surface preparation? State and test the relevant hypotheses using a significance level of 05 . What are you assuming about the shear strength distributions?

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 \(\sigma_{2}=20 \mathrm{psi}\) 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 \(99 \% \mathrm{CI}\) 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?

Is someone who switches brands because of a financial inducement less likely to remain loyal than someone who switches without inducement? Let \(p_{1}\) and \(p_{2}\) denote the true proportions of switchers to a certain brand with and without inducement, respectively, who subsequently make a repeat purchase. Test \(H_{0}: p_{1}-p_{2}=0\) versus \(H_{a}: p_{1}-p_{2}<0\) using \(\alpha=.01\) and the following data: \(m=200 \quad\) number of success \(=30\) \(n=600\) number of success \(=180\) (Similar data is given in "Impact of Deals and Deal Retraction on Brand Switching," J. of Marketing, 1980: 62-70.)

Cushing's disease is characterized by muscular weakness due to adrenal or pituitary dysfunction. To provide effective treatment, it is important to detect childhood Cushing's disease as early as possible. Age at onset of symptoms and age at diagnosis (months) for 15 children suffering from the disease were given in the article "Treatment of Cushing's Disease in Childhood and Adolescence by Transphenoidal Microadenomectomy" (New Engl. J. of Med., 1984:889). Here are the values of the differences between age at onset of symptoms and age at diagnosis: \(\begin{array}{llllllll}-24 & -12 & -55 & -15 & -30 & -60 & -14 & -21 \\ -48 & -12 & -25 & -53 & -61 & -69 & -80 & \end{array}\) a. Does the accompanying normal probability plot cast strong doubt on the approximate normality of the population distribution of differences? b. Calculate a lower \(95 \%\) confidence bound for the population mean difference, and interpret the resulting bound. c. Suppose the (age at diagnosis) - (age at onset) differences had been calculated. What would be a \(95 \%\) upper confidence bound for the corresponding population mean difference?

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.