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Chapter 9: Problem 10

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 high-purity 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 commercial-purity 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 \(\mu_{1}-\mu_{2}=6\)

Short Answer

Expert verified
Reject the null hypothesis; the test is very powerful ( \beta\) is near zero).

Step by step solution

01

Define the Hypotheses

We need to establish the null and alternative hypotheses. The null hypothesis (H_0) states that the fracture toughness difference between the high-purity steel and the commercial steel is not greater than 5 units, so H_0: \( \mu_1 - \mu_2 = 5 \ ). The alternative hypothesis (H_a) suggests that the fracture toughness of the high-purity steel exceeds that of the commercial steel by more than 5 units, so H_a: e H_a: \mu_1 - \mu_2 > 5 \).
02

Determine the Test Statistic

Given that both populations are normally distributed with known variances, we use the Z-test for two means. The test statistic z is given by:\[ z = \frac{(\bar{x} - \bar{y}) - D}{\sqrt{\frac{\sigma_1^2}{m} + \frac{\sigma_2^2}{n}}} \]Where \(\bar{x} = 65.6\), \(\bar{y} = 59.8\), \(D = 5\), \(\sigma_1 = 1.2\), \(\sigma_2 = 1.1\), \(m = 32\), and \(n = 38\).
03

Calculate the Test Statistic

Substitute the values into the formula:\[ z = \frac{(65.6 - 59.8) - 5}{\sqrt{\frac{1.2^2}{32} + \frac{1.1^2}{38}}} = \frac{0.8}{\sqrt{\frac{1.44}{32} + \frac{1.21}{38}}}\]Simplifying further,\[ z = \frac{0.8}{\sqrt{0.045 + 0.032}} = \frac{0.8}{0.094} \approx 8.51\]
04

Make a Decision Using the Critical Value Approach

For a one-tailed test at a significance level of \(\alpha = 0.001\), the critical z-value from the standard normal distribution table is approximately 3.09. Since our calculated z-value (\(8.51\)) is greater than 3.09, we reject the null hypothesis.
05

Compute the Power of the Test (\(\beta\))

\(\beta\) is the probability of failing to reject the null hypothesis when the true difference in means is 6. The test statistic under this condition is:\[ z_\beta = \frac{6 - 5}{\sqrt{\frac{1.44}{32} + \frac{1.21}{38}}} = \frac{1}{0.094} \approx 10.64 \]Find the power (\(1-\beta\)) corresponding to this z-value. The z-value of 10.64 is extremely high, indicating that \(\beta\) is near zero. Therefore, the power of the test is nearly 1.

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Key Concepts

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

Fracture Toughness
Fracture toughness is a critical property in materials science, particularly when assessing the durability and reliability of materials under stress. It measures a material's ability to resist crack propagation. When a material is subject to an external load, cracks may start to appear and grow. Fracture toughness quantifies how well a material can endure such crack growth before it fractures. Consider it a form of resilience in materials. A high fracture toughness is desirable in engineering applications as it indicates that the material can withstand significant stress and deformation鈥攁n important characteristic for structural materials used in demanding environments. In the original exercise, we examined fracture toughness in two types of steel: high-purity and commercial-purity steel. Understanding their toughness helps determine which material is more suitable for specific applications, especially when higher toughness is necessary.
Z-test
The Z-test is a statistical method used to determine if there is a significant difference between the means of two samples, particularly when the variance is known and the sample size is large. ### How the Z-test Works - It compares the observed difference between sample means to the difference expected if the null hypothesis were true. - It assumes the data is normally distributed鈥攚hich, in the exercise, was stated for both samples. - The calculated z-value helps us decide whether to reject the null hypothesis. ### Applying to Fracture Toughness In the exercise, we used the Z-test to determine if the fracture toughness of high-purity steel was significantly greater than that of commercial-purity steel by more than 5 units. This involved calculating the test statistic using means and known standard deviations of the samples. The calculation showed a z-value of 8.51, exceeding the critical value of 3.09 at the 0.001 significance level, leading to rejection of the null hypothesis.
Power of a Test
The power of a test is a fundamental concept in hypothesis testing. It represents the probability that the test correctly rejects the null hypothesis when the alternative hypothesis is true. In simpler terms, it鈥檚 the test's ability to detect a real effect when there is one. ### Why is Power Important? - High power is desirable as it means a test is likely to detect meaningful differences when they exist. - Power depends on the significance level, sample size, and the true difference in means. ### Application in the Exercise In our fracture toughness test, the power was calculated for an actual difference in means of 6. The calculated power was nearly 1, indicating an extremely high probability of detecting a true difference when it exists. This high power assures confidence in the decision to reject the null hypothesis.
Standard Normal Distribution
The standard normal distribution is a key concept in statistics. It is a normal distribution with a mean of 0 and a standard deviation of 1. This distribution is used in many statistical tests, including the Z-test, as it allows for easy determination of probabilities and critical values. ### Characteristics of Standard Normal Distribution - It is symmetric around the mean. - Nearly all data falls within three standard deviations of the mean, with about 68% within one standard deviation. - Its properties simplify the computation of probabilities using z-values. ### Role in Hypothesis Testing When conducting a Z-test, we use the standard normal distribution to find critical values that determine whether to reject the null hypothesis. In the exercise, a critical z-value of approximately 3.09 corresponds to the 0.001 significance level. Since our calculated z-value was higher, it confirmed that the difference in fracture toughness was statistically significant. Understanding this distribution is crucial for comprehending how statistical tests are evaluated, ensuring accurate inference from data.

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Most popular questions from this chapter

Fusible interlinings are being used with increasing frequency to support outer fabrics and improve the shape and drape of various pieces of clothing. The article "Compatibility of Outer and Fusible Interlining Fabrics in Tailored Garments" (Textile Res. \(J_{.}, 1997: 137-142\) ) gave the accompanying data on extensibility (\%) at \(100 \mathrm{gm} / \mathrm{cm}\) for both high-quality (H) fabric and poor-quality (P) fabric specimens. \(\begin{array}{rrrrrrrrrr}\mathrm{H} & 1.2 & .9 & .7 & 1.0 & 1.7 & 1.7 & 1.1 & .9 & 1.7 \\ & 1.9 & 1.3 & 2.1 & 1.6 & 1.8 & 1.4 & 1.3 & 1.9 & 1.6 \\ & 8 & 2.0 & 1.7 & 1.6 & 2.3 & 2.0 & & & \\ \mathrm{P} & 1.6 & 1.5 & 1.1 & 2.1 & 1.5 & 1.3 & 1.0 & 2.6 & \end{array}\) a. Construct normal probability plots to verify the plausibility of both samples having been selected from normal population distributions. b. Construct a comparative boxplot. Does it suggest that there is a difference between true average extensibility for high-quality fabric specimens and that for poorquality specimens? c. The sample mean and standard deviation for the highquality sample are \(1.508\) and \(.444\), respectively, and those for the poor-quality sample are \(1.588\) and 530 . Use the two-sample \(t\) test to decide whether true average extensibility differs for the two types of fabric.

The article "The Effects of a Low-Fat, Plant-Based Dietary Intervention on Body Weight, Metabolism, and Insulin Sensitivity in Postmenopausal Women" (Amer. J. of Med., \(2005: 991-997\) ) reported on the results of an experiment in which half of the individuals in a group of 64 postmenopausal overweight women were randomly assigned to a particular vegan diet, and the other half received a diet based on National Cholesterol Education Program guidelines. The sample mean decrease in body weight for those on the vegan diet was \(5.8\) \(\mathrm{kg}\), and the sample SD was \(3.2\), whereas for those on the control diet, the sample mean weight loss and standard deviation were \(3.8\) and \(2.8\), respectively. Does it appear the true average weight loss for the vegan diet exceeds that for the control diet by more than \(1 \mathrm{~kg}\) ? Carry out an appropriate test of hypotheses at significance level \(.05\) based on calculating a \(P\)-value.

Quantitative noninvasive techniques are needed for routinely assessing symptoms of peripheral neuropathies, such as carpal tunnel syndrome (CTS). The article "A Gap Detection Tactility Test for Sensory Deficits Associated with Carpal Tunnel Syndrome" (Ergonomics, 1995: 2588-2601) reported on a test that involved sensing a tiny gap in an otherwise smooth surface by probing with a finger; this functionally resembles many work-related tactile activities, such as detecting scratches or surface defects. When finger probing was not allowed, the sample average gap detection threshold for \(m=8\) normal subjects was \(1.71 \mathrm{~mm}\), and the sample standard deviation was 53 ; for \(n=10\) CTS subjects, the sample mean and sample standard deviation were \(2.53\) and \(.87\), respectively. Does this data suggest that the true average gap detection threshold for CTS subjects exceeds that for normal subjects? State and test the relevant hypotheses using a significance level of \(.01\).

It is thought that the front cover and the nature of the first question on mail surveys influence the response rate. The article "The Impact of Cover Design and First Questions on Response Rates for a Mail Survey of Skydivers" (Leisure Sciences, 1991: 67-76) tested this theory by experimenting with different cover designs. One cover was plain; the other used a picture of a skydiver. The researchers speculated that the return rate would be lower for the plain cover. \begin{tabular}{lcc} Cover & Number Sent & Number Returned \\ \hline Plain & 207 & 104 \\ Skydiver & 213 & 109 \\ \hline \end{tabular} Does this data support the researchers' hypothesis? Test the relevant hypotheses using \(\alpha=.10\) by first calculating a \(P\)-value.

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.?
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