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Chapter 8: Problem 69

One method for straightening wire before coiling it to make a spring is called "roller straightening." The article "The Effect of Roller and Spinner Wire Straightening on Coiling Performance and Wire Properties" (Springs, 1987: 27-28) reports on the tensile properties of wire. Suppose a sample of 16 wires is selected and each is tested to determine tensile strength \(\left(\mathrm{N} / \mathrm{mm}^{2}\right)\). The resulting sample mean and standard deviation are 2160 and 30, respectively. a. The mean tensile strength for springs made using spinner straightening is \(2150 \mathrm{~N} / \mathrm{mm}^{2}\). What hypotheses should be tested to determine whether the mean tensile strength for the roller method exceeds 2150 ? b. Assuming that the tensile strength distribution is approximately normal, what test statistic would you use to test the hypotheses in part (a)? c. What is the value of the test statistic for this data? d. What is the \(P\)-value for the value of the test statistic computed in part (c)? e. For a level \(.05\) test, what conclusion would you reach?

Short Answer

Expert verified
We lack evidence to conclude roller straightening's mean strength exceeds 2150 N/mm虏 at \\.05 significance.

Step by step solution

01

Define Hypotheses

We need to establish the null and alternative hypotheses. In this case, we're testing if the mean tensile strength using roller straightening exceeds 2150 N/mm虏. Hence,- Null Hypothesis (H_0): \( \mu = 2150 \)- Alternative Hypothesis (H_a): \( \mu > 2150 \)
02

Choose the Test Statistic

Since the sample size is 16 (which is less than 30), and the population standard deviation is unknown, we use the t-test statistic for hypothesis testing with a single mean in a normally distributed population.The formula for the test statistic is:\[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \]where:- \( \bar{x} \) is the sample mean (2160)- \( \mu_0 \) is the population mean under the null hypothesis (2150)- \( s \) is the sample standard deviation (30)- \( n \) is the sample size (16)
03

Calculate the Test Statistic

Using the test statistic formula, we substitute the given values:\[ t = \frac{2160 - 2150}{30/\sqrt{16}} = \frac{10}{7.5} \approx 1.333 \]
04

Determine the P-Value

To find the p-value, we reference the t-distribution table. With 15 degrees of freedom (n-1), we look for t = 1.333 in a one-tailed test. The p-value represents the probability of observing a test statistic as extreme as, or more extreme than, the observed value under the null hypothesis. Given t = 1.333, the p-value is between 0.1 and 0.05.
05

Make a Conclusion

At a significance level of 0.05, we compare the p-value from Step 4: - Since the p-value (between 0.1 and 0.05) is not less than 0.05, we fail to reject the null hypothesis. - Therefore, there is not enough statistical evidence to conclude that the mean tensile strength of wire using roller straightening exceeds 2150 N/mm虏.

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Key Concepts

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

t-test statistic
The t-test statistic is a crucial tool used in hypothesis testing when dealing with small sample sizes and unknown population standard deviations. The purpose of the t-test is to compare the sample mean to a known value or another sample mean to assess if there are significant differences. In our scenario, the t-test helps us determine if the mean tensile strength using roller straightening exceeds the benchmark value of 2150 N/mm虏.

The formula for the t-test statistic is given by: \[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \]
  • \( \bar{x} \) represents the sample mean, which is 2160 N/mm虏 in this case.
  • \( \mu_0 \) is the mean under the null hypothesis, here 2150 N/mm虏.
  • \( s \) denotes the sample standard deviation, given as 30.
  • \( n \) is the sample size, which is 16 here.
This approach is particularly suitable when the sample size is less than 30, highlighting the power of the t-test in situations where samples are limited in number.
tensile strength
Tensile strength is an essential measure in material science used to define the maximum stress that a material can withstand while being stretched or pulled before breaking. It is particularly important for industries involved in manufacturing springs, cables, and other products requiring strong, durable materials.

In this exercise, the focus is on the tensile properties of wires that have undergone roller straightening. We're examining if this method results in a mean tensile strength that is better than spinner straightening. The strength is measured in units of force per unit area such as N/mm虏.

Tensile tests involve subjecting material samples to controlled tension until failure. This provides engineers and scientists with valuable data on the elasticity, plasticity, and breakage point of materials, supporting decisions on their use in various applications.
p-value
The p-value in hypothesis testing is a probability that helps us understand the strength of the evidence against the null hypothesis. A smaller p-value indicates stronger evidence in favor of the alternative hypothesis.

In this context, the p-value reflects the likelihood of observing a test statistic as extreme as, or more extreme than, the calculated t-value, assuming the null hypothesis is true. Given our calculated t-value of approximately 1.333 with 15 degrees of freedom, the p-value is between 0.1 and 0.05 for a one-tailed test.
  • A p-value closer to 0 implies strong evidence against the null hypothesis, suggesting that the alternative hypothesis might be true.
  • If the p-value is large, it suggests that there is insignificant evidence against the null hypothesis.
Understanding p-values is vital for making informed decisions about the hypotheses in any empirical research, ensuring the results are statistically sound.
significance level
The significance level, often denoted by \( \alpha \), is a threshold chosen by researchers to decide when to reject the null hypothesis. It represents the probability of incorrectly rejecting the null hypothesis, known as a Type I error.
  • A common choice for significance levels is 0.05 or 5%, meaning that there is a 5% risk of concluding that a difference exists when there is none.
  • By selecting the significance level before performing the test, researchers set a clear benchmark for decision-making.
In the exercise, the significance level is set at 0.05. When comparing this level to the p-value obtained (between 0.1 and 0.05), we find that the p-value is not lower than 0.05, thus failing to reject the null hypothesis.

This conclusion indicates that there is not enough statistical evidence to assert that the roller straightening method results in higher tensile strength than the spinner method.

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Most popular questions from this chapter

A sample of 12 radon detectors of a certain type was selected, and each was exposed to \(100 \mathrm{pCi} / \mathrm{L}\) of radon. The resulting readings were as follows: \(\begin{array}{rrrrrr}105.6 & 90.9 & 91.2 & 96.9 & 96.5 & 91.3 \\ 100.1 & 105.0 & 99.6 & 107.7 & 103.3 & 92.4\end{array}\) a. Does this data suggest that the population mean reading under these conditions differs from 100 ? State and test the appropriate hypotheses using \(\alpha=.05\). b. Suppose that prior to the experiment a value of \(\sigma=7.5\) had been assumed. How many determinations would then have been appropriate to obtain \(\beta=.10\) for the alternative \(\mu=95\) ?

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from an exponential distribution with parameter \(\lambda\). Then it can be shown that \(2 \lambda \sum X_{i}\) has a chi-squared distribution with \(\nu=2 n\) (by first showing that \(2 \lambda X_{i}\) has a chi-squared distribution with \(v=2\) ). a. Use this fact to obtain a test statistic and rejection region that together specify a level a test for \(H_{0}: \mu=\mu_{0}\) versus each of the three commonly encountered alternatives. [Hint: \(E\left(X_{i}\right)=\mu=1 / \lambda\), so \(\mu=\mu_{0}\) is equivalent to \(\left.\lambda=1 / \mu_{0} \cdot\right]\) b. Suppose that ten identical components, each having exponentially distributed time until failure, are tested. The resulting failure times are \(\begin{array}{llllllllll}95 & 16 & 11 & 3 & 42 & 71 & 225 & 64 & 87 & 123\end{array}\) Use the test procedure of part (a) to decide whether the data strongly suggests that the true average lifetime is less than the previously claimed value of 75 .

An article in the Nov. 11,2005 , issue of the San Luis Obispo Tribune reported that researchers making random purchases at California Wal-Mart stores found scanners coming up with the wrong price \(8.3 \%\) of the time. Suppose this was based on 200 purchases. The National Institute for Standards and Technology says that in the long run at most two out of every 100 items should have incorrectly scanned prices. a. Develop a test procedure with a significance level of (approximately) .05, and then carry out the test to decide whether the NIST benchmark is not satisfied. b. For the test procedure you employed in (a), what is the probability of deciding that the NIST benchmark has been satisfied when in fact the mistake rate is \(5 \%\) ?

The desired percentage of \(\mathrm{SiO}_{2}\) in a certain type of aluminous cement is \(5.5\). To test whether the true average percentage is \(5.5\) for a particular production facility, 16 independently obtained samples are analyzed. Suppose that the percentage of \(\mathrm{SiO}_{2}\) in a sample is normally distributed with \(\sigma=.3\) and that \(\bar{x}=5.25\). a. Does this indicate conclusively that the true average percentage differs from \(5.5\) ? Carry out the analysis using the sequence of steps suggested in the text. b. If the true average percentage is \(\mu=5.6\) and a level \(\alpha=.01\) test based on \(n=16\) is used, what is the probability of detecting this departure from \(H_{0}\) ? c. What value of \(n\) is required to satisfy \(\alpha=.01\) and \(\beta(5.6)=.01 ?\)

Two different companies have applied to provide cable television service in a certain region. Let \(p\) denote the proportion of all potential subscribers who favor the first company over the second. Consider testing \(H_{0}: p=.5\) versus \(H_{\mathrm{a}}: p \neq .5\) based on a random sample of 25 individuals. Let \(X\) denote the number in the sample who favor the first company and \(x\) represent the observed value of \(X\). a. Which of the following rejection regions is most appropriate and why? $$ \begin{aligned} &R_{1}=\\{x: x \leq 7 \text { or } x \geq 18\\}, R_{2}=\\{x: x \leq 8\\} \\ &R_{3}=\\{x: x \geq 17\\} \end{aligned} $$ b. In the context of this problem situation, describe what the type I and type II errors are. c. What is the probability distribution of the test statistic \(X\) when \(H_{0}\) is true? Use it to compute the probability of a type I error. d. Compute the probability of a type II error for the selected region when \(p=.3\), again when \(p=.4\), and also for both \(p=.6\) and \(p=.7\). e. Using the selected region, what would you conclude if 6 of the 25 queried favored company 1 ?
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