/*! 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 10 An article in Industrial Enginee... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An article in Industrial Engineer (September 2012\()\) reported on a study of potential sources of injury to equine veterinarians conducted at a university veterinary hospital. Forces on the hand were measured for several common activities that veterinarians engage in when examining or treating horses. We will consider the forces on the hands for two tasks. lifting and using ultrasound. Assume that both sample sizes are \(6,\) the sample mean force for lifting was 6.0 pounds with standard deviation 1.5 pounds, and the sample mean force for using ultrasound was 6.2 pounds with standard deviation 0.3 pounds (data read from graphs in the article). Assume that the standard deviations are known. Is there evidence to conclude that the two activities result in significantly different forces on the hands?

Short Answer

Expert verified
No, there's not enough evidence to conclude a significant difference in forces.

Step by step solution

01

Define the Hypotheses

We need to determine if there is a significant difference in the mean forces of lifting and using ultrasound. - Null Hypothesis \( H_0 \): \( \mu_1 = \mu_2 \) (The mean forces are equal) - Alternative Hypothesis \( H_1 \): \( \mu_1 eq \mu_2 \) (The mean forces are not equal)
02

Specify the Significance Level

The significance level \( \alpha \) is typically chosen as 0.05 for a two-tailed test. This indicates a 5% chance of committing a Type I error, i.e., rejecting the null hypothesis when it is actually true.
03

Calculate the Test Statistic

For comparing two means with known standard deviations, use the following formula for the test statistic:\[z = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\]where:- \( \bar{x}_1 = 6.0 \), \( \bar{x}_2 = 6.2 \) (sample means)- \( \sigma_1 = 1.5 \), \( \sigma_2 = 0.3 \) (standard deviations)- \( n_1 = n_2 = 6 \) (sample sizes)Plug these values into the formula:\[z = \frac{(6.0 - 6.2)}{\sqrt{\frac{1.5^2}{6} + \frac{0.3^2}{6}}} = \frac{-0.2}{\sqrt{\frac{2.25}{6} + \frac{0.09}{6}}} = \frac{-0.2}{\sqrt{0.375 + 0.015}} = \frac{-0.2}{\sqrt{0.39}} \approx \frac{-0.2}{0.6245} \approx -0.32\]
04

Determine the Critical Value and Decision Rule

Since this is a two-tailed test at \( \alpha = 0.05 \), look up the z-table or use a standard normal distribution to find the critical z-values. For \( \alpha = 0.05 \), the critical values are approximately \( z = \pm 1.96 \).Decision Rule: If the calculated z-value lies outside of \([-1.96, 1.96]\), reject the null hypothesis; otherwise, do not reject it.
05

Conclusion

The calculated z-value is approximately \(-0.32\), which is within the range \([-1.96, 1.96]\). Therefore, we fail to reject the null hypothesis. At the 0.05 significance level, there is not enough evidence to conclude that the forces on the hand are significantly different for the two activities.

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.

Significance Level
In hypothesis testing, the significance level, often denoted as \( \alpha \), plays a critical role in determining the threshold for rejecting the null hypothesis. It is the probability of committing a Type I error, which occurs when we incorrectly reject a true null hypothesis. A common choice for the significance level is 0.05, meaning there is a 5% risk of concluding that a difference exists when there is none. When setting the significance level, consider the context of the problem and the potential consequences of errors. A lower \( \alpha \) reduces the risk of a Type I error but increases the chance of a Type II error (failing to reject a false null hypothesis).
  • Choosing \( \alpha = 0.05 \) means you're willing to accept a 5% chance of incorrectly thinking a difference exists.
  • A more stringent \( \alpha = 0.01 \) provides more confidence but might miss true differences.
Typically, \( \alpha \) is chosen before the test starts, providing a benchmark for evaluating the test results. In the exercise provided, a significance level of 0.05 was used for the two-tailed test.
Critical Value
The critical value is a cut-off point that determines the boundaries of the rejection region for a statistical test. It indicates the threshold beyond which we will reject the null hypothesis. It depends on the chosen significance level and the distribution of the test statistic.In a z-test, the critical values are found using the standard normal distribution (z-table). For a two-tailed test with \( \alpha = 0.05 \), the critical values are typically \( z = \pm 1.96 \). This means that if the calculated z-value falls outside the interval \([-1.96, 1.96]\), we reject the null hypothesis.
  • The critical values indicate the values where the null hypothesis is unlikely to be true.
  • These values, combined with the calculated statistic, help determine whether the effect observed is statistically significant.
In our exercise context, the calculated z-value of \(-0.32\) was within the critical value range, leading to the decision not to reject the null hypothesis.
Z-Test
A z-test is a type of statistical test used to determine whether there is a significant difference between sample means or between a sample mean and a known population mean. It is particularly useful when sample sizes are large enough for the Central Limit Theorem to hold or when population variances are known.The formula for the z-test involves the difference between sample means, the standard deviations, and the sample sizes. Here, it's applied as:\[z = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\]
  • Use it when comparing the means of two samples, where the variances are known.
  • Provides a test statistic to compare against critical values to determine statistical significance.
In the exercise, the z-test with a calculated z-value of \(-0.32\) indicated no significant difference between forces on the hand in the two activities of lifting and using ultrasound, within the chosen significance threshold.

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

In a series of tests to study the efficacy of ginkgo biloba on memory, Solomon et al. first looked at differences in memory tests of people six weeks before and after joining the study ["Ginkgo for Memory Enhancement: A Randomized Controlled Trial," Journal of the American Medical Association (2002, Vol. \(288,\) pp. \(835-840\) ) ]. For 99 patients receiving no medication, the average increase in category fluency (number of words generated in one minute) was 1.07 words with a standard deviation of 3.195 words. Researchers wanted to know whether the mean number of words recalled was positive. (a) Is this a one- or two-sided test? (b) Perform a hypothesis test to determine whether the mean increase is zero. (c) Why can this be viewed as a paired \(t\) -test? (d) What does the conclusion say about the importance of including placebos in such tests?

Consider the hypothesis test \(H_{0}: \sigma_{1}^{2}=\sigma_{2}^{2}\) against \(H_{1}: \sigma_{1}^{2}>\sigma_{2}^{2},\) Suppose that the sample sizes are \(n_{1}=20\) and \(n_{2}=8,\) and that \(s_{1}^{2}=4.5\) and \(s_{2}^{2}=2.3 .\) Use \(\alpha=0.01 .\) Test the hypothesis and explain how the test could be conducted with a confidence interval on \(\sigma_{1} / \sigma_{2}\),

Consider the hypothesis test \(H_{0}: \mu_{1}=\mu_{2}\) against \(H_{1}: \mu_{1}=\mu_{2} .\) Suppose that sample sizes \(n_{1}=15\) and \(n_{2}=15,\) that \(\bar{x}_{1}=6.2\) and \(\bar{x}_{2}=7.8,\) and that \(s_{1}^{2}=4\) and \(s_{2}^{2}=6.25 .\) Assume that \(\sigma_{1}^{2}=\sigma_{2}^{2}\) and that the data are drawn from normal distributions. Use \(\alpha=0.05\) (a) Test the hypothesis and find the \(P\) -value. (b) Explain how the test could be conducted with a confidence interval. (c) What is the power of the test in part (a) if \(\mu_{1}\) is 3 units less than \(\mu_{2}\) ? (d) Assume that sample sizes are equal. What sample size should be used to obtain \(\beta=0.05\) if \(\mu_{1}\) is 2.5 units less than \(\mu_{2}\) ? Assume that \(\alpha=0.05 .\)

An article in Electronic Components and Technology Conference \((2001,\) Vol. \(52,\) pp. \(1167-1171)\) compared single versus dual spindle saw processes for copper metallized wafers. A total of 15 devices of each type were measured for the width of the backside chipouts, \(\bar{x}_{\text {single }}=66.385, s_{\text {single }}=7.895\) and \(\bar{x}_{\text {double }}=45.278, s_{\text {double }}=8.612\). (a) Do the sample data support the claim that both processes have the same chip outputs? Use \(\alpha=0.05\) and assume that both populations are normally distributed and have the same variance. Find the \(P\) -value for the test. (b) Construct a \(95 \%\) two-sided confidence interval on the mean difference in spindle saw process. Compare this interval to the results in part (a). (c) If the \(\beta\) -error of the test when the true difference in chip outputs is 15 should not exceed \(0.1,\) what sample sizes must be used? Use \(\alpha=0.05 .\)

Two companies manufacture a rubber material intended for use in an automotive application. The part will be subjected to abrasive wear in the field application, so you decide to compare the material produced by each company in a test. Twenty-five samples of material from each company are tested in an abrasion test, and the amount of wear after 1000 cycles is observed. For company \(1,\) the sample mean and standard deviation of wear are \(\bar{x}_{1}=20\) milligrams/ 1000 cycles and \(s_{1}=2\) milligrams/1000 cycles, and for company 2, you obtain \(\bar{x}_{2}=15\) milligrams/ 1000 cycles and \(s_{2}=8\) milligrams \(/ 1000\) cycles. (a) Do the data support the claim that the two companies produce material with different mean wear? Use \(\alpha=0.05,\) and assume that each population is normally distributed but that their variances are not equal. What is the \(P\) -value for this test? (b) Do the data support a claim that the material from company 1 has higher mean wear than the material from company \(2 ?\) Use the same assumptions as in part (a). (c) Construct confidence intervals that will address the questions in parts (a) and (b) above.

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.