91Ó°ÊÓ

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)?

Step by step solution

01

Define Hypotheses

For this hypothesis test, we need to check if the mean tensile strength of the roller method exceeds 2150 N/mm². We set the null hypothesis as \( H_0: \mu = 2150 \) and the alternative hypothesis as \( H_a: \mu > 2150 \). This is a one-tailed test.

02

Choose the Test Statistic

Given that the sample size is small ( = 16) and the population standard deviation is unknown (only the sample standard deviation is available), we should use the t-test for a single sample mean. The t-test statistic is calculated as: \( t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \), where \( \bar{x} \) is the sample mean, \( \mu_0 \) is the population mean under null hypothesis, \( s \) is the sample standard deviation, and \( n \) is the sample size.

03

Calculate Test Statistic

Using the formula for the test statistic \( t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \), we substitute the given values: \( \bar{x} = 2160 \), \( \mu_0 = 2150 \), \( s = 30 \), and \( n = 16 \). Thus, the test statistic is calculated as: \( t = \frac{2160 - 2150}{30/\sqrt{16}} = \frac{10}{7.5} \approx 1.33 \).

04

Determine P-Value

To determine the p-value for a t-test with \( t = 1.33 \), we look up this value in the t-distribution table or use statistical software, with degrees of freedom \( df = n-1 = 15 \). A t-score of 1.33 corresponds to a one-tailed p-value approximately equal to 0.10.

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Key Concepts

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

T-test

The T-test is a statistical test that helps us determine if there is a significant difference between the means of two groups. In our scenario of testing wire tensile strength, we applied a one-sample T-test. This is an appropriate method since we want to understand how the tensile strength from the roller method compares to a known mean value from the spinner method.

• The basic idea is to see if our sample mean (from the roller method) significantly surpasses the known population mean (from the spinner method).
• Given that our sample size is relatively small and only the sample standard deviation is available, the T-test becomes the apt choice.

In mathematical terms, the T-test helps us understand if the differences we observe could occur due to random sampling variability. This makes it a fundamental tool for hypothesis testing in statistics. The formula we use for the T-test is: \[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \]where:\( \bar{x} \) = sample mean,\( \mu_0 \) = hypothesized population mean,\( s \) = sample standard deviation,\( n \) = sample size. This formula helps us compute the test statistic, which is pivotal in determining statistical significance.

Tensile Strength

Tensile strength refers to the maximum stress a material can withstand while being stretched or pulled before breaking. It is a crucial property of materials, notably metals like the wire in our example, as it dictates the material's resistance to breaking under tension.

• Measured in Newton per square millimeter ( /mm²), tensile strength provides insight into the durability and performance of materials under load.
• In engineering applications, understanding tensile strength is vital, as it influences material selections and treatments to ensure safety and functionality.

In the context of this exercise, we are comparing the tensile strengths achieved through different straightening methods - roller vs. spinner - to find a method that enhances performance. Understanding tensile strength provides engineers with essential information to predict how materials behave under stress. Thus, ensuring optimal usage in various applications.

Null and Alternative Hypotheses

The formulation of null and alternative hypotheses is a critical step in hypothesis testing. It sets the frame for the statistical test we perform.

• The null hypothesis ( _0) typically represents a statement of no effect or no difference. In our wire test, this implies that there is no increase in tensile strength due to the roller method. Mathematically, this is expressed as: \( H_0: \mu = 2150 \).
• The alternative hypothesis ( _a) stands for what you want to test or prove. Here, it suggests that the roller method results in a tensile strength greater than 2150, expressed as: \( H_a: \mu > 2150 \).

This construction is pivotal as it directly influences the direction and type of the test employed. A one-tailed T-test, as applied here, focuses on whether the mean is greater than a certain value, hence aligning with the formulated hypothesis.Establishing clear null and alternative hypotheses aids in making logical conclusions based on the statistical evidence gathered.

P-value

The P-value serves as a crucial indicator of the statistical significance in hypothesis testing. It reflects the probability of obtaining the observed test results, or more extreme, given that the null hypothesis is true.

• This probability value helps us decide whether to reject the null hypothesis in favor of the alternative hypothesis.
• A lower P-value indicates stronger evidence against the null hypothesis, thereby supporting the alternative hypothesis.

In the exercise, for our calculated test statistic of \( t = 1.33 \), we determined the P-value using the t-distribution with 15 degrees of freedom. The resulting P-value of approximately 0.10 suggests a weak evidence against the null hypothesis.When analyzing P-values, researchers often use a threshold level called the significance level (commonly 0.05). If the P-value is less than this threshold, the null hypothesis is typically rejected.Understanding P-values is essential for interpreting the results of hypothesis tests and making informed decisions based on statistical data.