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)? e. For a level \(.05\) test, what conclusion would you reach?

Step by step solution

01

Define Hypotheses

We need to determine if the mean tensile strength of wires using the roller method is greater than 2150. Hence, our null hypothesis is that the mean tensile strength is 2150, while the alternative hypothesis is that it is greater than 2150. The hypotheses can be expressed as:- Null Hypothesis: \( H_0: \mu = 2150 \)- Alternative Hypothesis: \( H_1: \mu > 2150 \)

02

Choose Test Statistic

Since the sample size is 16, which is relatively small, and we assume the distribution to be normal with an unknown population variance, we use the t-test statistic for one sample mean. The formula for the t-statistic is:\[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \]where \( \bar{x} \) is the sample mean, \( \mu_0 \) is the population mean under the null hypothesis, \( s \) is the sample standard deviation, and \( n \) is the sample size.

03

Calculate t-statistic

Substitute the given values into the t-statistic formula:- \( \bar{x} = 2160 \)- \( \mu_0 = 2150 \)- \( s = 30 \)- \( n = 16 \)\[ t = \frac{2160 - 2150}{30 / \sqrt{16}} = \frac{10}{7.5} = 1.3333 \]

04

Determine P-value

To find the P-value, look at the t-distribution with \( n-1 = 15 \) degrees of freedom. The P-value is the probability that the t-distribution with 15 df is greater than 1.3333. Using a t-table or statistical software, find that the P-value is approximately 0.101 or 10.1%.

05

Conclusion for Hypothesis Test

At a significance level of \( \alpha = 0.05 \), we compare the P-value with \( \alpha \). Here, \( \text{P-value} = 0.101 > 0.05 \), indicating that we do not reject the null hypothesis. Thus, there is not enough evidence to conclude that the mean tensile strength for the roller method exceeds 2150.

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Key Concepts

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

T-test

A T-test is a statistical tool used to analyze the mean across one or two groups and determine if there is a significant difference between them. In the context of our exercise, the T-test helps us evaluate if the mean tensile strength of wires that underwent roller straightening exceeds 2150 N/mm². There are a few important characteristics about the T-test:

• It is typically used when the sample size is relatively small (n < 30).
• The data is assumed to be approximately normally distributed.
• The test uses the T-distribution, which accounts for the added uncertainty present when estimating the population standard deviation from a sample.To conduct a T-test, you must first establish your hypotheses. The null hypothesis (\(H_0\)) typically states that there is no effect or difference, while the alternative hypothesis (\(H_1\)) indicates that there is an effect or a difference. In our case:- Null Hypothesis: \(H_0: \mu = 2150\)- Alternative Hypothesis: \(H_1: \mu > 2150\)We then calculate the test statistic using the formula:

  \[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \]

  Where:- \(\bar{x}\) is the sample mean,- \(\mu_0\) is the mean under the null hypothesis,- \(s\) is the sample standard deviation, and - \(n\) is the sample size.Under our scenario, if the computed T-value is large, it suggests that our sample mean significantly exceeds our hypothesized population mean, supporting the alternative hypothesis.

P-value

The P-value is a critical component in the realm of hypothesis testing. It helps us determine the strength of the evidence against the null hypothesis, providing us with a probability measure indicating how strange the observed data would be if the null hypothesis were true.

• A smaller P-value suggests that the observed data is unlikely given the null hypothesis.
• A larger P-value indicates that the observed data is consistent with the null hypothesis.

In our step-by-step solution, the P-value was calculated by referencing the t-distribution with the calculated T-test statistic and degrees of freedom (n - 1). Given our setup with a sample size of 16, our degrees of freedom are 15. The P-value obtained in the exercise was approximately 0.101, suggesting that there is a 10.1% probability of observing a sample mean as extreme as 2160 N/mm² under the assumption that the true mean is 2150 N/mm². Understanding the P-value allows us to determine the statistical significance of our results. When the P-value is low, it suggests that the null hypothesis is unlikely. However, a higher P-value tells us that the data does not provide sufficient evidence to reject it.

Significance Level

The significance level, often denoted by \(\alpha\), is a threshold set by the researcher to determine how extreme the results must be in order to reject the null hypothesis. It is a pivotal concept in hypothesis testing that helps researchers decide how much risk they are willing to take for rejecting a true null hypothesis.

• Common significance levels include 0.05, 0.01, and 0.10.
• It represents the probability of making a Type I error, which is rejecting the null hypothesis when it is actually true.

In our given problem, the significance level was set at 0.05. This means we accept a 5% risk of wrongly rejecting the null hypothesis if it were true. When we found the P-value (0.101) in the context of this analysis, we compared it to our significance level. Since the P-value was greater than our established \(\alpha = 0.05\), we concluded that the data did not provide strong enough evidence to reject the null hypothesis.This guides the conclusion of hypothesis tests: if the P-value is less than or equal to the significance level, the null hypothesis can be rejected. But if it is greater, like in our case, we don't reject it, suggesting that more data or a larger effect might be necessary to confidently state a difference.