91Ó°ÊÓ

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 we 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, while for company 2 we 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 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.

Step by step solution

01

Define Hypotheses for Part (a)

We want to test if the two companies produce material with different mean wear. The null hypothesis is that the means are equal: \( H_0: \mu_1 = \mu_2 \). The alternative hypothesis is that the means are different: \( H_a: \mu_1 eq \mu_2 \). This is a two-tailed test.

02

Choose the Appropriate Test

Since the population variances are not equal and the sample size is small, we will use the two-sample t-test assuming unequal variances (Welch's t-test).

03

Calculate the Test Statistic for Part (a)

The test statistic \( t \) is calculated using the formula: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] Substitute \( \bar{x}_1 = 20 \), \( \bar{x}_2 = 15 \), \( s_1 = 2 \), \( s_2 = 8 \), \( n_1 = 25 \), and \( n_2 = 25 \) into the formula.

04

Compute the Degrees of Freedom

The degrees of freedom for the test statistic is calculated using: \[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1-1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2-1}} \] Substitute the variances and sample sizes to find the appropriate degrees of freedom.

05

Determine the Critical Value and P-value for Part (a)

Using the computed \( t \) statistic and degrees of freedom, find the two-tailed P-value from the t-distribution table. Compare the P-value to \( \alpha = 0.05 \) to decide if you reject the null hypothesis.

06

Define Hypotheses for Part (b)

We test if company 1 has higher mean wear than company 2. The null hypothesis is \( H_0: \mu_1 \leq \mu_2 \). The alternative hypothesis is \( H_a: \mu_1 > \mu_2 \). This is a one-tailed test.

07

Calculate the Test Statistic for Part (b)

Use the same test statistic calculated for Part (a), as the data and assumptions have not changed; however, interpret in a one-tailed context.

08

Determine P-value for Part (b)

Using the calculated \( t \) statistic, determine the one-tailed P-value and compare it to \( \alpha = 0.05 \) to decide if you reject the null hypothesis.

09

Construct Confidence Interval for Part (a)

Calculate the 95% confidence interval for the difference in means using:\[ (\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2, df} \times \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \] Find the critical value \( t_{\alpha/2, df} \) from the t-distribution table.

10

Construct Confidence Interval for Part (b)

For Part (b), calculate the one-sided confidence interval if necessary. Often, for one-sided tests, presenting the two-sided interval suffices for interpretation. Interpret this interval in the context of the direction of the test.

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Key Concepts

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

Two-Sample t-Test

When comparing two different samples, like the material wear from two companies, the two-sample t-test is a handy tool. It helps determine if there's a significant difference between the means of these two groups. This test is particularly useful when sample sizes are small, say less than 30, and you want to see if one sample mean is statistically different from another.
In the case of the exercise, we want to know if the rubber materials from two companies show a difference in wear after being put through a cycle test. By performing this two-sample t-test, we're checking if the observed difference in the means of wear from company 1 and company 2 could have happened by random chance.

• If the test points out that the difference is too large to be random, then we conclude a significant difference exists.

P-Value

The P-value is an essential output of the two-sample t-test. It tells us how likely it is that our observed data could have happened under the null hypothesis (that there is no real difference between the groups).
In our exercise, a small P-value (typically less than our significance level, \( \alpha = 0.05 \), suggests that the observed differences in wear are rare under the hypothesis that the means are equal. This leads us to question the null hypothesis and consider the alternative that a true difference in the means exists.
The P-value interpretation is crucial for hypothesis testing. It’s our decisive tool for understanding the statistical significance of our test results.

Confidence Interval

Confidence intervals provide a range of values which likely contain the true difference between the two sample means. In other words, it's a way to estimate the uncertainty around our measured differences.
In our scenario, for a 95% confidence interval, we are fairly sure that the true difference in material wear, say from company 1 minus company 2, lies within this computed range.

• It's important to note that if zero is within this interval, it suggests that there could be no significant difference between the two companies' means.

This interval gives us a practical check beyond just the P-value, offering a sense of scale for the difference in performance between the two companies' materials.

Unequal Variances

Not all datasets will have equal variances, which is a measure of the spread or dispersion of the data points. When variances aren’t equal, it influences the choice of t-test. Welcome, the Welch's t-test!
This version of the t-test allows for unequal variances, providing a more reliable result in such cases. In our exercise with the wear of rubber materials, the variances of company 1 and company 2 were different. Using Welch's t-test accounts for this difference, leading to more accurate conclusions.
Using the typical t-test when variances are unequal can result in inaccurate results, ultimately affecting the decision on whether significant differences exist. Therefore, it’s crucial to determine variance equality before proceeding with analysis.