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Chapter 9: Problem 8

Tensile-strength tests were carried out on two different grades of wire rod ('"Fuidized Bed Patenting of Wire Rods, Wire \(J .\) J June 1977: \(56-61\) ), resulting in the accompanying data. \begin{tabular}{lccc} Grade & Sample Size & Sample Mean \(\left(\mathrm{kg} / \mathrm{mm}^{2}\right)\) & Sample SD \\ \hline AISI 1064 & \(m=129\) & \(\bar{x}=107.6\) & \(s_{1}=1.3\) \\ AISI 1078 & \(n=129\) & \(\bar{y}=123.6\) & \(s_{2}=2.0\) \\ \hline \end{tabular} a. Does the data provide compelling evidence for concluding that true average strength for the 1078 grade exceeds that for the 1064 grade by more than \(10 \mathrm{~kg} /\) \(\mathrm{mm}^{2}\) ? Test the appropriate hypotheses using a significance level of \(.01\). b. Estimate the difference between true average strengths for the two grades in a way that provides information about precision and reliability.

Short Answer

Expert verified
Yes, the data supports it with 99% confidence. The difference is between 15.46 and 16.54 kg/mm虏.

Step by step solution

01

Formulate Hypotheses

We want to test if the true average strength of grade AISI 1078 exceeds that of grade AISI 1064 by more than 10 kg/mm虏. The null hypothesis is that the difference in means, \( \mu_2 - \mu_1 \leq 10 \), and the alternative hypothesis is \( \mu_2 - \mu_1 > 10 \). Where \( \mu_1 \) is the mean strength of grade 1064, and \( \mu_2 \) is the mean strength of grade 1078.
02

Determine Critical Value and Test Statistic

We'll use a two-sample t-test. Since the sample sizes and variances are given, the test statistic is calculated using:\[ t = \frac{(\bar{y} - \bar{x}) - d_0}{\sqrt{\frac{s_1^2}{m} + \frac{s_2^2}{n}}} \]where \(d_0 = 10\), \(\bar{x} = 107.6\), \(\bar{y} = 123.6\), \(s_1 = 1.3\), \(s_2 = 2.0\), \(m = 129\), and \(n = 129\). The critical value for a significance level of 0.01 is needed from t-distribution table for df = 256.
03

Calculate the Test Statistic

Substitute the values into the formula:\[ t = \frac{(123.6 - 107.6) - 10}{\sqrt{\frac{1.3^2}{129} + \frac{2.0^2}{129}}} \]\[ t = \frac{6}{\sqrt{\frac{1.69}{129} + \frac{4}{129}}} \]\[ t = \frac{6}{\sqrt{0.0131 + 0.0310}} \]\[ t = \frac{6}{\sqrt{0.0441}} \]\[ t \approx \frac{6}{0.2099} \]\[ t \approx 28.6 \]
04

Decision Making

The calculated t-value is approximately 28.6, which is much larger than the critical t-value for an upper one-tailed test at 0.01 significance level, indicating statistical significance. Thus, we reject the null hypothesis.
05

Estimate the Difference with Confidence Interval

To estimate the difference, calculate a confidence interval using the formula:\[ (\bar{y} - \bar{x}) \pm t^* \times \sqrt{\frac{s_1^2}{m} + \frac{s_2^2}{n}} \]where \(t^*\) is the critical t-value for 99% confidence level. The calculation is:\[ 16 \pm t^* \times 0.2099 \]Assuming \(t^*\) from the t-table for df = 256 is about 2.58,\[ 16 \pm 2.58 \times 0.2099 \]\[ 16 \pm 0.542 \]The interval is approximately \( (15.46, 16.54) \).
06

Conclusion

The evidence supports that the average strength of grade 1078 significantly exceeds that of grade 1064 by more than 10 kg/mm虏. The estimated difference is between 15.46 and 16.54 kg/mm虏 with 99% confidence, suggesting high precision and reliability.

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Key Concepts

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

Two-Sample T-Test
The two-sample t-test is a statistical method used to determine if there is a significant difference between the means of two independent groups. In the given problem, it is applied to check if the true average tensile strength of the AISI 1078 grade wire rod exceeds that of the AISI 1064 grade by more than 10 kg/mm虏. This test is particularly useful when you cannot assume the variances of the two populations are equal. The test works by comparing the difference between the sample means to what would be expected if the null hypothesis is true. Here, the null hypothesis states that the difference is less than or equal to 10 kg/mm虏. On the contrary, the alternative hypothesis, which we are testing, suggests that the mean tensile strength of 1078 grade exceeds that of 1064 by more than the specified threshold. By calculating the test statistic using the formula and comparing it with a critical value from a t-distribution, one can make an informed decision about the hypothesis.
Significance Level
When conducting hypothesis testing, the significance level is a crucial factor. It is denoted by alpha (\(\alpha\)) and represents the probability of rejecting the null hypothesis when it is actually true. Essentially, it is the risk you are willing to take to make a type I error in your testing.In this problem, a significance level of 0.01 is chosen, meaning there is a 1% chance of incorrectly concluding that the 1078 grade wire rod is stronger by more than 10 kg/mm虏. A lower significance level means you require stronger evidence to reject the null hypothesis, thereby reducing the chance of a false positive. By comparing the calculated t-value with our critical value derived from the t-distribution at this level, we maintain control over the probability of making an incorrect decision.
Confidence Interval
A confidence interval provides a range of values within which we believe the true difference between the two means lies, with a certain level of confidence. For this problem, a 99% confidence interval is calculated to estimate the difference in tensile strengths between the two grades of wire rods, offering an insight into the precision and reliability of this estimate. The calculation considers both the standard deviations and the sample sizes to determine the interval around the difference in sample means. The resulting interval tells us that we are 99% confident that the true difference in tensile strength is between 15.46 and 16.54 kg/mm虏. This means that not only can we be quite sure that the strength difference exceeds 10 kg/mm虏, it also gives us a specific range that, with very high confidence, includes the actual difference.
T-Distribution
The t-distribution is a statistical distribution that is used extensively when performing t-tests, especially when dealing with smaller sample sizes or unknown population standard deviations. It is similar in shape to the normal distribution but has "fatter tails," which account for the increased variability expected with smaller sample sizes or uncertain data. In this exercise, the t-distribution is used to find the critical value necessary to make our decision about the hypothesis at the given significance level. With a degrees of freedom (df) calculated as the sum of the sample sizes minus two, the t-distribution helps determine the range of t-values that would lead us to reject or not reject the null hypothesis. We used the t-distribution because our sample sizes are large enough to approximate the normal distribution, yet still rely on the t-distribution due to variability considerations. The critical value derived from this distribution ensures that our hypothesis testing accounts for potential errors in estimation, thus making our conclusions more robust.

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Most popular questions from this chapter

In an experiment to compare bearing strengths of pegs inserted in two different types of mounts, a sample of 14 observations on stress limit for red oak mounts resulted in a sample mean and sample standard deviation of \(8.48 \mathrm{MPa}\) and . \(79 \mathrm{MPa}\), respectively, whereas a sample of 12 observations when Douglas fir mounts were used gave a mean of \(9.36\) and a standard deviation of \(1.52\) ('Bearing Strength of White Oak Pegs in Red Oak and Douglas Fir Timbers," \(J\). of Testing and Evaluation, 1998, 109-114). Consider testing whether or not true average stress limits are identical for the two types of mounts. Compare df's and \(P\)-values for the unpooled and pooled \(t\) tests.

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