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Chapter 10: Problem 98

The brcaking strength of yarn supplicd by two manufacturers is being investigated. You know from experience with the manufacturers' processes that \(\sigma_{1}=5\) psi and \(\sigma_{2}=4\) psi. A random sample of 20 test specimens from each manufacturer results in \(\bar{x}_{1}=88\) psi and \(\bar{x}_{2}=91\) psi, respectively. (a) Using a \(90 \%\) confidence interval on the difference in mean breaking strength, comment on whether or not there is cvidence to support the claim that manufacturer 2 produces yarn with higher mean breaking strength. (b) Using a \(98 \%\) confidence interval on the difference in mean breaking strength, comment on whether or not there is evidence to support the claim that manufacturer 2 produces yarn with higher mean breaking strength. (c) Comment on why the results from parts (a) and (b) are different or the same. Which would you choose to make your decision and why?

Short Answer

Expert verified
Manufacturer 2 has a higher mean breaking strength, supported by both 90% and 98% confidence intervals.

Step by step solution

01

Understand the Problem

We need to calculate the confidence interval for the difference in mean breaking strength between two manufacturers. We have sample means, population variances, and sample sizes for both manufacturers. We will use this information to form confidence intervals at 90% and 98% levels.
02

Determine the Formula

We use the confidence interval for the difference between two population means with known variances:\[CI = \left( (\bar{x}_1 - \bar{x}_2) \mp Z_{\alpha/2} \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} \right)\]where \(Z_{\alpha/2}\) is the critical value from the Z-distribution, \(\sigma_1\) and \(\sigma_2\) are the known standard deviations, and \(n_1\) and \(n_2\) are the sample sizes.
03

Calculate the 90% Confidence Interval

For a 90% confidence interval, \(\alpha = 0.10\), so \(\alpha/2 = 0.05\). The critical value \(Z_{0.05} = 1.645\).\[CI = \left( (88 - 91) \mp 1.645 \sqrt{\frac{5^2}{20} + \frac{4^2}{20}} \right)\]Calculate the standard error:\[\sqrt{\frac{5^2}{20} + \frac{4^2}{20}} = \sqrt{1.25} \approx 1.118\]Calculate the CI bounds:\[-3 \mp 1.645 \times 1.118 = (-3 \mp 1.839)\]Thus, the 90% CI is approximately \((-4.839, -1.161)\).
04

Interpret the 90% Confidence Interval

The 90% confidence interval for the difference in means is \((-4.839, -1.161)\). Since the interval is entirely below zero, it supports the claim that Manufacturer 2 has a higher mean breaking strength.
05

Calculate the 98% Confidence Interval

For a 98% confidence interval, \(\alpha = 0.02\), so \(\alpha/2 = 0.01\). The critical value \(Z_{0.01} = 2.33\).\[CI = \left( (88 - 91) \mp 2.33 \sqrt{\frac{5^2}{20} + \frac{4^2}{20}} \right)\]Using the same standard error \(1.118\) calculated earlier:\[-3 \mp 2.33 \times 1.118 = (-3 \mp 2.605)\]Thus, the 98% CI is approximately \((-5.605, -0.395)\).
06

Interpret the 98% Confidence Interval

The 98% confidence interval for the difference in means is \((-5.605, -0.395)\). Since the interval is entirely below zero, it still supports the claim that Manufacturer 2 has a higher mean breaking strength.
07

Compare and Choose the Result

Both confidence intervals are below zero, indicating Manufacturer 2 has a higher breaking strength. The intervals are different because a 98% CI provides a wider range (more conservative) than a 90% CI due to higher confidence level. I would choose the 98% CI as it offers more confidence in the claim.

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Key Concepts

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

Difference in Means
In statistics, the difference in means focuses on comparing the average values from two different groups. In our case, these groups are the yarns from two manufacturers. We calculate how much one average deviates from the other. This numerical difference gives insight into whether one group consistently performs better or worse than the other. When comparing manufacturer 1 and manufacturer 2, the mean breaking strengths are 88 psi and 91 psi, respectively. Thus, the initial difference in means is 91 - 88 = 3 psi favoring manufacturer 2.
Breaking Strength
Breaking strength refers to the ability of a yarn to withstand pulling forces without breaking. It's a crucial quality measure for yarn manufacturers. Testing this involves applying force to a sample until it breaks, noting the maximum force endured. In our exercise, manufacturer 1's yarn has an average breaking strength of 88 psi, whereas manufacturer 2's is 91 psi. This slight difference might seem negligible, but even small improvements can significantly impact overall material performance.
Z-distribution
The Z-distribution, also known as the standard normal distribution, describes data that clusters around a mean of 0 with a standard deviation of 1. It is particularly useful for hypothesis testing when the population variance is known, as in this exercise. We transform test data into a Z-score, which allows us to determine the probability of observing such data under the normal distribution. For our confidence intervals, critical Z-values (like 1.645 for 90% confidence and 2.33 for 98% confidence) help us understand what range of values is statistically expected if no true difference exists between the two means.
Critical Value
Critical values define the boundaries of a confidence interval, representing the threshold beyond which we consider an observed outcome to be statistically significant. These thresholds differentiate between common random variation and noteworthy deviations that suggest a real effect or difference. Critical values correspond to specific confidence levels, which are associated with specific probabilities in the Z-distribution. In our exercise, critical values of 1.645 and 2.33 were selected for the 90% and 98% confidence intervals, respectively, indicating the level of statistical confidence we have in concluding that manufacturer 2 has superior yarn breaking strength.

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Most popular questions from this chapter

The concentration of active ingredient in a liquid laundry detergent is thought to be affected by the type of catalyst used in the process. The standard deviation of active concentration is known to be 3 grams per liter regardless of the catalyst type. Ten observations on concentration are taken with each catalyst, and the data follow: $$\begin{array}{l}\text { Catalyst } 1: 57.9,66.2,65.4,65.4,65.2,62.6,67.6,63.7, \\\\\quad 67.2,71.0\end{array}$$ Catalyst 2: 66.4,71.7,70.3,69.3,64.8,69.6,68.6,69.4 $$65.3,68.8$$ (a) Find a \(95 \%\) confidence interval on the difference in mean active concentrations for the two catalysts. Find the \(P\) -value. (b) Is there any evidence to indicate that the mean active concentrations depend on the choice of catalyst? Base your answer on the results of part (a). (c) Suppose that the true mean difference in active concentration is 5 grams per liter. What is the power of the test to detect this difference if \(\alpha=0.05 ?\) (d) If this difference of 5 grams per liter is really important, do you consider the sample sizes used by the experimenter to be adequate? Does the assumption of normality seem reasonable for both samples?

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 \(\sigma_{1} / \sigma_{2}\),

Two different formulations of an oxygenated motor fuel are being tested to study their road octane numbers. The variance of road octane number for formulation 1 is \(\sigma_{1}^{2}=1.5,\) and for formulation, 2 it is \(\sigma_{2}^{2}=1.2 .\) Two random samples of size \(n_{1}=15\) and \(n_{2}=20\) are tested, and the mean road octane numbers observed are \(\bar{x}_{1}=89.6\) and \(\bar{x}_{2}=92.5 .\) Assume normality. (a) If formulation 2 produces a higher road octane number than formulation \(1,\) the manufacturer would like to detect it. Formulate and test an appropriate hypothesis using \(\alpha=0.05 .\) What is the \(P\) -value? (b) Explain how the question in part (a) could be answered with a \(95 \%\) confidence interval on the difference in mean road octane number. (c) What sample size would be required in each population if you wanted to be \(95 \%\) confident that the error in estimating the difference in mean road octane number is less than \(1 ?\)

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.

The burning rates of two different solid-fuel propellants used in air crew escape systems are being studied. It is known that both propellants have approximately the same standard deviation of burning rate; that is \(\sigma_{1}=\sigma_{2}=3\) centimeters per second. Two random samples of \(n_{1}=20\) and \(n_{2}=20\) specimens are tested; the sample mean burning rates are \(\bar{x}_{1}=18\) centimeters per second and \(\bar{x}_{2}=24\) centimeters per second. (a) Test the hypothesis that both propellants have the same mean burning rate. Use \(\alpha=0.05 .\) What is the \(P\) -value? (b) Construct a \(95 \%\) confidence interval on the difference in means \(\mu_{1}-\mu_{2} .\) What is the practical meaning of this interval? (c) What is the \(\beta\) -error of the test in part (a) if the true difference in mean burning rate is 2.5 centimeters per second? (d) Assume that sample sizes are equal. What sample size is needed to obtain power of 0.9 at a true difference in means of \(14 \mathrm{~cm} / \mathrm{s} ?\)
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