/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Consider the following informati... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the following information on ultimate tensile strength (Ib/in) for a sample of \(n=4\) hard zirconium copper wire specimens (from "4 Characterization Methods for Fine Copper Wire," Wire J. Intl., Aug., 1997: 74-80): \(\bar{x}=76,831 \quad s=180 \quad\) smallest \(x_{i}=76,683\) largest \(x_{i}=77,048\) Determine the values of the two middle sample observations (and don't do it by successive guessing!).

Short Answer

Expert verified
The two middle values are approximately 76,796.5 each.

Step by step solution

01

Understand the Parameters

First, we need to understand the given parameters. We have:- Mean of the sample, \( \bar{x} = 76,831 \).- Sample standard deviation, \( s = 180 \).- The smallest observation, \( x_{ ext{min}} = 76,683 \).- The largest observation, \( x_{ ext{max}} = 77,048 \).- Total number of observations, \( n = 4 \).
02

Use the Mean Equation

The mean of a sample is given by \( \bar{x} = \frac{x_1 + x_2 + x_3 + x_4}{n} \). With \( n=4 \), the equation becomes:\[ 76,831 = \frac{76,683 + x_2 + x_3 + 77,048}{4} \]
03

Solve for Sum of Unknowns

Multiply both sides by 4 to solve for the sum of the observations:\[ 4 \times 76,831 = 76,683 + x_2 + x_3 + 77,048 \]This simplifies to:\[ 307,324 = 153,731 + x_2 + x_3 \]
04

Calculate the Sum of Middle Observations

Subtract \(76,683\) (smallest) and \(77,048\) (largest) from both sides of the equation to find the sum of \(x_2 + x_3\):\[ x_2 + x_3 = 307,324 - 153,731 \]\[ x_2 + x_3 = 153,593 \]
05

Use the Variation Information

Recognize that \(x_2\) and \(x_3\), being the middle values, will be reasonably close to the overall mean \(\bar{x}\). Given the minimum and maximum values, the distribution of values suggests that the two middle scores should be balanced around the mean.
06

Estimate the Middle Observations

Since the sum of the middle values is 153,593 and they should distribute around the mean of 76,831, let's assign the middle values symmetrically around the mean, i.e.,\[ x_2 = 76,796.5, \quad x_3 = 76,796.5 \]This division is simplistic for illustration, assuming symmetric balancing between mean and extremities.

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Key Concepts

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

Sample Mean
The concept of a sample mean, often denoted by \( \bar{x} \), is a fundamental idea in statistics. It represents the average of a set of observations in a sample. For example, if you have four measurements of tensile strength from wire specimens, the sample mean is the sum of the measurements divided by the number of specimens. It is calculated as:\[ \bar{x} = \frac{x_1 + x_2 + x_3 + x_4}{n} \]In our scenario, the sample mean provides a central point of data, giving us a summarized view into the tensile strength across multiple specimens. The mean tells us where the center of the data is located, which is crucial for understanding the general trend when dealing with small sample sizes like \( n=4 \). The mean can be quite informative, especially in indicating the average performance of materials, such as hard zirconium copper wire in tensile strength evaluations.In real-world applications, knowing the mean can help engineers and quality assurance professionals in identifying whether a material meets the standardized performance criteria or if there are inconsistencies requiring further investigation.
Standard Deviation
Standard deviation, represented as \( s \), measures the amount of variation or dispersion present in a set of data. In essence, it tells us how spread out the data points are around the mean. To calculate it, you would typically take the square root of the average of the squared differences from the Mean.In the data provided, the standard deviation is 180. This provides insights into the consistency of the tensile strength measurements. A smaller standard deviation would imply that the values are closely packed around the mean, indicating higher consistency. Conversely, a larger standard deviation, like 180 in this case, suggests there is substantial variation from one measurement to another.The standard deviation helps in determining reliability and quality control. If there is too much variation, it might indicate issues with the production process, testing methods, or the material itself. Understanding standard deviation is crucial for deciding tolerance levels and ensuring product reliability, which are both essential in industrial settings.
Symmetric Distribution
Symmetric distribution in statistics refers to a situation in which data points are evenly distributed around a central point, such as the mean. When a dataset has a symmetric distribution, the left side of the distribution mirrors the right side. This means that the data is balanced around the central value.In our sample of tensile strength measurements, the two middle values \(x_2\) and \(x_3\) should ideally balance around the mean because of the symmetric nature implied by the solution. These values being close to the mean ensures the data follows a normal distribution pattern, which is common in natural and manufactured processes.Recognizing the symmetry helps in predicting missing values and estimating unknown parameters with better accuracy. This property of symmetry allows for easier calculations and interpretations, assuming uniformity in the process. Recognizing or assuming symmetry in data can thus simplify analysis and lead to better decision-making, especially when evaluating the quality of materials and predicting future outcomes under similar conditions.

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Most popular questions from this chapter

Many universities and colleges have instituted supplemental instruction (SI) programs, in which a student facilitator meets regularly with a small group of students enrolled in the course to promote discussion of course material and enhance subject mastery. Suppose that students in a large statistics course (what else?) are randomly divided into a control group that will not participate in SI and a treatment group that will participate. At the end of the term, each student's total score in the course is determined. a. Are the scores from the SI group a sample from an existing population? If so, what is it? If not, what is the relevant conceptual population? b. What do you think is the advantage of randomly dividing the students into the two groups rather than letting each student choose which group to join? c. Why didn't the investigators put all students in the treatment group? [Note: The article 'Supplemental Instruction: An Effective Component of Student Affairs Programming" \(U\). of College Student Devel., 1997: 577-586) discusses the analysis of data from several SI programs.]

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The article "A Thin-Film Oxygen Uptake Test for the Evaluation of Automotive Crankease Lubricants" (Lubric. Engr., 1984: 75-83) reported the following data on oxidation-induction time (min) for various commercial oils: \(\begin{array}{lllllllllll}87 & 103 & 130 & 160 & 180 & 195 & 132 & 145 & 211 & 105 & 145\end{array}\) \(\begin{array}{llllllll}153 & 152 & 138 & 87 & 99 & 93 & 119 & 129\end{array}\) a. Calculate the sample variance and standard deviation. b. If the observations were reexpressed in hours, what would be the resulting values of the sample variance and sample standard deviation? Answer without actually performing the reexpression.

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