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Chapter 1: Problem 66

a. For what value of \(c\) is the quantity \(\sum\left(x_{i}-c\right)^{2}\) minimized? [Hint: Take the derivative with respect to \(c\), set equal to 0 , and solve.] b. Using the result of part (a), which of the two quantities \(\sum\left(x_{i}-\bar{x}\right)^{2}\) and \(\sum\left(x_{i}-\mu\right)^{2}\) will be smaller than the other (assuming that \(\bar{x} \neq \mu) ?\)

Short Answer

Expert verified
a. The minimizing value is \(c = \bar{x}\). b. \(\sum(x_i - \bar{x})^2\) is smaller.

Step by step solution

01

Understand the Objective

We are tasked with determining the value of \(c\) that minimizes the quantity \(\sum\left(x_{i}-c\right)^{2}\). This is a common optimization problem where you need to find the value of \(c\) that minimizes the sum of squared deviations from \(c\).
02

Write the Function to Minimize

The function to minimize is given by \(f(c) = \sum(x_i - c)^2\). This involves the sum of squared distances of \(x_i\) from \(c\).
03

Differentiate with Respect to c

To find the minimum, we need to take the derivative of \(f(c)\) with respect to \(c\). The derivative is \(f'(c) = \sum -2(x_i - c) = -2 \sum (x_i - c)\).
04

Set the Derivative to Zero and Solve

Setting the derivative equal to zero gives us \( -2 \sum (x_i - c) = 0\). Simplifying this leads to \(\sum (x_i - c) = 0\), which gives \(c = \bar{x}\), where \(\bar{x}\) is the mean of the data points.
05

Statement of Result for Part (a)

The value of \(c\) that minimizes \(\sum(x_i - c)^2\) is \(c = \bar{x}\), the mean of the data points.
06

Analyze Part (b)

Next, we compare \(\sum(x_i - \bar{x})^2\) and \(\sum(x_i - \mu)^2\). Given \(\bar{x} eq \mu\), and knowing that the sum of squared deviations is minimized at the mean, \(\sum(x_i - \bar{x})^2\) is smaller than \(\sum(x_i - \mu)^2\).
07

Statement of Result for Part (b)

Since \(\bar{x} eq \mu\) and deviations are minimized at the mean, \(\sum(x_i - \bar{x})^2\) will be smaller than \(\sum(x_i - \mu)^2\).

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Key Concepts

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

Sum of Squared Deviations
When we talk about the "Sum of Squared Deviations", we're dealing with how spread out a set of data points are relative to a particular value. This concept is usually used to measure the variability in data.
  • The squared deviations are calculated by taking each data point \(x_i\), subtracting a chosen value \(c\), and squaring the result.
  • The formula for the sum of these squared deviations is \(\sum(x_i - c)^2\).
  • This measure is important because squaring the deviations penalizes larger deviations, giving more weight to outliers.
When we want to minimize this sum, we often compare each data point to some central value, like the mean (average) of the data set. This helps us find the best representation of the data, focusing on minimizing variability around it. By doing so, we identify the mean as the value that results in the smallest possible sum of squared deviations.
Derivative in Optimization
In optimization, derivatives play a crucial role, especially when finding the minimum or maximum of a function. Let's explore how derivatives help in optimizing the "Sum of Squared Deviations".
  • A derivative represents the rate of change of a function with respect to one of its variables.
  • By taking the derivative of the function \(f(c) = \sum(x_i - c)^2\) with respect to \(c\), we measure how changes in \(c\) affect the value of the function.
  • The derivative \(f'(c) = \sum -2(x_i - c) = -2 \sum (x_i - c)\) tells us whether the function is increasing or decreasing as \(c\) changes.
To find the minimum, we set this derivative equal to zero. Mathematically, this means finding where the slope of the function is zero, indicating a potential minimum or maximum point.
  • Solving \(-2\sum(x_i - c) = 0\) simplifies to \(\sum(x_i - c) = 0\).
  • This step confirms that \(c = \bar{x}\), the mean, is the point where the deviations are minimized.
Using derivatives allows us to efficiently identify key points in the optimization process.
Data Analysis
Data analysis involves examining data to extract meaningful insights. In the context of this exercise, we use concepts like the mean to better understand and interpret our data set.
  • One key task in data analysis is summarizing data, often using measures of central tendency such as the mean \(\bar{x}\) and median.
  • Another task is determining how data points deviate from these summary stats, which can be expressed as the variance or standard deviation.
Through this exercise, we understood that minimizing the sum of squared deviations helps in identifying the mean as the most accurate summary measure for any given data set.
  • This is because the mean minimizes overall discrepancies between the actual data points and the central value.
  • Additionally, comparing different sums of squared deviations can reveal insights; for instance, \(\sum(x_i - \bar{x})^2\) being smaller than \(\sum(x_i - \mu)^2\) suggests that the data is centered around the mean rather than any other hypothetical central value.
Overall, data analysis is invaluable in extracting meaningful conclusions, driving decisions, and furthering our understanding of data-driven processes.

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Most popular questions from this chapter

The following data on HC and CO emissions for one particular vehicle was given in the chapter introduction. \(\begin{array}{lllll}H C \text { (g/mile) } & 13.8 & 18.3 & 32.2 & 32.5 \\\ \text { CO }(\text { g/mile }) & 118 & 149 & 232 & 236\end{array}\) a. Compute the sample standard deviations for the HC and CO observations. Does the widespread belief appear to be justified? b. The sample coefficient of variation \(s / \bar{x}\) (or \(100 \cdot s / \bar{x}\) ) assesses the extent of variability relative to the mean. Values of this coefficient for several different data sets can be compared to determine which data sets exhibit more or less variation. Carry out such a comparison for the given data.

The article "A Thin-Fim Oxygen Uptake Test for the Evaluation of Automotive Crankcase Lubricants" (Lubric. Engrg., 1984: 75-83) reported the following data on oxidation-induction time (min) for various commercial oils: \(\begin{array}{lllllllllll}87 & 103 & 130 & 160 & 180 & 195 & 132 & 145 & 211 & 105 & 145 \\ 153 & 152 & 138 & 87 & 99 & 93 & 119 & 129 & & & \end{array}\) 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.

In 1997 a woman sued a computer keyboard manufacturer, charging that her repetitive stress injuries were caused by the keyboard (Genessy v. Digital Equipment Corp.). The jury awarded about \(\$ 3.5\) million for pain and suffering, but the court then set aside that award as being unreasonable compensation. In making this determination, the court identified a "normative" group of 27 similar cases and specified a reasonable award as one within two standard deviations of the mean of the awards in the 27 cases. The 27 awards were (in \(\$ 1000\) s) 37,60 , \(75,115,135,140,149,150,238,290,340\), \(410,600,750,750,750,1050,1100,1139\), \(1150,1200,1200,1250,1576,1700,1825\), and 2000 , from which \(\sum x_{i}=20,179, \sum x_{i}^{2}=\) \(24,657,511\). What is the maximum possible amount that could be awarded under the twostandard- deviation rule?

Calculate and interpret the values of the sample median, sample mean, and sample standard deviation for the following observations on fracture strength (MPa, read from a graph in "Heat-Resistant Active Brazing of Silicon Nitride: Mechanical Evaluation of Braze Joints,"Welding \(J .\), Aug. 1997): \(\begin{array}{llllllllll}87 & 93 & 96 & 98 & 105 & 114 & 128 & 131 & 142 & 168\end{array}\)

The minimum injection pressure (psi) for injection molding specimens of high amylose corn was determined for eight different specimens (higher pressure corresponds to greater processing difficulty), resulting in the following observations (from "Thermoplastic Starch Blends with a Polyethylene-Co-Vinyl Alcohol: Processability and Physical Properties,"Polymer Engrg. \& Sci., 1994: 17-23): \(\begin{array}{llllllll}15.0 & 13.0 & 18.0 & 14.5 & 12.0 & 11.0 & 8.9 & 8.0\end{array}\) a. Determine the values of the sample mean, sample median, and \(12.5 \%\) trimmed mean, and compare these values. b. By how much could the smallest sample observation, currently \(8.0\), be increased without affecting the value of the sample median? c. Suppose we want the values of the sample mean and median when the observations are expressed in kilograms per square inch (ksi) rather than psi. Is it necessary to reexpress each observation in ksi, or can the values calculated in part (a) be used directly? [Hint: \(\mathrm{l} \mathrm{kg}=2.2 \mathrm{lb} .]\)
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