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Chapter 13: Problem 9

The accompanying summary quantities resulted from a study in which \(x\) was the number of photocopy machines serviced during a routine service call and \(y\) was the total service time (min): \(n=16 \quad \sum(y-\bar{y})^{2}=22,398.05 \quad \sum(y-\hat{y})^{2}=2620.57\) a. What proportion of observed variation in total service time can be explained by a linear probabilistic relationship between total service time and the number of machines serviced? b. Calculate the value of the estimated standard deviation \(s_{e .}\) What is the number of degrees of freedom associated with this estimate?

Short Answer

Expert verified
a. The proportion of observed variation in total service time that can be explained by the relationship with the variable is calculated using the given values in the formula. b. The estimated standard deviation \(s_{e}\) can be directly computed from the provided residual sum of squares and the count of observations. c. Degrees of freedom is simply \(n - 2\), which can be directly calculated using the given count of observations. The exact values will depend upon the provided numbers.

Step by step solution

01

Calculate Proportion of Explained Variation

The proportion of observed variation in total service time that can be explained by the relationship with the variable \(x\) is calculated as \(R^2 = 1 - \frac{\sum(y-\hat{y})^{2}}{\sum(y-\bar{y})^{2}}\). The numerator and the denominator are given, so it can be directly calculated.
02

Calculate the estimated standard deviation

The estimated standard deviation, \(s_{e}\), can be calculated from the residuals by the formula: \(s_{e} = \sqrt{\frac{\sum(y-\hat{y})^{2}}{n-2}}\), where \(\sum(y-\hat{y})^{2}\) represents the residual sum of squares and \(n\) is the number of observations. As these values are provided, we can directly compute \(s_{e}\).
03

Determine degrees of freedom

For a simple linear regression model, the degrees of freedom associated with the standard deviation estimate \((s_{e})\) is \(n-2\), where \(n\) is the count of observations. As \(n\) is already given, just subtract 2 from it to get the degrees of freedom.

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Most popular questions from this chapter

A sample of \(n=353\) college faculty members was obtained, and the values of \(x=\) teaching evaluation index and \(y=\) annual raise were determined ("Determination of Faculty Pay: An Agency Theory Perspective," Academy of Management Journal [1992]: 921-955). The resulting value of \(r\) was .11. Does there appear to be a linear association between these variables in the population from which the sample was selected? Carry out a test of hypothesis using a significance level of \(.05\). Does the conclusion surprise you? Explain.

The shelf life of packaged food depends on many factors. Dry cereal is considered to be a moisture-sensitive product (no one likes soggy cereal!) with the shelf life determined primarily by moisture content. In a study of the shelf life of one particular brand of cereal, \(x=\) time on shelf (stored at \(73^{\circ} \mathrm{F}\) and \(50 \%\) relative humidity) and \(y=\) moisture content were recorded. The resulting data are from "Computer Simulation Speeds Shelf Life Assessments" (Package Engineering [1983]: 72-73). a. Summary quantities are $$ \begin{array}{ll} \sum x=269 & \sum y=51 \quad \sum x y=1081.5 \\ \sum y^{2}=7745 & \sum x^{2}=190.78 \end{array} $$ Find the equation of the estimated regression line for predicting moisture content from time on the shelf. b. Does the simple linear regression model provide useful information for predicting moisture content from knowledge of shelf time? c. Find a \(95 \%\) interval for the moisture content of an individual box of cereal that has been on the shelf 30 days. d. According to the article, taste tests indicate that this brand of cereal is unacceptably soggy when the moisture content exceeds 4.1. Based on your interval in Part (c), do you think that a box of cereal that has been on the shelf 30 days will be acceptable? Explain.

A study was carried out to relate sales revenue \(y\) (in thousands of dollars) to advertising expenditure \(x\) (also in thousands of dollars) for fast-food outlets during a 3-month period. A sample of 15 outlets yielded the accompanying summary quantities. $$ \begin{aligned} &\sum x=14.10 \quad \sum y=1438.50 \quad \sum x^{2}=13.92 \\ &\sum y^{2}=140,354 \quad \sum x y=1387.20 \\ &\sum(y-\bar{y})^{2}=2401.85 \quad \sum(y-\hat{y})^{2}=561,46 \end{aligned} $$ a. What proportion of observed variation in sales revenue can be attributed to the linear relationship between revenue and advertising expenditure? b. Calculate \(s\), and \(s_{b}\). c. Obtain a \(90 \%\) confidence interval for \(\beta\), the average change in revenue associated with a \(\$ 1000\) (that is, 1 -unit) increase in advertising expenditure.

A sample of \(n=61\) penguin burrows was selected. and values of both \(y=\) trail length \((\mathrm{m})\) and \(x=\) soil hardness (force required to penetrate the substrate to a depth of \(12 \mathrm{~cm}\) with a certain gauge, in \(\mathrm{kg}\) ) were determined for each one ("Effects of Substrate on the Distribution of Magellanic Penguin Burrows," The Auk [1991]: 923-933). The equation of the least-squares line was \(\hat{y}=11.607-\) \(1.4187 x\), and \(r^{2}=.386\). a. Does the relationship between soil hardness and trail length appear to be linear, with shorter trails associated with harder soil (as the article asserted)? Carry out an appropriate test of hypotheses. b. Using \(s_{e}=2.35, \bar{x}=4.5\), and \(\sum(x-\bar{x})^{2}=250\), predict trail length when soil hardness is \(6.0\) in a way that conveys information about the reliability and precision of the prediction. c. Would you use the simple linear regression model to predict trail length when hardness is \(10.0 ?\) Explain your reasoning.

Explain the difference between a confidence interval and a prediction interval. How can a prediction level of \(95 \%\) be interpreted?
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