/*! 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 9 The accompanying summary quantit... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

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 can be explained by a linear relationship is 88.29%. b. The estimated standard deviation is 13.28 and the degrees of freedom associated with this estimate is 14.

Step by step solution

01

Calculate \(R^2\)

The coefficient of determination \(R^2\) is found by 1 - (\(\sum(y-\hat{y})^{2}\) / \(\sum(y-\bar{y})^{2}\)). Substituting the given values in, we get1 - (2620.57 / 22398.05) = 0.8829 or 88.29%.
02

Calculate estimated standard deviation \(s_{e}\)

The estimate for standard deviation can be found by taking the square root of SSE/(n-2), where n represents the number of observations. So we get\(\sqrt{2620.57 / (16-2)} = 13.28\).
03

Determine degrees of freedom

The degrees of freedom in this case would be \(n-2\), due to us estimating both intercept and the slope in a simple linear regression, thus using up two degrees of freedom. Therefore, the degrees of freedom in this instance is \(16-2=14\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Coefficient of Determination
The Coefficient of Determination, commonly denoted as \( R^2 \), is a key metric in regression analysis that helps us understand how well data fits a statistical model. In simpler terms, \( R^2 \) tells us what fraction of the variance in the dependent variable can be explained by the independent variable in the model. This value ranges from 0 to 1.

A value of \( R^2 = 0 \) suggests that the model does not explain any of the variability in the response data around its mean. On the other hand, an \( R^2 = 1 \) means that the model perfectly explains all the variability. In the given exercise, the \( R^2 \) value calculated is 0.8829, or 88.29%.

This implies that 88.29% of the observed variation in the total service time can be explained by the number of photocopy machines serviced. The closer \( R^2 \) is to 1, the better the model explains the variation.
  • Formula: \( R^2 = 1 - \frac{SS_{res}}{SS_{tot}} \)
  • \( SS_{res} = \sum(y - \hat{y})^2 \) (Sum of squares due to error)
  • \( SS_{tot} = \sum(y - \bar{y})^2 \) (Total sum of squares)
Standard Deviation in Regression
The Standard Deviation in Regression, denoted as \( s_e \), provides an estimate of the typical distance that the observed values fall from the regression line. It essentially measures the spread of the residuals, helping us understand the model's accuracy.

In the context of our exercise, the estimated standard deviation \( s_e \) was calculated to be 13.28 minutes. A smaller \( s_e \) value indicates that the data points are closely clustered around the regression line, which means the model's predictions are relatively accurate.

The formula used to find \( s_e \) is:
  • \( s_e = \sqrt{\frac{SS_{res}}{n-2}} \)
  • Where \( SS_{res} = \sum(y - \hat{y})^2 \) is the sum of squares due to errors (2620.57 in this case)
  • \( n \) represents the number of observations

By using the formula, we gain insight into how much the actual service time data fluctuates around the predicted values by the model.
Degrees of Freedom in Regression
Degrees of Freedom (DF) in Regression convey the number of independent quantities that can vary in the calculation of a statistic. This is important because it affects the variability and reliability of our estimations.

In the study case mentioned, we have a simple linear regression, which involves estimating two parameters: the slope and the y-intercept. Due to these estimations, two degrees of freedom are used, and so the degrees of freedom for the error term is \( n - 2 \).

Thus, for the given number of observations \( n = 16 \), the degrees of freedom is 14:
  • DF formula: \( DF = n - p \)
  • Where \( p = 2 \) (the number of parameters estimated: slope and intercept)

Degrees of freedom play a critical role in various statistical tests, including the determination of confidence intervals and hypothesis testing.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Occasionally an investigator may wish to compute a confidence interval for \(\alpha\), the \(y\) intercept of the true regression line, or test hypotheses about \(\alpha .\) The estimated \(y\) intercept is simply the height of the estimated line when \(x=0\), since \(a+b(0)=a .\) This implies that \(s_{a}\) the estimated standard deviation of the statistic \(a\), results from substituting \(x^{\prime \prime}=0\) in the formula for \(s_{a+b x^{+}} .\) The desired confidence interval is then \(a \pm(t\) critical value \() s_{a}\) \(-\) and a test statistic is 1 $$ t=\frac{a-\text { hypothesized value }}{s_{a}} $$ a. The article "Comparison of Winter-Nocturnal Geostationary Satellite Infrared-Surface Temperature with Shelter-Height Temperature in Florida" (Remote Sensing of the Emvironment \([1983]: 313-327\) ) used the simple linear regression model to relate surface temperature as measured by a satellite \((y)\) to actual air temperature \((x)\) as determined from a thermocouple placed on a traversing vehicle. Selected data are given (read from a scatterplot in the article). \(\begin{array}{rrrrrrrr}x & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ y & -3.9 & -2.1 & -2.0 & -1.2 & 0.0 & 1.9 & 0.6 \\ x & 5 & 6 & 7 & & & & \\ y & 2.1 & 1.2 & 3.0 & & & & \end{array}\) Estimate the true regression line. b. Compute the estimated standard deviation \(s_{a}\). Carry out a test at level of significance \(.05\) to see whether the \(y\) intercept of the true regression line differs from zero. c. Compute a \(95 \%\) confidence interval for \(\alpha\). Does the result indicate that \(\alpha=0\) is plausible? Explain.

'The article "Photocharge Effects in Dye Sensitized \(\mathrm{Ag}[\mathrm{Br}, \mathrm{I}]\) Emulsions at Millisecond Range Exposures" (Photographic Science and Engineering [1981]: \(138-144\) ) gave the accompanying data on \(x=\%\) light absorption and \(y=\) peak photovoltage. \(\begin{array}{rrrrrrrrrr}x & 4.0 & 8.7 & 12.7 & 19.1 & 21.4 & 24.6 & 28.9 & 29.8 & 30.5 \\ y & 0.12 & 0.28 & 0.55 & 0.68 & 0.85 & 1.02 & 1.15 & 1.34 & 1.29\end{array}\) \(\sum x=179.7 \quad \sum x^{2}=4334.41\) \(\sum y=7.28 \quad \sum y^{2}=7.4028 \quad \sum x y=178.683\) a. Construct a scatterplot of the data. What does it suggest? b. Assuming that the simple linear regression model is appropriate, obtain the equation of the estimated regression line. c. How much of the observed variation in peak photovoltage can be explained by the model relationship? d. Predict peak photovoltage when percent absorption is 19.1, and compute the value of the corresponding residual. e. The authors claimed that there is a useful linear relationship between the two variables. Do you agree? Carry out a formal test. f. Give an estimate of the average change in peak photovoltage associated with a \(1 \%\) increase in light absorption. Your estimate should convey information about the precision of estimation. g. Give an estimate of true average peak photovoltage when percentage of light absorption is 20 , and do so in a way that conveys information about precision.

Exercise \(5.48\) described a regression situation in which \(y=\) hardness of molded plastic and \(x=\) amount of time elapsed since termination of the molding process. Summary quantities included \(n=15\), SSResid \(=\) \(1235.470\), and SSTo \(=25,321.368\). a. Calculate a point estimate of \(\sigma\). On how many degrees of freedom is the estimate based? b. What percentage of observed variation in hardness can be explained by the simple linear regression model relationship between hardness and elapsed time?

The accompanying data on \(x=\) advertising share and \(y=\) market share for a particular brand of cigarettes during 10 randomly selected years are from the article "Testing Alternative Econometric Models on the Existence of Advertising Threshold Effect" (Journal of Marketing Research \([1984]: 298-308)\). \(\begin{array}{lllllllllll}x & .103 & .072 & .071 & .077 & .086 & .047 & .060 & .050 & .070 & .052\end{array}\) \(\begin{array}{rlllllllll}y & .135 & .125 & .120 & .086 & .079 & .076 & .065 & .059 & .051 & .039\end{array}\) a. Construct a scatterplot for these data. Do you think the simple linear regression model would be appropriate for describing the relationship between \(x\) and \(y ?\) b. Calculate the equation of the estimated regression line and use it to obtain the predicted market share when the advertising share is . 09 . c. Compute \(r^{2}\). How would you interpret this value? d. Calculate a point estimate of \(\sigma .\) On how many degrees of freedom is your estimate based?

Exercise \(5.46\) presented data on \(x=\) squawfish length and \(y=\) maximum size of salmonid consumed, both in \(\mathrm{mm}\). Use the accompanying MINITAB output along with the values \(\bar{x}=343.27\) and \(S_{x x}=69,112.18\) to answer the following questions. The regression equation is Size \(=-89.1=0.729\) length \(\begin{array}{lrrrr}\text { Predictor } & \text { Coef } & \text { Stdev } & \text { t-ratio } & p \\ \text { Constant } & 89.09 & 16.83 & 5.29 & 0.000 \\\ \text { length } & 0.72907 & 0.04778 & 15.26 & 0.000 \\ s=12.56 & R-s q=96.3 \% & R-s q(a d j)=95.9 \% \text { Analysis of } V a\end{array}\) Variance \(\begin{array}{lrrrrr}\text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { p } \\ \text { Regression } & 1 & 36736 & 36736 & 232.87 & 0.000 \\ \text { Error } & 9 & 1420 & 158 & & \\\ \text { Total } & 10 & 38156 & & & \end{array}\) a. Does there appear to be a useful linear relationship between length and size? b. Does it appear that the average change in maximum size associated with a 1 -mm increase in length is less than \(.8 \mathrm{~mm}\) ? State and test the appropriate hypotheses. c. Estimate average maximum size when length is 325 \(\mathrm{mm}\) in a way that conveys information about the precision of estimation. d. How would the estimate when length is \(250 \mathrm{~mm}\) compare to the estimate of Part (c)? Answer without actually calculating the new estimate.
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.