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Chapter 12: Problem 8

Professor Isaac Asimov was one of the most prolific writers of all time. Prior to his death, he wrote nearly 500 books during a 40-year career. In fact, as his career progressed, he became even more productive in terms of the number of books written within a given period of time. \({ }^{1}\) The data give the time in months required to write his books in increments of 100 : $$ \begin{array}{l|ccccc} \text { Number of Books, } x & 100 & 200 & 300 & 400 & 490 \\ \hline \text { Time in Months, } y & 237 & 350 & 419 & 465 & 507 \end{array} $$ a. Assume that the number of books \(x\) and the time in months \(y\) are linearly related. Find the least-squares line relating \(y\) to \(x\). b. Plot the time as a function of the number of books written using a scatterplot, and graph the leastsquares line on the same paper. Does it seem to provide a good fit to the data points? c. Construct the ANOVA table for the linear regression.

Short Answer

Expert verified
b. Does the least-squares line appear to provide a good fit to the data points? c. What is the F-value in the ANOVA table for the linear regression?

Step by step solution

01

Calculate the least-squares line

To find the least-squares line, we need to compute the slope and the y-intercept. The formula for the slope, \(m\), is given by: \(m = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2}\) For the intercept, \(b\), the formula is: \(b = \frac{\sum y - m\sum x}{n}\) Calculate the required sums: \(\sum x = 100 + 200 + 300 + 400 + 490 = 1490\) \(\sum y = 237 + 350 + 419 + 465 + 507 = 1978\) \(\sum xy = (100\cdot237) + (200\cdot350) + (300\cdot419) + (400\cdot465) + (490\cdot507) = 353270\) \(\sum x^2 = 100^2 + 200^2 + 300^2 + 400^2 + 490^2 = 486900\) Now, plug in the values into the slope and intercept formulas: \(m = \frac{5\cdot353270 - 1490\cdot1978}{5\cdot486900 - (1490)^2} = 1.014\) \(b = \frac{1978 - 1.014\cdot1490}{5} = 50.616\) So, the least-squares line is given by the equation \(y = 1.014x + 50.616\).
02

Plot the scatterplot and least-squares line

Using any graphing software (e.g., Excel, Desmos, or GeoGebra), plot the given data points and the least-squares line \(y = 1.014x + 50.616\). Remember to label the axes with the appropriate variables and units.
03

Determine if the line provides a good fit

Once the scatterplot and the least-squares line are graphed, analyze the graph by observing how closely the line follows the data points. If the line appears to pass through or near the data points, it indicates a good fit. Assessing the goodness of the fit may be subjective but can provide a useful indication of how well the line represents the data.
04

Construct the ANOVA table

Constructing an ANOVA table requires several steps, including calculating the Sum of Squares Regression (SSR), Sum of Squares Error (SSE), Sum of Squares Total (SST), and Mean Squares (MS) for regression and error. Then, calculate the F-value. Note: There are 5 sample points, so the degrees of freedom for the Regression (dfR) is 1, and for Error (dfE) = (5 - 2) = 3 . 1. Calculate the mean of \(y\) values: \(\bar{y} = \frac{\sum y}{n} = \frac{1978}{5} = 395.6\) 2. Calculate the Sum of Squares Regression (SSR), Sum of Squares Error (SSE), and Sum of Squares Total (SST): \(SSR = \sum(y_{predicted} - \bar{y})^2 = \sum(1.014x + 50.616 - 395.6)^2\) \(SSE = \sum(y_{observed} - y_{predicted})^2 = \sum(y - (1.014x + 50.616))^2\) \(SST = \sum(y_{observed} - \bar{y})^2 = \sum(y - 395.6)^2\) 3. Calculate the Mean Squares (MS) for Regression and Error: \(MS_{R} = \frac{SSR}{dfR} = \frac{SSR}{1}\) \(MS_{E} = \frac{SSE}{dfE} = \frac{SSE}{3}\) 4. Calculate the F-value: Note that the F-value represents the ratio between the MS Regression and MS Error. \(F = \frac{MS_{R}}{MS_{E}}\) 5. Organize the values in the ANOVA table: | Source | df | Sum of Squares | Mean Square | F | |----------------|----|---------------|-------------|---------| | Regression | 1 | SSR | MS_R | F-value | | Error | 3 | SSE | MS_E | | | Total | 4 | SST | | | Substitute the calculated values into the ANOVA table to complete it.

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Key Concepts

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

Least-squares line
The least-squares line is a central concept in linear regression analysis. It is the line that minimizes the sum of the squared vertical distances (residuals) from the data points to the line itself. This method ensures the best possible fit through the scatter of points in a manner that is most 'fair' to all points.

To determine the least-squares line, we calculate the slope (\(m\)) and the y-intercept (\(b\)) using specific formulas. The slope indicates the change in the dependent variable (in our exercise, the time in months) for each unit change in the independent variable (the number of books). A higher slope means a steeper line, reflecting a greater change in time per book written. The y-intercept is the expected value of the dependent variable when the independent variable is zero. In the context of our problem, the y-intercept indicates an estimate for how long writing would take if Professor Asimov had written no books 鈥� a hypothetical scenario.

For example, if we have the slope (\(m\)) as 1.014 and the y-intercept (\(b\)) as 50.616, then our least-squares line equation would be written as \(y = 1.014x + 50.616\). When applying these calculations to the real-world trend in Asimov's productivity, this line allows us to estimate the month required for any number of books.
Scatterplot
A scatterplot is an essential visualization tool in statistics, used to represent the relationship between two variables. Each point on the scatterplot corresponds to one observation in the dataset. The position of a point is determined by the values of the two variables: one variable is represented along the x-axis, and the other along the y-axis.

In our Asimov example, if we were to create a scatterplot, we would plot the number of books on the x-axis and the time in months (\(y\)) on the y-axis. We would have points at coordinates representing each combination of books written and months required. Upon plotting these points, which represent actual data, we also graph the least-squares line obtained from our calculations. If the model is a good fit, most points should be close to this line. However, it is normal for some points to diverge due to natural variability in real-world data.
ANOVA table

Understanding the ANOVA Table

Analysis of variance (ANOVA) is a technique that allows us to compare the mean differences between groups and determine if any of those differences are statistically significant. When applied to regression, the ANOVA table breaks down the total variability of the data into two parts: variation that the model explains and variation due to random error.

The ANOVA table has several components, including the Degrees of Freedom (df), Sum of Squares (SS), Mean Square (MS), and the F-statistic. Degrees of Freedom represent the number of independent pieces of information used in the calculation. Sum of Squares measures the total variation and is split into the Regression Sum of Squares (SSR), which measures how much of the data's movement is explained by the line, and the Residual (Error) Sum of Squares (SSE), which measures the movement of the data around the line. The Mean Square is the Sum of Squares divided by the respective Degrees of Freedom.

Finally, the F-statistic measures the ratio of the variance explained by the model to the variance due to error. A large F-value suggests that the model is a good fit for the data as it can explain a significant amount of variation. Constructing an ANOVA table is an integral part of analyzing regression models, as seen in our exercise.
Probability and Statistics

Exploring Probability and Statistics

Probability and statistics are mathematical fields that quantify uncertainty and analyze data, respectively. Probability dives into the likelihood of various outcomes occurring, while statistics harnesses this and other principles to collect, analyze, interpret, and present empirical data.

In the context of linear regression, statistical methods are used to create models, make inferences, and check the validity of those models. In our exercise, we use statistical measures to fit a least-squares line to a dataset, plot this relationship in a scatterplot, and use an ANOVA table to determine the fit's quality. All of these actions are grounded in probability and statistics. Understanding these concepts helps us interpret the results of regression analysis and draw meaningful conclusions about the relationship between the variables in question.

For example, the slope鈥檚 significance in the least-squares line tells us if the change in the number of books is significantly associated with the time taken, which is a question of inference. Meanwhile, principles from probability aid in understanding concepts like the expectations inherent in regression coefficients. Together, probability and statistics are the backbone of good decision-making with data.

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Most popular questions from this chapter

Some varieties of nematodes, roundworms that live in the soil and frequently are so small as to be invisible to the naked eye, feed on the roots of lawn grasses and other plants. This pest, which is particularly troublesome in warm climates, can be treated by the application of nematicides. Data collected on the percent kill of nematodes for various rates of application (dosages given in pounds per acre of active ingredient) are as follows: $$ \begin{array}{l|l|l|l|l} \text { Rate of Application, } x & 2 & 3 & 4 & 5 \\ \hline \text { Percent Kill, } y & 50,56,48 & 63,69,71 & 86,82,76 & 94,99,97 \end{array} $$ Use an appropriate computer printout to answer these questions: a. Calculate the coefficient of correlation \(r\) between rates of application \(x\) and percent kill \(y\) b. Calculate the coefficient of determination \(r^{2}\) and interpret. c. Fit a least-squares line to the data. d. Suppose you wish to estimate the mean percent kill for an application of 4 pounds of the nematicide per acre. What do the diagnostic plots generated by MINITAB tell you about the validity of the regression assumptions? Which assumptions may have been violated? Can you explain why?

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Athletes and others suffering the same type of injury to the knee often require anterior and posterior ligament reconstruction. In order to determine the proper length of bone-patellar tendonbone grafts, experiments were done using three imaging techniques to determine the required length of the grafts, and these results were compared to the actual length required. A summary of the results of a simple linear regression analysis for each of these three methods is given in the following table. \({ }^{15}\) $$ \begin{array}{llrcc} \text { Imaging Technique } & \text {Coeffcient of Determination, } r^{2} & \text { Intercept } & \text { Slope } & p \text { -value } \\ \hline \text { Radiographs } & 0.80 & -3.75 & 1.031 & <0.0001 \\ \text { Standard MRI } & 0.43 & 20.29 & 0.497 & 0.011 \\ \text { 3-dimensional MRI } & 0.65 & 1.80 & 0.977 & <0.0001 \end{array} $$ a. What can you say about the significance of each of the three regression analyses? b. How would you rank the effectiveness of the three regression analyses? What is the basis of your decision? c. How do the values of \(r^{2}\) and the \(p\) -values compare in determining the best predictor of actual graft lengths of ligament required?

What diagnostic plot can you use to determine whether the data satisfy the normality assumption? What should the plot look like for normal residuals?

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