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Use the data given in Exercises 6-7 (Exercises 17-18, Section 12.1). Construct the ANOVA table for a simple linear regression analysis, showing the sources, degrees of freedom, sums of squares, and mean sauares. $$\begin{array}{l|llllll}x & 1 & 2 & 3 & 4 & 5 & 6 \\\\\hline y & 5.6 & 4.6 & 4.5 & 3.7 & 3.2 & 2.7\end{array}$$

Short Answer

Expert verified
Answer: The Sum of Squares Regression (SSR) is 8.709959, the Sum of Squares Error (SSE) is 0.141841, and the Mean Squares Error/Residual (MSE) is 0.03546.

Step by step solution

01

Compute the estimated regression equation

First, let's calculate the means of x and y. Using these means, calculate the Pearson correlation coefficient between \(x\) and \(y\), \(b_1\), and \(b_0\). Using these values, we can determine the estimated linear regression equation: \(ŷ = b_0 + b_1x\). Mean of \(x\): \(\bar{x} = \frac{1+2+3+4+5+6}{6} = 3.5\) Mean of \(y\): \(\bar{y} = \frac{5.6+4.6+4.5+3.7+3.2+2.7}{6} = 4.05\) Pearson correlation coefficient, \(r = -0.986187\) \(b_1 = r\frac{S_y}{S_x} = -0.986187\frac{\sqrt{1.309392}}{\sqrt{2.916}}=-1.247\) \(b_0 = \bar{y} - b_1\bar{x} = 4.05 - (-1.247)(3.5) = 8.415\) Estimated Regression Equation: \(\displaystyle ŷ = 8.415 - 1.247x\)
02

Calculate the Sum of Squares

Calculate the Sum of Squares Regression (SSR), Sum of Squares Error (SSE), and Sum of Squares Total (SST). 1. \(SSR = \sum_{i=1}^{6}(ŷ_i- \bar{y})^2\) 2. \(SSE = \sum_{i=1}^{6}(y_i- ŷ_i)^2\) 3. \(SST = \sum_{i=1}^{6}(y_i- \bar{y})^2\) \(SSR = 8.709959\) \(SSE = 0.141841\) \(SST = 8.8518\)
03

Compute the Degrees of Freedom

For Regression: \(DF_{Regression} = 1\) (one independent variable) For Errors/Residuals: \(DF_{Errors} = n - 2 = 6 - 2 = 4\) Total: \(DF_{Total} = n - 1 = 6 - 1 = 5\)
04

Calculate Mean Squares

Mean Squares Regression (MSR) = \(\frac{SSR}{DF_{Regression}} = \frac{8.709959}{1} = 8.709959\) Mean Squares Error/Residual (MSE) = \(\frac{SSE}{DF_{Errors}} = \frac{0.141841}{4} = 0.03546\)
05

Construct the ANOVA table

Below is the complete ANOVA table for the given problem: $$\begin{array}{l|l|l|l|l}Source & DF & Sum\ of\ Squares & Mean\ Square \\\\ \hline Regression & 1 & 8.709959 & 8.709959 \\\\ Error/Residual & 4 & 0.141841 &0.03546 \\\\ \hline Total & 5 & 8.8518 & \end{array} $$

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Key Concepts

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

Pearson Correlation Coefficient
The Pearson correlation coefficient, denoted as \(r\), measures the strength and direction of the linear relationship between two variables on a scatterplot. The value of \(r\) ranges from -1 to 1, where -1 indicates a perfect negative linear correlation, 1 indicates a perfect positive linear correlation, and 0 means no linear correlation exists.

In the context of simple linear regression, the Pearson correlation coefficient is used to quantify how well the independent variable predicts the dependent variable. Calculating \(r\) Involves comparing each pair of values for the two variables. A higher absolute value of \(r\) signifies a stronger linear relationship. As demonstrated in our exercise solution, a very high or low Pearson coefficient (such as -0.986) indicates that the regression line is a good fit for the data points.
Sum of Squares
Understanding the sum of squares is essential for analyzing variance in data. The sum of squares measures the total variation within a set of data points, broken down into components related to the regression and the residuals (errors).

There are three types of sum of squares:
  • Sum of Squares Total (SST), which quantifies the total variance in the dependent variable,
  • Sum of Squares Regression (SSR), which measures the variation explained by the regression line, and
  • Sum of Squares Error (SSE), which represents the unexplained variation or the variation due to error.
The SST is the sum of the SSR and SSE, which follows from the principle that the total variation is the sum of the explained variation and the unexplained variation.
Degrees of Freedom
Degrees of Freedom (DF) in statistics refers to the number of values that are free to vary given certain constraints or the amount of independent information within a dataset. Calculating degrees of freedom is crucial for various statistical tests to ensure accurate results.

In a simple linear regression analysis, the degrees of freedom are split between the regression and the error component. Specifically:
  • The DF for Regression is typically 1 because there is only one independent variable,
  • while the DF for Errors is one less than the number of observations minus the number of parameters estimated (usually \(n-2\)).
In our provided exercise, with 6 data points and estimating two parameters (the intercept and slope), the residual degrees of freedom would be 4. The total degrees of freedom is \(n - 1\), which accounts for the overall variability in the dataset.
Mean Squares
Mean Squares represent the average of the squares of the deviations (differences) and are a fundamental part of variance analysis in an ANOVA table. These are obtained by dividing the sum of squares by their corresponding degrees of freedom.

The Mean Squares are of two kinds:
  • Mean Square Regression (MSR), which represents the average variation explained by the independent variable(s), and
  • Mean Square Error (MSE), which represents the average variation that is not explained by the regression model.
The larger the MSR compared to the MSE, the more significant is the regression. If these mean squares are used in an F-test, it can determine whether the regression model provides a better fit to the data than a model that does not include the independent variable.

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Most popular questions from this chapter

Use the data set below to answer the questions. $$ \begin{array}{r|rrrrr} x & -2 & -1 & 0 & 1 & 2 \\ \hline y & 1 & 1 & 3 & 5 & 5 \end{array} $$ Find a \(90 \%\) prediction interval for some value of \(y\) to be observed in the future when \(x=1\).

Use the data set and the MINITAB output (Exercise I8, Section 12.1) below to answer the questions. $$ \begin{array}{l|llllll} x & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline y & 5.6 & 4.6 & 4.5 & 3.7 & 3.2 & 2.7 \end{array} $$ Find a \(95 \%\) prediction interval for some value of \(y\) to be observed in the future when \(x=2\).

The following data were obtained in an experiment relating the dependent variable, \(y\) (texture of strawberries), with \(x\) (coded storage temperature). $$\begin{array}{l|rrrrr}x & -2 & -2 & 0 & 2 & 2 \\\\\hline y & 4.0 & 3.5 & 2.0 & 0.5 & 0.0\end{array}$$ a. Find the least-squares line for the data. b. Plot the data points and graph the least-squares line as a check on your calculations. c. Construct the ANOVA table.

Of two personnel evaluation methods, the first requires a two-hour test interview while the second can be completed in less than an hour. The scores for each of the 15 individuals who took both tests are given in the next table. $$\begin{array}{ccc}\hline \text { Applicant } & \text { Test } 1(x) & \text { Test } 2(y) \\\\\hline 1 & 75 & 38 \\\2 & 89 & 56 \\\3 & 60 & 35 \\\4 & 71 & 45 \\\5 & 92 & 59 \\\6 & 105 & 70 \\\7 & 55 & 31 \\\8 & 87 & 52 \\\9 & 73 & 48 \\\10 & 77 & 41\end{array}$$ $$\begin{array}{ccc}\hline \text { Applicant } & \text { Test } 1(x) & \text { Test } 2(y) \\\\\hline 11 & 84 & 51 \\\12 & 91 & 58 \\\13 & 75 & 45 \\\14 & 82 & 49 \\\15 & 76 & 47 \\\\\hline\end{array}$$ a. Construct a scatterplot for the data. Does the assumption of linearity appear to be reasonable? b. Find the least-squares line for the data. c. Use the regression line to predict the score on the second test for an applicant who scored 85 on Test 1 . d. Construct the ANOVA table for the linear regression relating \(y\) to \(x\).

The data points given in Exercises \(6-7\) were formed by reversing the slope of the lines in Exercises 4 - 5. Plot the points on graph paper and calculater and \(r^{2}\). Notice the change in the sign of \(r\) and the relationship between the values of \(r^{2}\) compared to Exercises \(4-5 .\) By what percentage was the sum of squares of deviations reduced by using the least-squares predictor \(\hat{y}=a+b x\) rather than \(\bar{y}\) as a predictor of \(y\) ? $$\begin{array}{l|llllll}x & 1 & 2 & 3 & 4 & 5 & 6 \\\\\hline y & 0 & 2 & 3 & 5 & 5 & 7\end{array}$$

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