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Chapter 12: Problem 18

Find the least-squares line for the data. Plot the points and graph the line on the same graph. Does the line appear to provide a good fit to the data points? $$\begin{array}{c|cccccc}x & 1 & 2 & 3 & 4 & 5 & 6 \\\\\hline y & 5.6 & 4.6 & 4.5 & 3.7 & 3.2 & 2.7\end{array}$$

Short Answer

Expert verified
Based on the given data points, we have found the least-squares line with the equation $y = -0.45x + 5.625$. After plotting the data points and the least-squares line on the same graph, we can conclude that the line appears to provide a reasonable fit to the data points. However, to further determine the goodness of fit, we can calculate the coefficient of determination (R-squared value) in a more advanced analysis.

Step by step solution

01

Calculate the mean of x and y values

First, we need to calculate the mean (average) of the x and y values: $$\bar{x} = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5$$ $$\bar{y} = \frac{5.6 + 4.6 + 4.5 + 3.7 + 3.2 + 2.7}{6} = \frac{24.3}{6} = 4.05$$
02

Calculate the slope (m)

Next, we will find the slope (m) of the least-squares line using the following formula: $$m = \frac{\sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})}{\sum_{i=1}^n (x_i - \bar{x})^2}$$ Plugging in the given data points and means of x and y values, we get: $$m = \frac{[(1-3.5)(5.6-4.05) + (2-3.5)(4.6-4.05) + (3-3.5)(4.5-4.05) + (4-3.5)(3.7-4.05) + (5-3.5)(3.2-4.05) + (6-3.5)(2.7-4.05) ]}{[(1-3.5)^2 + (2-3.5)^2 + (3-3.5)^2 + (4-3.5)^2 + (5-3.5)^2 + (6-3.5)^2]}$$ Calculating the numerator and denominator, we get: $$m = \frac{-7.875}{17.5} = -0.45$$
03

Calculate the y-intercept (b)

Now, we will find the y-intercept (b) using the following formula: $$b = \bar{y} - m\bar{x}$$ Plugging in the mean of y values, slope, and mean of x values, we get: $$b = 4.05 - (-0.45)(3.5) = 4.05 + 1.575 = 5.625$$
04

Write the equation of the least-squares line

With values of m and b calculated, we can write the equation of the least-squares line: $$y = -0.45x + 5.625$$
05

Plot the data points and the least-squares line

Plot the given data points and the least-squares line on the same graph using any graphing tool. The graph should show data points and a line that appears to provide a reasonable fit. However, a visual inspection might not capture how well the line fits the data. To determine if the line provides a good fit, we can calculate the coefficient of determination (R-squared value) that measures the proportion of the variance in the y values that is predictable from the x values. But, in the scope of the current exercise, a discussion about the visual appearance of the line and the data points is enough. From the plot, the line appears to provide a reasonable fit to the data points.

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Key Concepts

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

Mean Calculation
Understanding the mean, or average, of a set of data points is crucial when crafting statistical models like the least-squares regression line. The mean captures the central tendency of your data, serving as an anchor point for further calculations. To compute the mean of x and y values, simply add all the x values together, then divide the sum by the number of x values. Repeat the process for the y values.

In the given exercise, the means were calculated as follows: for the x values, you sum up 1, 2, 3, 4, 5, and 6, then divide by the total count which is 6, giving you a mean of 3.5. The y values follow the same pattern, summing up to 24.3 and, divided by 6, result in a mean of 4.05. These averages are fundamental baselines for the upcoming steps.
Slope Determination
The slope of the regression line is a measure of how steeply the line rises or falls as you move along the x-axis. It quantifies the relationship between the x and y variables; in other words, it tells us by how much y changes for a unit change in x. The formula for the slope, denoted as m, involves summing the product of the differences between each x value and the mean of x, and the differences between each corresponding y value and the mean of y. This sum is then divided by the sum of the squares of the differences between each x value and the mean of x.

The provided exercise gives a practical demonstration of this calculation and results in a slope of -0.45. This indicates that for every one unit increase in x, the predicted value of y decreases by 0.45 units, suggesting an inverse relationship between x and y within this dataset.
Y-intercept Calculation
After determining the slope, the y-intercept, or b, is the next crucial element of the least-squares regression line equation. The y-intercept represents the point where the line crosses the y-axis; this is the value of y when x is zero. To find the y-intercept, the product of the slope and the mean of x values is subtracted from the mean of y values.

Our example calculates the y-intercept as follows: by taking the previously computed mean of y (4.05) and subtracting the product of the slope (-0.45) and the mean of x (3.5), yielding a y-intercept of 5.625. This forms the basis for constructing the full equation of the regression line.
Coefficient of Determination
The coefficient of determination, commonly denoted as R-squared, is a statistical measure that represents the proportion of the variance for the dependent variable (y) that's explained by the independent variable (x) in a regression model. It takes a value between 0 and 1, where a higher value indicates a better fit of the line to the data. A value close to 1 suggests that the model explains a large portion of the variance in the response variable, while a value close to 0 indicates the opposite.

In the context of this exercise, the coefficient of determination is not calculated directly. However, it is an essential concept that students should understand to evaluate the strength of the relationship depicted by the regression line. If computed, it would provide a numerical value confirming how well the line fits the data points, supplementing the visual fit observed in the graph.

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Most popular questions from this chapter

A Chemical Experiment A chemist measured SET the peak current generated (in microamperes) DS1205 when a solution containing a given amount of nickel (in parts per billion) is added to a buffer: $$\begin{array}{cc}\hline x=\mathrm{Ni}(\mathrm{ppb}) & y=\text { Peak } \text { Current }(\mathrm{mA}) \\\\\hline 19.1 & .095 \\\38.2 & .174 \\\57.3 & .256 \\\76.2 & .348 \\\95 & .429 \\\114 & .500 \\\131 & .580 \\\150 & .651 \\\170 & .722 \\\\\hline\end{array}$$ a. Use the data entry method for your calculator to calculate the preliminary sums of squares and crossproducts, \(S_{x x}, S_{y},\) and \(S_{x y}\) b. Calculate the least-squares regression line. c. Plot the points and the fitted line. Does the assumption of a linear relationship appear to be reasonable? d. Use the regression line to predict the peak current generated when a solution containing 100 ppb of nickel is added to the buffer. e. Construct the ANOVA table for the linear regression.

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 \(x=2\).

Independent and Dependent Variables Identify which of the two variables in Exercises \(10-14\) is the independent variable \(x\) and which is the dependent variable \(y .\) Number of hours spent studying and grade on a history test.

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. $$\begin{aligned}&\text { Six points have these coordinates: }\\\&\begin{array}{l|llllll}x & 1 & 2 & 3 & 4 & 5 & 6 \\\\\hline y & 9.7 & 6.5 & 6.4 & 4.1 & 2.1 & 1.0\end{array}\end{aligned}$$ a. Find the least-squares line for the data. b. Plot the six points and graph the line. Does the line appear to provide a good fit to the data points? c. Use the least-squares line to predict the value of \(y\) when \(x=3.5\) d. Fill in the missing entries in the MS Exce/ analysis of variance table.

Plot the data points given in Exercises 4-5. Based on the graph, what will be the sign of the correlation coefficient? Then calculate the correlation coefficient, \(r\), and the coefficient of determination, \(r^{2} .\) Is the sign of \(r\) as you expected? $$\begin{array}{l|llllll}x & 1 & 2 & 3 & 4 & 5 & 6 \\\\\hline y & 7 & 5 & 5 & 3 & 2 & 0\end{array}$$
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