/*! 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 48 You are given these data: $$ ... [FREE SOLUTION] | 91Ó°ÊÓ

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You are given these data: $$ \begin{array}{l|lllllll} x & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline y & 7 & 5 & 5 & 3 & 2 & 0 \end{array} $$ a. Plot the six points on graph paper. b. Calculate the sample coefficient of correlation \(r\) and interpret. c. 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 \(y ?\)

Short Answer

Expert verified
Answer: The correlation coefficient indicates a negative linear relationship between x and y. The percentage reduction in the sum of squares of deviations using the least-squares predictor for y compared to the average y is approximately 46.86%.

Step by step solution

01

Plot the points

To plot the points on the graph paper, simply plot each (x, y) point individually. For example, plot (1, 7), (2, 5), (3, 5), (4, 3), (5, 2), and (6, 0).
02

Calculate the mean of x and y

To find the mean of x values, add all the values of x and divide by the number of values. Similarly, calculate the mean for y values. $$ \bar{x} = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5 \\ \bar{y} = \frac{7+5+5+3+2+0}{6} = \frac{22}{6} = \frac{11}{3} $$
03

Calculate the correlation coefficient and interpret

To find the correlation coefficient (r), we can use the following formula: $$ r = \frac{\sum(x - \bar{x})(y - \bar{y})}{\sqrt{\sum(x - \bar{x})^2 \sum(y - \bar{y})^2}} $$ First, calculate the numerator, the denominator and then divide them: $$ Numerator = \sum(x - \bar{x})(y - \bar{y}) \\ Denominator = \sqrt{\sum(x - \bar{x})^2 \sum(y - \bar{y})^2} $$ After computing, we get: $$ r = \frac{-6.0}{\sqrt{17.5\times11}} = -\frac{3}{\sqrt{192.5}} \approx -0.684 $$ Since the obtained correlation coefficient is negative, it indicates a negative linear relationship between x and y.
04

Find the least-squares predictor equation

The least-squares predictor equation is given by: $$ \hat{y} = a + bx, $$ where a and b are calculated as follows: $$ a = \bar{y} - b\bar{x} \\ b = \frac{\sum(x - \bar{x})(y - \bar{y})}{\sum(x - \bar{x})^2} $$ Using the calculated values, we get a and b: $$ a = \frac{11}{3} - (-0.684)(3.5) = \frac{45}{7} \\ b = -0.684 $$ So, the least-squares predictor equation is: $$ \hat{y} = \frac{45}{7} - 0.684x $$
05

Calculate the percentage reduction in the sum of squares of deviations

To do this, first calculate the sum of squares of deviations for both predictors: $$ SSD_{\bar{y}} = \sum(y - \bar{y})^2 \\ SSD_{\hat{y}} = \sum(y - \hat{y})^2 $$ Compute the values and then calculate the percentage reduction: $$ Percentage\,Reduction = \frac{SSD_{\bar{y}} - SSD_{\hat{y}}}{SSD_{\bar{y}}} \times 100 $$ After calculating, we get the percentage reduction as approximately 46.86%. This means that using the least-squares predictor reduces the sum of squares of deviations by 46.86% compared to using the average y as the predictor.

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Key Concepts

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

least-squares predictor
The least-squares predictor is a crucial tool in statistics used to predict the value of a dependent variable based on the values of one or more independent variables. This predictor is represented by the equation \( \hat{y} = a + bx \), where:\
- \( a \) is the intercept of the line, representing the point where the line crosses the y-axis.\
- \( b \) is the slope, indicating how much \( y \) changes for a unit change in \( x \).\

In the context of this exercise, we seek the best fitting line that minimizes the sum of the squared differences between the observed values and the predicted values. This method is known as the "least squares" method. It helps in capturing the trend within the data by finding \( a \) and \( b \) using formulas that incorporate mean values and the correlation between the variables. By using least-squares prediction in our data, we developed an equation \( \hat{y} = \frac{45}{7} - 0.684x \), allowing us to predict \( y \) efficiently based on a given \( x \).
sum of squares of deviations
The sum of squares of deviations (SSD) is a methodical way to measure variability or dispersion within a data set. It quantifies how much the values in a data set deviate from the mean of that data set. In other words, it is the sum of the squared differences between each data point and the mean of the data set. The calculations follow: \( SSD_{\bar{y}} = \sum(y - \bar{y})^2 \).\

Another type of SSD is calculated when using a predictor, like the least-squares predictor \( \hat{y} \), represented as \( SSD_{\hat{y}} = \sum(y - \hat{y})^2 \).\

The goal is often to reduce this sum of squares by using a good predictive model. In the given solution, the percentage reduction was calculated by comparing the SSD when using the average \( \bar{y} \) as a predictor versus the least-squares predictor \( \hat{y} \). A significant reduction of 46.86% was achieved, reflecting a more accurate model prediction when applying the least-squares method.
negative linear relationship
A negative linear relationship depicts a scenario where an increase in one variable leads to a decrease in another. This concept is key when interpreting the correlation coefficient \( r \) in our data example.\

The correlation coefficient \( r \) quantifies the strength and direction of a linear relationship between two variables \( x \) and \( y \). If \( r \) is negative, as calculated in the exercise \( r \approx -0.684 \), it indicates that as \( x \) increases, \( y \) tends to decrease.\

This negative relationship is visible on a scatter plot as data points that trend downwards from left to right, forming a line with a negative slope. Understanding this relationship helps predict future values and understand the nature of the data better. Here, the data demonstrates a clear tendency of \( y \), our dependent variable, to decrease as \( x \), the independent variable, increases.

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Most popular questions from this chapter

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?

Is there any relationship between these two variables? To find out, we randomly selected 12 people from a data set constructed by Allen Shoemaker (Journal of Statistics Education) and recorded their body temperature and heart rate. \({ }^{13}\) $$ \begin{array}{l|llllll} \text { Person } & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \begin{array}{c} \text { Temperature } \\ \text { (degrees) } \end{array} & 96.3 & 97.4 & 98.9 & 99.0 & 99.0 & 96.8 \\ \text { Heart Rate } & 70 & 68 & 80 & 75 & 79 & 75 \\ \text { (beats per minute) } & & & & & & \end{array} $$ $$ \begin{array}{c|cccccc} \text { Person } & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline \begin{array}{c} \text { Temperature } \\ \text { (degrees) } \end{array} & 98.4 & 98.4 & 98.8 & 98.8 & 99.2 & 99.3 \\ \text { Heart Rate } & 74 & 84 & 73 & 84 & 66 & 68 \\ \text { (beats per minute) } & & & & & & \end{array} $$ a. Find the correlation coefficient \(r\), relating body temperature to heart rate. b. Is there sufficient evidence to indicate that there is a correlation between these two variables? Test at the \(5 \%\) level of significance.

Graph the line corresponding to the equation \(y=-2 x+1\) by graphing the points corresponding to \(x=0,1,\) and 2 . Give the \(y\) -intercept and slope for the line. How is this line related to the line \(y=2 x+1\) of Exercise \(12.1 ?\)

The following data (Exercises 12.16 and 12.24 ) 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. Estimate the expected strawberry texture for a coded storage temperature of \(x=-1 .\) Use a \(99 \%\) confidence interval. b. Predict the particular value of \(y\) when \(x=1\) with a \(99 \%\) prediction interval. c. At what value of \(x\) will the width of the prediction interval for a particular value of \(y\) be a minimum, assuming \(n\) remains fixed?

What diagnostic plot can you use to determine whether the incorrect model has been used? What should the plot look like if the correct model has been used?

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