/*! 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 293 For the following exercises, con... [FREE SOLUTION] | 91Ó°ÊÓ

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For the following exercises, consider the data in Table 2.18, which shows the percent of unemployed in a city of people 25 years or older who are college graduates is given below, by year. $$\begin{array}{|c|c|c|c|c|c|}\hline \text { Year } & {2000} & {2002} & {2005} & {2007} & {2010} \\ \hline \text { Percent Graduates } & {6.5} & {7.0} & {7.4} & {8.2} & {9.0} \\ \hline\end{array}$$ Based on the set of data given in Table 2.20, calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient to three decimal places. $$\begin{array}{|c|c|c|c|c|c|}\hline x & {10} & {12} & {15} & {18} & {20} \\\ \hline y & {36} & {34} & {30} & {28} & {22} \\ \hline\end{array}$$

Short Answer

Expert verified
Regression line: \( y = -0.8x + 42 \), Correlation coefficient: \( r = -0.964 \).

Step by step solution

01

Assign Variables and Reformat Data

First, we need to assign variables to the data in the tables. For the data given, set year 2000 as 0 and increment by 1 for each subsequent two-year period, which means the years will be {0, 1, 2, 3, 4} corresponding to the {2000, 2002, 2005, 2007, 2010} years respectively in Table 2.18. The data from Table 2.20 will be used as provided.
02

Calculate the Means

Calculate the means of both sets of data. For Table 2.20, calculate the mean of x's and y's:\[ \bar{x} = \frac{10+12+15+18+20}{5} = 15 \]\[ \bar{y} = \frac{36+34+30+28+22}{5} = 30 \]
03

Compute Required Summations

Calculate the necessary summations for the least squares regression line:\[ \sum{x} = 75, \sum{y} = 150, \sum{xy} = 2036, \sum{x^2} = 1194 \]
04

Determine the Slope and Intercept

Using the formulas for the slope (m) and the y-intercept (b) of the line:\[ m = \frac{n \sum{xy} - (\sum{x})(\sum{y})}{n \sum{x^2} - (\sum{x})^2} = \frac{5(2036) - (75)(150)}{5(1194) - (75)^2} = -0.8 \]\[ b = \bar{y} - m \bar{x} = 30 - (-0.8) \cdot 15 = 42 \]Therefore, the regression line equation is: \[ y = -0.8x + 42 \]
05

Calculate the Correlation Coefficient

To find the correlation coefficient (r), we use the formula:\[ r = \frac{n \sum{xy} - (\sum{x})(\sum{y})}{\sqrt{[n\sum{x^2} - (\sum{x})^2][n\sum{y^2} - (\sum{y})^2]}} \]First calculate:\[ \sum{y^2} = 5432 \]\[ r = \frac{5(2036) - (75)(150)}{\sqrt{[5(1194) - (75)^2][5(5432) - (150)^2]}} = -0.964 \]
06

Conclusion

The regression line for the data provided in Table 2.20 is \( y = -0.8x + 42 \), and the correlation coefficient is \( r = -0.964 \).

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Key Concepts

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

Least Squares Regression
Least squares regression is a fundamental statistical approach used to identify the line of best fit through a set of points in a scatter plot. The main objective is to minimize the sum of the squares of the vertical distances (residuals) between the observed values and the values anticipated by the linear model. In essence, this technique seeks to find the most accurate line that represents the relationship between two variables (x and y).
  • Residuals: The difference between observed values and predicted values from the regression line.
  • Sum of squared residuals: The sum of the squares of residuals, which least squares regression tries to minimize.
  • Line of best fit: The line that minimizes the sum of squared residuals.
While calculating the least squares regression, you'll derive two key components: the slope (m) and the y-intercept (b) of the line, leading to the regression equation of the form: \[ y = mx + b \] Finding the values of m and b involves understanding various summations and averages within your data set.
Correlation Coefficient
The correlation coefficient, denoted by the symbol \( r \), is a statistical measure that describes the degree and direction of a linear relationship between two variables. Its value ranges between -1 and 1. When it comes to interpreting \( r \):
  • \( r = 1 \): Perfect positive linear relationship.
  • \( r = -1 \): Perfect negative linear relationship.
  • \( r = 0 \): No linear correlation.
  • Positive \( r \): As one variable increases, the other tends to increase.
  • Negative \( r \): As one variable increases, the other tends to decrease.
In practice, a correlation coefficient of \( -0.964 \) suggests a strong negative linear relationship between the variables. This means that as one variable increases, the other tends to decrease considerably at a predictable rate. The closer \( r \) is to -1 or 1, the stronger the correlation. Calculating \( r \) involves using a specific formula that requires knowledge of the means and summations of the squares of the variables involved.
Linear Regression Equation
The linear regression equation is a mathematical representation of the linear relationship between two variables. It is typically expressed in the form \( y = mx + b \), where:
  • \( y \): The dependent variable or the response variable.
  • \( x \): The independent variable or the predictor variable.
  • \( m \): The slope of the line, indicating the change in \( y \) for a unit change in \( x \).
  • \( b \): The y-intercept, representing the value of \( y \) when \( x \) is zero.
Using data, the linear regression equation can help predict the expected value of \( y \) for any given \( x \). In the provided exercise, the line is expressed as \( y = -0.8x + 42 \), where \( -0.8 \) is the slope indicating a decrease in \( y \) as \( x \) increases, and \( 42 \) is the starting value of \( y \) when \( x \) is zero. This equation is particularly useful in forecasting and making informed predictions based on the given data.

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Most popular questions from this chapter

A town’s population increases at a constant rate. In 2010 the population was 55,000. By 2012 the population had increased to 76,000. If this trend continues, predict the population in 2016.

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