/*! 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 3 For a fixed confidence level, ho... [FREE SOLUTION] | 91Ó°ÊÓ

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For a fixed confidence level, how does the length of the confidence interval for predicted values of \(y\) change as the corresponding \(x\) values become further away from \(\bar{x} ?\)

Short Answer

Expert verified
The confidence interval length for \( y \) increases as \( x \) moves further from \( \bar{x} \).

Step by step solution

01

Understanding the Problem

We need to consider how the confidence interval for predicted values of \( y \) changes as \( x \) moves away from the mean of \( x \), \( \bar{x} \), while keeping the confidence level fixed.
02

Analyzing Confidence Interval Formula

The formula for the confidence interval in linear regression is often calculated as: \[ \hat{y} \pm t_{\alpha/2, n-2} \cdot SE(\hat{y}) \] where \( SE(\hat{y}) = s \sqrt{\frac{1}{n} + \frac{(x - \bar{x})^2}{S_{xx}}} \). Notice the term \( \frac{(x - \bar{x})^2}{S_{xx}} \), which increases as \( x \) moves away from \( \bar{x} \).
03

Identifying Changes in Standard Error

Since \( SE(\hat{y}) \) depends on \( \frac{(x - \bar{x})^2}{S_{xx}} \), as \( x \) gets further from \( \bar{x} \), \( SE(\hat{y}) \) increases. Thus, the width of the confidence interval, \( 2 \cdot t_{\alpha/2, n-2} \cdot SE(\hat{y}) \), also increases.
04

Conclusion

As the \( x \) values become further from \( \bar{x} \), the length of the confidence interval for \( y \) increases due to the increase in the standard error as predicted by the term \( \frac{(x - \bar{x})^2}{S_{xx}} \).

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Key Concepts

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

Linear Regression
Linear regression is a fundamental statistical method used to model the relationship between two variables by fitting a linear equation to the observed data. The primary aim is to predict the value of a dependent variable, often denoted as \( y \), based on the value of an independent variable, \( x \). The linear equation can be expressed in the form: \( y = \beta_0 + \beta_1 x + \epsilon \). Here, \( \beta_0 \) is the intercept, \( \beta_1 \) is the slope, and \( \epsilon \) represents the error term.
The slope, \( \beta_1 \), indicates how much \( y \) is expected to change with a one-unit increase in \( x \). Linear regression assumes a constant relationship between \( x \) and \( y \), meaning the slope remains the same across all values of \( x \). It's crucial for understanding how changes in \( x \) impact the predicted value of \( y \).
The model employs the least squares method to find the best-fitting line by minimizing the sum of squared differences between observed and predicted values. Linear regression helps in forecasting and finding relationships but requires assumptions like linearity, independence, and normal distribution of residuals.
Standard Error
The standard error is a measure of variability or dispersion in statistical terms. In the context of linear regression, it refers to the standard error of the predicted value of \( y \), denoted as \( SE(\hat{y}) \). This quantifies the uncertainty associated with predictions made using the regression line. The smaller the standard error, the closer the observed values are to the regression line, indicating a more precise prediction model.
The standard error of the predicted value \( SE(\hat{y}) \) is calculated using the equation: \[ SE(\hat{y}) = s \sqrt{\frac{1}{n} + \frac{(x - \bar{x})^2}{S_{xx}}} \] where \( s \) is the standard deviation of the residuals, \( n \) is the sample size, \( x \) is the independent variable's value, \( \bar{x} \) is the mean of \( x \), and \( S_{xx} \) is the sum of squares of \( x \).
A critical aspect here is \( \frac{(x - \bar{x})^2}{S_{xx}} \), which increases when \( x \) moves away from \( \bar{x} \). As this term grows, \( SE(\hat{y}) \) increases, making predictions less precise and widening the confidence interval.
Mean Value
A mean value, typically represented as \( \bar{x} \) or \( \mu \), is the average of a set of numbers and provides a central value of a dataset. It's a measure of central tendency and is calculated as the sum of all values divided by the number of values. In the context of linear regression, \( \bar{x} \) represents the average of the independent variable \( x \) values and plays a crucial role in the model's predictions.
The mean value serves as a reference point in regression analysis. When calculating the confidence interval for predictions, deviations from this mean value significantly affect the width of the interval. Greater deviations from the mean typically result in increased uncertainty in predictions, as indicated by the formula for standard error.
Understanding the mean value is vital because it impacts how we interpret the relationship between \( x \) and \( y \). It offers insight into how varying \( x \) might impact the predictability of \( y \), influencing how we approach data analysis and decision-making in real-world applications.

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Most popular questions from this chapter

In the least squares line \(\hat{y}=5+3 x,\) what is the marginal change in \(\hat{y}\) for each unit change in \(x ?\)

Use appropriate multiple regression software of your choice and enter the data. Note that the data are also available for download at the Companion Sites for this text. Medical: Blood Pressure The systolic blood pressure of individuals is thought to be related to both age and weight. For a random sample of 11 men, the following data were obtained: $$\begin{array}{ccc|ccc} \hline \begin{array}{c} \text { Systolic } \\ \text { Blood Pressure } \end{array} & \begin{array}{c} \text { Age } \\ \text { (years) } \end{array} & \begin{array}{c} \text { Weight } \\ \text { (pounds) } \end{array} & \begin{array}{c} \text { Systolic } \\ \text { Blood Pressure } \end{array} & \begin{array}{c} \text { Age } \\ \text { (years) } \end{array} & \begin{array}{c} \text { Weight } \\ \text { (pounds) } \end{array} \\ \hline x_{1} & x_{2} & x_{3} & x_{1} & x_{2} & x_{3} \\ \hline 132 & 52 & 173 & 137 & 54 & 188 \\ 143 & 59 & 184 & 149 & 61 & 188 \\ 153 & 67 & 194 & 159 & 65 & 207 \\ 162 & 73 & 211 & 128 & 46 & 167 \\ 154 & 64 & 196 & 166 & 72 & 217 \\ 168 & 74 & 220 & & & \\ \hline \end{array}$$ (a) Generate summary statistics, including the mean and standard deviation of each variable. Compute the coefficient of variation (see Section 3.2) for each variable. Relative to its mean, which variable has the greatest spread of data values? Which variable has the smallest spread of data values relative to its mean? (b) For each pair of variables, generate the sample correlation coefficient \(r\) Compute the corresponding coefficient of determination \(r^{2}\). Which variable (other than \(x_{1}\) ) has the greatest influence (by itself) on \(x_{1} ?\) Would you say that both variables \(x_{2}\) and \(x_{3}\) show a strong influence on \(x_{1}\) ? Explain your answer. What percent of the variation in \(x_{1}\) can be explained by the corresponding variation in \(x_{2}\) ? Answer the same question for \(x_{3}\) (c) Perform a regression analysis with \(x_{1}\) as the response variable. Use \(x_{2}\) and \(x_{3}\) as explanatory variables. Look at the coefficient of multiple determination. What percentage of the variation in \(x_{1}\) can be explained by the corresponding variations in \(x_{2}\) and \(x_{3}\) taken together? (d) Look at the coefficients of the regression equation. Write out the regression equation. Explain how each coefficient can be thought of as a slope. If age were held fixed, but a person put on 10 pounds, what would you expect for the corresponding change in systolic blood pressure? If a person kept the same weight but got 10 years older, what would you expect for the corresponding change in systolic blood pressure? (e) Test each coefficient to determine if it is zero or not zero. Use level of significance \(5 \% .\) Why would the outcome of each test help us determine whether or not a given variable should be used in the regression model? (f) Find a \(90 \%\) confidence interval for each coefficient. (g) Suppose Michael is 68 years old and weighs 192 pounds. Predict his systolic blood pressure, and find a \(90 \%\) confidence range for your prediction (if your software produces prediction intervals).

Over the past decade, there has been a strong positive correlation between teacher salaries and prescription drug costs. (a) Do you think paying teachers more causes prescription drugs to cost more? Explain. (b) What lurking variables might be causing the increase in one or both of the variables? Explain.

When drawing a scatter diagram, along which axis is the explanatory variable placed? Along which axis is the response variable placed?

In the least-squares line \(\hat{y}=5-2 x,\) what is the value of the slope? When \(x\) changes by 1 unit, by how much does \(\hat{y}\) change?

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