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Ch. 9 - Correlation and Regression
Larson - Elementary Statistics: Picturing the World 8th Edition
Larson8th EditionElementary Statistics: Picturing the WorldISBN: 9780137493470Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoLarson 8th EditionCh. 9 - Correlation and RegressionProblema 9.3.24
Capitolo 9, Problema 9.3.24

"Constructing and Interpreting a Prediction Interval In Exercises 21-30, construct the indicated prediction interval and interpret the results.
24. Trees Construct a 90% prediction interval for the trunk diameter of a tree in Exercise 14 when the height is 80 feet."

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1
Identify the regression equation from Exercise 14, which relates tree height to trunk diameter. This equation typically has the form: \(\hat{y}\) = b+mx, where \(\hat{y}\) is the predicted trunk diameter and x is the height of the tree.
Calculate the predicted trunk diameter for a tree height of 80 feet by substituting x = 80 into the regression equation.
Determine the standard error of the prediction, which accounts for both the variability of the estimate of the mean response and the variability of individual observations. The formula for the standard error of prediction is: SE_{pred} = s \(\sqrt{1 + \frac{1}{n}\) + \(\frac{(x_0 - \bar{x}\))^2}{\(\sum\) (x_i - \(\bar{x}\))^2}}, where s is the standard error of the estimate, n is the sample size, x_0 is the value 80, and \(\bar{x}\) is the mean of the observed heights.
Find the critical t-value for a 90% prediction interval with degrees of freedom equal to n - 2. This value comes from the t-distribution table and reflects the desired confidence level.
Construct the prediction interval using the formula: \(\hat{y}\) \(\pm\) t_{\(\alpha\)/2} \(\times\) SE_{pred}. This interval estimates the range in which the trunk diameter of a single tree with height 80 feet is likely to fall with 90% confidence. Finally, interpret this interval in the context of the problem.

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Prediction Interval

A prediction interval estimates the range within which a single new observation is expected to fall, with a specified level of confidence. Unlike confidence intervals for the mean, prediction intervals account for both the uncertainty in estimating the mean and the variability of individual data points.
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Prediction Intervals

Linear Regression and Prediction

Linear regression models the relationship between an independent variable (e.g., tree height) and a dependent variable (e.g., trunk diameter). Using the regression equation, we can predict the dependent variable's value for a given independent variable and construct intervals around this prediction.
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Using Regression Lines to Predict Values

Confidence Level and Interpretation

The confidence level (e.g., 90%) indicates the proportion of such intervals that would contain the true value if the experiment were repeated many times. Interpreting a 90% prediction interval means we are 90% confident the actual trunk diameter for a tree 80 feet tall falls within the calculated range.
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Introduction to Confidence Intervals
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Two variables have a bivariate normal distribution. Explain what this means.

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3. What does the sample correlation coefficient r measure? Which value indicates a stronger correlation: r =0.918 or r =- 0.932? Explain your reasoning.

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"In Exercises 7-10, use the value of the correlation coefficient r to calculate the coefficient of determination r^2. What does this tell you about the explained variation of the data about the regression line? about the unexplained variation?

7. r =0.465"

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"In Exercises 7-12, match the description in the left column with its symbol(s) in the right column.

10. y-intercept

a. \(\hat{y}\)_i

b. y_i

c. b

d. (\(\bar{x}\), \(\bar{y}\))

e. m

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In Exercise 24, remove the data for the student who is 57 inches tall and scored 128 on the IQ test. Describe how this affects the correlation coefficient r.

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"In Exercises 7-10, use the value of the correlation coefficient r to calculate the coefficient of determination r^2. What does this tell you about the explained variation of the data about the regression line? about the unexplained variation?

8.r =- 0.328"

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