10The variance of the prediction error is,a41.8479 b43.1422 c44.4765 d45.8520 Questions 1-17 arerelated to the followingBelow are the values for two variables x and y obtained from a sample of size 25. We want to build a regression equationbased the sample data.ý = bo + bịxbo =b1 =yX418.321.64.93223.5419.24.829.97.622.88.727.45.846.612.229.58276.941.49.935.712.938.49.226.78.634.710.119.54.943.69.535.78.851.711.738.66.844.79.4407.228.610.733.47.127.2723.26.432.47.735.89.728.57.540.58.37.29.718.44.550.88.636.58.8

For the given data The variance of the prediction error is = ?

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10 The variance of the prediction error is, a 41.8479 b 43.1422 c 44.4765 d 45.8520

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)

6th Edition

ISBN:9781337111348

Author:Bruce Crauder, Benny Evans, Alan Noell

Publisher:Bruce Crauder, Benny Evans, Alan Noell

Chapter3: Straight Lines And Linear Functions

Section3.4: Linear Regression

Problem 12SBE: Find the equation of the regression line for the following data set. x 1 2 3 y 0 3 4

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For the following exercises, use Table 4 which shows the percent of unemployed persons 25 years or older who are college graduates in a particular city, by year. Based on the set of data given in Table 5, calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient. Round to three decimal places of accuracy
Olympic Pole Vault The graph in Figure 7 indicates that in recent years the winning Olympic men’s pole vault height has fallen below the value predicted by the regression line in Example 2. This might have occurred because when the pole vault was a new event there was much room for improvement in vaulters’ performances, whereas now even the best training can produce only incremental advances. Let’s see whether concentrating on more recent results gives a better predictor of future records. (a) Use the data in Table 2 (page 176) to complete the table of winning pole vault heights shown in the margin. (Note that we are using x=0 to correspond to the year 1972, where this restricted data set begins.) (b) Find the regression line for the data in part ‚(a). (c) Plot the data and the regression line on the same axes. Does the regression line seem to provide a good model for the data? (d) What does the regression line predict as the winning pole vault height for the 2012 Olympics? Compare this predicted value to the actual 2012 winning height of 5.97 m, as described on page 177. Has this new regression line provided a better prediction than the line in Example 2?
XYZ Corporation Stock Prices The following table shows the average stock price, in dollars, of XYZ Corporation in the given month. Month Stock price January 2011 43.71 February 2011 44.22 March 2011 44.44 April 2011 45.17 May 2011 45.97 a. Find the equation of the regression line. Round the regression coefficients to three decimal places. b. Plot the data points and the regression line. c. Explain in practical terms the meaning of the slope of the regression line. d. Based on the trend of the regression line, what do you predict the stock price to be in January 2012? January 2013?
For the following exercises, consider the data in Table 5, which shows the percent of unemployed in a city ofpeople25 years or older who are college graduates is given below, by year. 41. Based on the set of data given in Table 7, calculatethe regression line using a calculator or othertechnology tool, and determine the correlationcoefficient to three decimal places.
For the following exercises, consider the data in Table 5, which shows the percent of unemployed ina city of people 25 years or older who are college graduates is given below, by year. 40. Based on the set of data given in Table 6, calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient to three decimal places.
Using the lengths (in.), chest sizes (in.), and weights (lb) of bears from a data set, the resulting regression equation is Weight=-274 +0.426 Length + 12.1 Chest Size. The P-value is 0.000 and the adjusted R² value is 0.925. If an additional predictor variable of neck size (in.) is included, the P-value becomes 0.000 and the adjusted R² becomes 0.933. Why is it better to use values of adjusted R² instead of simply using values of R²? Choose the correct answer below. C O A. The unadjusted R² can only be calculated for regression equations with two or fewer predictor variables, while the adjusted R² can be calculated for regression equations with any number of predictor variables. O B. The unadjusted R² increases or remains the same as more variables are included, but the adjusted R² is adjusted for the number of variables and sample size. OC. The unadjusted R² can only be calculated for regression equations with P-values greater than 0, while the adjusted R² can be calculated for…
Using the lengths (in.), chest sizes (in.), and weights (lb) of bears from a data set, the resulting regression equation is Weight= -274 +0.426 Length + 12.1 Chest Size. The P-value is 0.000 and the adjusted R² value is 0.925. If an additional predictor variable of neck size (in.) is included, the P-value becomes 0.000 and the adjusted R² becomes 0.933. Why is it better to use values of adjusted R² instead of simply using values of R²? Choose the correct answer below. C O A. The unadjusted R² can only be calculated for regression equations with P-values greater than 0, while the adjusted R² can be calculated for regression equations with any manner of P-value. O B. The unadjusted R² increases or remains the same as more variables included, but the adjusted R² is adjusted for the number of variables and sample size. OC. The unadjusted R² decreases or remains the same as more variables are included, but the adjusted R² is adjusted for the number of variables and sample size. O D. The…
Full data set High | Low High Low 21 33 36 29 31 22 59 42 91 72 62 48 86 71 64 52 66 54 85 81 61 43 36 25 64 44 85 70 59 43 41 35 42 35 52 40 33 21 68 49 26 17 34 24 57 45 52 41 45 35 61 53 42 34 35 23 70 47 85 77 85 71 28 15 82 74 45 35 52 37 48 38 85 72 81 64 86 75 28 8 44 38 36 27 59 49 45 41 35 24 85 68 80 64 44 33 85 66 49 36
A high school track & field coach wanted to asses the relationship between an athletes height and how far they can jump in the long jump event (both in inches). They collect data on each athletes height and how far they can jump. Let the height of the athlete be the explanatory variable (X) and the length of the jump be the response (Y). The scatterplot of the data is as follows: Scatter Plot distance 88 86 84 82 80 78 72 74 76 78 height A. Based on the scatterplot, is there a positive or negative association between height of athlete and length of jump? B. Based on the scatterplot, state why using a linear regression equation is justified to asses the relationship between height and length of jump.
Use the given information to find the coefficient of determination. 31) A regression equation is obtained for a collection of paired data. It is found that the total variation is 24.488, the explained variation is 15.405, and the unexplained variation is 9.083. Find the coefficient of determination.
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The regression equation is computed for a set of n = 18 pairs of X and Y values with a correlation of r = 0.50 and SSy = 48. Find the standard error of estimate for the regression equation. The standard error of estimate = How big would the standard error be if the sample size were n = 66? The standard error of estimate =