Answer: Interpreting a Computer Display. In Exercises 912, refer to the Minitab display

s e Notation Using Data Set 1 from Appendix B, if we let the predictor variable x represent heights of males and let the response variable y represent weights of males, the sample of 40 heights and weights results in s e = 17.5436 cm. In your own words, describe what that value of s e represents.

Prediction Interval Using the heights and weights described in Exercise 1, a height of 180 cm is used to find that the predicted weight is 88.0 kg, and the 95% prediction interval is displayed by Minitab as (50.7, 123.0). Write a statement that interprets that prediction interval. What is the major advantage of using a prediction interval instead of simply using the predicted weight of 88.0 kg? Why is the terminology of prediction interval used instead of confidence interval?

Coefficient of Determination Using the heights and weights described in Exercise 1, the linear correlation coefficient r is 0.356. Find the value of the coefficient of determination. What practical information does the coefficient of determination provide?

Standard Error of Estimate A sample of 12 different statistics textbooks is obtained and their weights are measured in kilograms and in pounds. Using the 12 paired weights (kg, lb), what is the value of s e ? For a Triola textbook that weighs 4.5 lb, the predicted weight in kilograms is 2.04 kg. What is the 95% prediction interval?

Interpreting the Coefficient of Determination. In Exercises 58, use the value of the linear correlation coefficient r to find the coefficient of determination and the percentage of the total variation that can be explained by the linear relationship between the two variables from the Appendix B data sets.. r = 0.933 ( x = weight of male , y = waist size of male )

Interpreting the Coefficient of Determination. In Exercises 58, use the value of the linear correlation coefficient r to find the coefficient of determination and the percentage of the total variation that can be explained by the linear relationship between the two variables from the Appendix B data sets.r = 0.963 ( x = chest size of a bear , y = weight of a bear )

Interpreting the Coefficient of Determination. In Exercises 58, use the value of the linear correlation coefficient r to find the coefficient of determination and the percentage of the total variation that can be explained by the linear relationship between the two variables from the Appendix B data sets.. r = 0.793 ( x = weight of a car , y = highway fuel consumption )

Interpreting the Coefficient of Determination. In Exercises 58, use the value of the linear correlation coefficient r to find the coefficient of determination and the percentage of the total variation that can be explained by the linear relationship between the two variables from the Appendix B data sets.r = 0.751 ( x = weight of discarded plastic , y = household size )

Interpreting a Computer Display. In Exercises 912, refer to the Minitab display obtained by using the paired data consisting of foot lengths (cm) and heights (cm) of the 40 people listed in Data Set 2 of Appendix B. (Unlike the examples in this section, these exercises use foot lengths instead of shoe print lengths.) Along with the paired sample data, Minitab was also given a foot length of 29.0 cm to be used for predicting height. MINITAB Testing for Correlation Use the information provided in the display to determine the value of the linear correlation coefficient. Given that there are 40 pairs of data, is there sufficient evidence to support a claim of a linear correlation between foot lengths of people and their heights?

Interpreting a Computer Display. In Exercises 912, refer to the Minitab display obtained by using the paired data consisting of foot lengths (cm) and heights (cm) of the 40 people listed in Data Set 2 of Appendix B. (Unlike the examples in this section, these exercises use foot lengths instead of shoe print lengths.) Along with the paired sample data, Minitab was also given a foot length of 29.0 cm to be used for predicting height. MINITAB. Identifying Total Variation What percentage of the total variation in height can be explained by the linear correlation between foot length and height?

Interpreting a Computer Display. In Exercises 912, refer to the Minitab display obtained by using the paired data consisting of foot lengths (cm) and heights (cm) of the 40 people listed in Data Set 2 of Appendix B. (Unlike the examples in this section, these exercises use foot lengths instead of shoe print lengths.) Along with the paired sample data, Minitab was also given a foot length of 29.0 cm to be used for predicting height. MINITAB Predicting Height If someone has a foot length of 29.0 cm, what is the single value that is the best predicted height?

Interpreting a Computer Display. In Exercises 912, refer to the Minitab display obtained by using the paired data consisting of foot lengths (cm) and heights (cm) of the 40 people listed in Data Set 2 of Appendix B. (Unlike the examples in this section, these exercises use foot lengths instead of shoe print lengths.) Along with the paired sample data, Minitab was also given a foot length of 29.0 cm to be used for predicting height. MINITAB Finding a Prediction Interval For someone with a foot length of 29.0 cm, identify the 95% prediction interval estimate of their height, and write a statement interpreting that interval.

Finding a Prediction Interval. In Exercises 1316, use the paired data consisting of foot lengths (cm) and heights (cm) of the 40 people listed in Data Set 2 of Appendix B. Let x represent foot length and let y represent the corresponding height. Use the given foot length and the given confidence level to construct a prediction interval estimate of height. (Hint: Use the results in the top half of the Minitab display that accompanies Exercises 912. Also, x = 25.68 c m , x = 1027.2 c m , and x 2 = 26 , 530.92 cm 2 . )Foot length: 25.0 cm 95% confidence

Finding a Prediction Interval. In Exercises 1316, use the paired data consisting of foot lengths (cm) and heights (cm) of the 40 people listed in Data Set 2 of Appendix B. Let x represent foot length and let y represent the corresponding height. Use the given foot length and the given confidence level to construct a prediction interval estimate of height. (Hint: Use the results in the top half of the Minitab display that accompanies Exercises 912. Also, x = 25.68 c m , x = 1027.2 c m , and x 2 = 26 , 530.92 cm 2 . )Foot length: 25.0 cm 99% confidence

Finding a Prediction Interval. In Exercises 1316, use the paired data consisting of foot lengths (cm) and heights (cm) of the 40 people listed in Data Set 2 of Appendix B. Let x represent foot length and let y represent the corresponding height. Use the given foot length and the given confidence level to construct a prediction interval estimate of height. (Hint: Use the results in the top half of the Minitab display that accompanies Exercises 912. Also, x = 25.68 c m , x = 1027.2 c m , and x 2 = 26 , 530.92 cm 2 . )Foot length: 22.0 cm 90% confidence

Finding a Prediction Interval. In Exercises 1316, use the paired data consisting of foot lengths (cm) and heights (cm) of the 40 people listed in Data Set 2 of Appendix B. Let x represent foot length and let y represent the corresponding height. Use the given foot length and the given confidence level to construct a prediction interval estimate of height. (Hint: Use the results in the top half of the Minitab display that accompanies Exercises 912. Also, x = 25.68 c m , x = 1027.2 c m , and x 2 = 26 , 530.92 cm 2 . )Foot length: 26.0 cm 95% confidence

Variation and Prediction Intervals. In Exercises 1720, find the (a) explained variation, (b) unexplained variation, and (c) indicated prediction interval. In each case, there is sufficient evidence to support a claim of a linear correlation, so it is reasonable to use the regression equation when making predictions Altitude and Temperature Listed below are altitudes (thousands of feet) and outside air temperatures ( F ) recorded by the author during Delta Flight 1053 from New Orleans to Atlanta. For the prediction interval, use a 95% confidence level with the altitude of 6327 ft (or 6.327 thousand feet). Altitude 3 10 14 22 28 31 33 Temperature 57 37 24 5 30 41 54

Variation and Prediction Intervals. In Exercises 1720, find the (a) explained variation, (b) unexplained variation, and (c) indicated prediction interval. In each case, there is sufficient evidence to support a claim of a linear correlation, so it is reasonable to use the regression equation when making predictions Town Courts Listed below are amounts of court income and salaries paid to the town justices (based on data from the Poughkeepsie Journal). All amounts are in thousands of dollars, and all of the towns are in Dutchess County, New York. For the prediction interval, use a 99% confidence level with a court income of $800,000. Court Income 65 404 1567 1131 272 252 111 154 32 Justice Salary 30 44 92 56 46 61 25 26 18

Variation and Prediction Intervals. In Exercises 1720, find the (a) explained variation, (b) unexplained variation, and (c) indicated prediction interval. In each case, there is sufficient evidence to support a claim of a linear correlation, so it is reasonable to use the regression equation when making predictions Galaxy Distances The table below lists measured amounts of redshift and the distances (billions of lightyears) to randomly selected clusters of galaxies (based on data from The Cosmic Perspective by Bennett et al., Benjamin Cummings). For the prediction interval, use a 90% confidence level with a redshift of the cluster Hydra of 0.0126. Redshift 0.0233 0.0539 0.0718 0.0395 0.0438 0.0103 Distance 0.32 0.75 1.00 0.55 0.61 0.14

Variation and Prediction Intervals. In Exercises 1720, find the (a) explained variation, (b) unexplained variation, and (c) indicated prediction interval. In each case, there is sufficient evidence to support a claim of a linear correlation, so it is reasonable to use the regression equation when making predictions Diamond Prices The table below lists weights (carats) and prices (dollars) of randomly selected diamonds. For the prediction interval, use a 95% confidence level with a diamond that weighs 0.8 carats. Weight 0.3 0.4 0.5 0.5 1.0 0.7 Price 510 1151 1343 1410 5669 2277

Confidence Intervals for 0 and 1 Confidence intervals for the yintercept 0 and slope 1 for a regression line ( y = 0 + 1 x ) can be found by evaluating the limits in the intervals below. b 0 E < 0 < b 0 + E where E = t / 2 s e 1 n + x 2 x 2 ( x ) 2 n b 1 E < 1 < b 1 + E where E = t / 2 s e x 2 ( x ) 2 n The yintercept b 0 and the slope b 1 are found from the sample data and t / 2 is found from Table A3 by using n 2 degrees of freedom. Using the 40 pairs of shoe print lengths (x) and heights (y) from Data Set 2 in Appendix B, find the 95% confidence interval estimates of 0 and 1

Confidence Interval for Mean Predicted Value Example 1 in this section illustrated the procedure for finding a prediction interval for an individual value of y. When using a specific value x 0 for predicting the mean of all values of y, the confidence interval is as follows: y ^ E < y < y ^ + E where E = t / 2 s e 1 n + n ( x 0 x ) 2 n ( x 2 ) ( x ) 2 The critical value t / 2 is found with n 2 degrees of freedom. Using the 40 pairs of shoe print lengths (x) and heights (y) from Data Set 2 in Appendix B, find a 95% confidence interval estimate of the mean height given that the shoe print length is 29.0 cm.