Introductory Statistics, Books a la Carte Plus NEW MyLab Statistics with Pearson eText -- Access Card Package (10th Edition)
10th Edition
ISBN: 9780134270364
Author: Neil A. Weiss
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Textbook Question
Chapter 15.2, Problem 61E
In Exercises 15.58–15.63, we repeat the information from Exercises 15.22–15.27. Presuming that the assumptions for regression inferences are met, decide at the specified significance level whether the data provide sufficient evidence to conclude that the predictor variable is useful for predicting the response variable.
15.61 Plant Emissions. Following are the data on plant weight and quantity of volatile emissions from Exercise 15.25. Use α = 0.05.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
The question that I need help with is attached.
thanks
Thank you for helping with this question.
Regression and Predictions. Exercises 13–28 use the same data sets as Exercises 13–28 in Section 10-1. In each case, find the regression equation, letting the first variable be the predictor (x) variable. Find the indicated predicted value by following the prediction procedure summarized in Figure 10-5 on page 493.
Crickets and Temperature Find the best predicted temperature at a time when a cricket chirps 3000 times in 1 minute. What is wrong with this predicted temperature?
Chapter 15 Solutions
Introductory Statistics, Books a la Carte Plus NEW MyLab Statistics with Pearson eText -- Access Card Package (10th Edition)
Ch. 15.1 - Suppose that x and y are predictor and response...Ch. 15.1 - Prob. 2ECh. 15.1 - Prob. 3ECh. 15.1 - Prob. 4ECh. 15.1 - Prob. 5ECh. 15.1 - In Exercises 15.315.6, assume that the variables...Ch. 15.1 - The difference between an observed value and a...Ch. 15.1 - Identify two graphs used in a residual analysis to...Ch. 15.1 - Which graph used in a residual analysis provides...Ch. 15.1 - Figure 15.8 shows three residual plots and a...
Ch. 15.1 - Figure 15.9 on the next page shows three residual...Ch. 15.1 - In Exercises 15.1215.21, we repeat the data and...Ch. 15.1 - In Exercises 15.1215.21, we repeat the data and...Ch. 15.1 - Prob. 14ECh. 15.1 - Prob. 15ECh. 15.1 - Prob. 16ECh. 15.1 - Prob. 17ECh. 15.1 - Prob. 18ECh. 15.1 - Prob. 19ECh. 15.1 - Prob. 20ECh. 15.1 - Prob. 21ECh. 15.1 - Prob. 22ECh. 15.1 - Prob. 23ECh. 15.1 - Prob. 24ECh. 15.1 - Prob. 25ECh. 15.1 - In Exercises 15.2215.27, we repeat the information...Ch. 15.1 - Prob. 27ECh. 15.1 - Prob. 28ECh. 15.1 - In Exercises 15.2815.33, a. compute the standard...Ch. 15.1 - Prob. 30ECh. 15.1 - In Exercises 15.2815.33, a. compute the standard...Ch. 15.1 - In Exercises 15.2815.33, a. compute the standard...Ch. 15.1 - In Exercises 15.2815.33, a. compute the standard...Ch. 15.1 - In Exercises 15.3415.43, use the technology of...Ch. 15.1 - In Exercises 15.3415.43, use the technology of...Ch. 15.1 - In Exercises 15.3415.43, use the technology of...Ch. 15.1 - In Exercises 15.3415.43, use the technology of...Ch. 15.1 - Prob. 38ECh. 15.1 - Prob. 39ECh. 15.1 - Prob. 40ECh. 15.1 - Prob. 41ECh. 15.1 - Prob. 42ECh. 15.1 - Prob. 43ECh. 15.2 - Explain why the predictor variable is useless as a...Ch. 15.2 - Prob. 45ECh. 15.2 - Prob. 46ECh. 15.2 - In this section, we used the statistic b1 as a...Ch. 15.2 - In Exercises 15.4815.57, we repeat the information...Ch. 15.2 - Prob. 49ECh. 15.2 - In Exercises 15.4815.57, we repeat the information...Ch. 15.2 - In Exercises 15.4815.57, we repeat the information...Ch. 15.2 - Prob. 52ECh. 15.2 - Prob. 53ECh. 15.2 - Prob. 54ECh. 15.2 - In Exercises 15.4815.57, we repeat the information...Ch. 15.2 - Prob. 56ECh. 15.2 - Prob. 57ECh. 15.2 - Prob. 58ECh. 15.2 - In Exercises 15.5815.63, we repeat the information...Ch. 15.2 - Prob. 60ECh. 15.2 - In Exercises 15.5815.63, we repeat the information...Ch. 15.2 - Prob. 62ECh. 15.2 - In Exercises 15.5815.63, we repeat the information...Ch. 15.2 - Prob. 64ECh. 15.2 - In each of Exercises 15.6415.69, apply Procedure...Ch. 15.2 - In each of Exercises 15.6415.69, apply Procedure...Ch. 15.2 - Prob. 67ECh. 15.2 - Prob. 68ECh. 15.2 - Prob. 69ECh. 15.2 - Prob. 70ECh. 15.2 - In Exercises 15.7015.80, use the technology of...Ch. 15.2 - In Exercises 15.7015.80, use the technology of...Ch. 15.2 - Prob. 73ECh. 15.2 - Prob. 74ECh. 15.2 - Prob. 75ECh. 15.2 - In Exercises 15.7015.80, use the technology of...Ch. 15.2 - Prob. 77ECh. 15.2 - Prob. 78ECh. 15.2 - In Exercises 15.7015.80, use the technology of...Ch. 15.2 - Prob. 80ECh. 15.3 - Without doing any calculations, fill in the blank....Ch. 15.3 - Prob. 82ECh. 15.3 - Prob. 83ECh. 15.3 - Prob. 84ECh. 15.3 - In Exercises 15.8215.91, we repeat the data from...Ch. 15.3 - Prob. 86ECh. 15.3 - Prob. 87ECh. 15.3 - In Exercises 15.8215.91, we repeat the data from...Ch. 15.3 - Prob. 89ECh. 15.3 - Prob. 90ECh. 15.3 - Prob. 91ECh. 15.3 - Prob. 92ECh. 15.3 - In Exercises 15.9215.97, presume that the...Ch. 15.3 - In Exercises 15.9215.97, presume that the...Ch. 15.3 - In Exercises 15.9215.9, presume that the...Ch. 15.3 - Prob. 96ECh. 15.3 - In Exercises 15.9215.97, presume that the...Ch. 15.3 - Prob. 98ECh. 15.3 - In Exercises 15.9815.108, use the technology of...Ch. 15.3 - In Exercises 15.9815.108, use the technology of...Ch. 15.3 - In Exercises 15.9815.108, use the technology of...Ch. 15.3 - In Exercises 15.9815.108, use the technology of...Ch. 15.3 - Prob. 103ECh. 15.3 - Prob. 104ECh. 15.3 - Prob. 105ECh. 15.3 - Prob. 106ECh. 15.3 - In Exercises 15.9815.108, use the technology of...Ch. 15.3 - Prob. 108ECh. 15.3 - Margin of Error in Regression. In Exercises 15.109...Ch. 15.3 - Refer to the confidence interval and prediction...Ch. 15.4 - Identify the statistic used to estimate the...Ch. 15.4 - Prob. 112ECh. 15.4 - Suppose that, for a sample of pairs of...Ch. 15.4 - Prob. 114ECh. 15.4 - Prob. 115ECh. 15.4 - Prob. 116ECh. 15.4 - Prob. 117ECh. 15.4 - Prob. 118ECh. 15.4 - Prob. 119ECh. 15.4 - Prob. 120ECh. 15.4 - Prob. 121ECh. 15.4 - Prob. 122ECh. 15.4 - Prob. 123ECh. 15.4 - Prob. 124ECh. 15.4 - Prob. 125ECh. 15.4 - Prob. 126ECh. 15.4 - Prob. 127ECh. 15.4 - Prob. 128ECh. 15.4 - Prob. 129ECh. 15.4 - Prob. 130ECh. 15.4 - Prob. 131ECh. 15.4 - Prob. 132ECh. 15.4 - Prob. 133ECh. 15.4 - In each of Exercises 15.13415.144, use the...Ch. 15.4 - In each of Exercises 15.13415.144, use the...Ch. 15.4 - Prob. 136ECh. 15.4 - Prob. 137ECh. 15.4 - Prob. 138ECh. 15.4 - Prob. 139ECh. 15.4 - Prob. 140ECh. 15.4 - In each of Exercises 15.13415.144, use the...Ch. 15.4 - Prob. 142ECh. 15.4 - Prob. 143ECh. 15.4 - Prob. 144ECh. 15 - Prob. 1RPCh. 15 - Suppose that x and y are two variables of a...Ch. 15 - What two plots did we use in this chapter to...Ch. 15 - Regarding analysis of residuals, decide in each...Ch. 15 - Suppose that you perform a hypothesis test for the...Ch. 15 - Prob. 6RPCh. 15 - Prob. 7RPCh. 15 - Prob. 8RPCh. 15 - Prob. 9RPCh. 15 - Identify the relationship between two variables...Ch. 15 - Graduation Rates. Graduation ratethe percentage of...Ch. 15 - Prob. 12RPCh. 15 - Prob. 13RPCh. 15 - For Problems 1417, presume that the variables...Ch. 15 - For Problems 1417, presume that the variables...Ch. 15 - For Problems 1417, presume that the variables...Ch. 15 - Prob. 17RPCh. 15 - In Problems 1820, use the technology of your...Ch. 15 - In Problems 1820, use the technology of your...Ch. 15 - In Problems 1820, use the technology of your...Ch. 15 - Recall from Chapter 1 (see page 34) that the Focus...Ch. 15 - At the beginning of this chapter, we presented...
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.Similar questions
- Sam Jones has 2 years of historical sales data for his company. He is applyingfor a business loan and must supply his projections of sales by month for thenext 2 years to the bank. a. Using the data from Table 6–12, provide a regression forecast for timeperiods 25 through 48.b. Does Sam’s sales data show a seasonal pattern?arrow_forwardRegression and Predictions. Exercises 13–28 use the same data sets as Exercises 13–28 in Section 10-1. In each case, find the regression equation, letting the first variable be the predictor (x) variable. Find the indicated predicted value by following the prediction procedure summarized in Figure 10-5 on page 493. Tips Using the bill/tip data, find the best predicted tip amount for a dinner bill of $100. What tipping rule does the regression equation suggest?arrow_forwardQ1) Interpret the following regression line y = 10.50 – 0.18xarrow_forward
- 10 – 11. Margaret, an archeologist, is conducting a test to determine if there is a positive linear relationship between the total height of a dinosaur and its leg length. Her random sample of 15 dinosaur total heights (in feet) and leg lengths (in feet) produced the results shown in the following TI calculator screen. Use the TI calculations in the screen shot to help you answer questions: 10 & 11. LinReg y=a+bx a=28.67845743 b=5.639892354 r=559696513 r=.7481286741 10. What would you predict for a dinosaur's total height (to 2 decimal places) in feet, if the leg length is 5.8 feet? a) 61.39 feet b) 28.68 feet c) 114.99 feet d) 61.33 feet e) 74.81 feet 11. What percent of variation in the dinosaur's total height can be accounted for by the variation in the dinosaur's leg length? a) 28.68% b) 5.64%% c) 55.97% d) 74.81% e) none of thesearrow_forward'1) Interpret the following regression line y = 10.50 – 0.18x 2) Interpret the following coefficient of determination r? = 0.69 3) Interpret the following coefficient of correlation r =-0.83arrow_forwardA logistic regression was used to investigate obesity and poor physical health while controlling for the following variables: age, gender, race, income, health status, education, current smoker, and diet/exercise status. Justify the use of a logistic regression.arrow_forward
- 13) Use computer software to find the multiple regression equation. Can the equation be used for prediction? An anti-smoking group used data in the table to relate the carbon monoxide( CO) of various brands of cigarettes to their tar and nicotine (NIC) content. 13). CO TAR NIC 15 1.2 16 15 1.2 16 17 1.0 16 6. 0.8 1 0.1 1 8. 0.8 8. 10 0.8 10 17 1.0 16 15 1.2 15 11 0.7 9. 18 1.4 18 16 1.0 15 10 0.8 9. 0.5 18 1.1 16 A) CO = 1.37 + 5.50TAR – 1.38NIC; Yes, because the P-value is high. B) CÓ = 1.37 - 5.53TAR + 1.33NIC; Yes, because the R2 is high. C) CO = 1.25 + 1.55TAR – 5.79NIC; Yes, because the P-value is too low. D) CO = 1.3 + 5.5TAR - 1.3NIC; Yes, because the adjusted R2 is high. %3Darrow_forward10) A regression was run to determine if there is a relationship between hours of TV watched per day (x) and number of situps a person can do (y).The results of the regression were:y=ax+b a=-0.767 b=31.009 r2=0.609961 r=-0.781 Use this to predict the number of situps a person who watches 7.5 hours of TV can do (to one decimal place)arrow_forwardA paper gives data on x = change in Body Mass Index (BMI, in kilograms/meter) and y = change in a measure of depression for patients suffering from depression who participated in a pulmonary rehabilitation program. The table below contains a subset of the data given in the paper and are approximate values read from a scatterplot in the paper. BMI Change (kg/m²) -0.5 0.7 0.5 0.1 0.8 1 1.5 1.2 1 0.4 0.4 Depression Score Change -1 4 4 8 13 14 16 18 12 14 The accompanying computer output is from Minitab. Fitted Line Plot Depression score change = 6.598 + 5.327 BMI change 20- 5.10254 R-Sq R-Sq (adj) 20.06% 27.32% 15- 10- 5- 0- -0.5 0.0 0.5 1.0 1.5 BMI change R-sq 5.10254 27.32% Coefficients Term Coef SE Coef T-Value P-Value VIF Constant 6.598 2.19 3.01 0.0132 BMI change 5.327 2.75 1.94 0.0812 1.00 Regression Equation Depression score change = 6.598 + 5.327 BMI change (a) What percentage of observed variation in depression score change can be explained by the simple linear regression model?…arrow_forward
- Define the ADL and GLS Estimators of Regression.arrow_forwardThe November 24, 2001, issue of The Economist published economic data for 15 industrialized nations. Included were the percent changes in gross domestic product (GDP), industrial production (IP), consumer prices (CP), and producer prices (PP) from Fall 2000 to Fall 2001, and the unemployment rate in Fall 2001 (UNEMP). An economist wants to construct a model to predict GDP from the other variables. A fit of the model GDP = , + P,IP + 0,UNEMP + f,CP + P,PP + € yields the following output: The regression equation is GDP = 1.19 + 0.17 IP + 0.18 UNEMP + 0.18 CP – 0.18 PP Predictor Coef SE Coef тР Constant 1.18957 0.42180 2.82 0.018 IP 0.17326 0.041962 4.13 0.002 UNEMP 0.17918 0.045895 3.90 0.003 CP 0.17591 0.11365 1.55 0.153 PP -0.18393 0.068808 -2.67 0.023 Predict the percent change in GDP for a country with IP = 0.5, UNEMP = 5.7, CP = 3.0, and PP = 4.1. a. b. If two countries differ in unemployment rate by 1%, by how much would you predict their percent changes in GDP to differ, other…arrow_forwardAssume we have data demonstrating a strong linear link between the amount of fertilizer applied to certain plants and their yield. Which is the independent variable in this research question?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Hypothesis Testing using Confidence Interval Approach; Author: BUM2413 Applied Statistics UMP;https://www.youtube.com/watch?v=Hq1l3e9pLyY;License: Standard YouTube License, CC-BY
Hypothesis Testing - Difference of Two Means - Student's -Distribution & Normal Distribution; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=UcZwyzwWU7o;License: Standard Youtube License