10. Run each listed regression for the data set and determine which model is best. Hours Studied Test Score 1 0 78 75 Linger Dog Faustion 3 1.5 2.75 90 89 97 1 85 0.5 81 2 80
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- Which of the following is the most appropriate equation to model the data? ŷ = –10.89x + 0.85 ŷ = –0.85x + 10.89 ŷ = 10.89(0.85)x ŷ = 0.85(10.89)xThe data show the chest size and weight of several bears. Find the regression equation, letting chest size be the independent (x) variable. Then find the best predicted weight of a bear with a chest size of 57 inches. Is the result close to the actual weight of 466 pounds? Use a significance level of 0.05. Chest size (inches) 58 48 51 58 44 60 Weight (pounds) 425 266 347 453 282 408 Click the icon to view the critical values of the Pearson correlation coefficient r. What is the regression equation? x (Round to one decimal place as needed.) + What is the best predicted weight of a bear with a chest size of 57 inches? The best predicted weight for a bear with a chest size of 57 inches is pounds. (Round to one decimal place as needed.) Is the result close to the actual weight of 466 pounds? A. This result is very close to the actual weight of the bear. B. This result is not very close to the actual weight of the bear. C. This result is close to the actual weight of the bear. D. This…Use the data in the table below to complete parts (a) through (d). x 37 34 40 46 42 50 62 56 51 y 22 20 25 32 27 30 30 25 28 Find the equation of the regression line. y=
- The data show the chest size and weight of several bears. Find the regression equation, letting chest size be the independent (x) variable. Then find the best predicted weight of a bear with a chest size of 63 inches. Is the result close to the actual weight of 522 pounds? Use a significance level of 0.05. Chest size (inches) 58 50 65 59 59 48 D 414 312 499 450 456 260 Weight (pounds) Click the icon to view the critical values of the Pearson correlation coefficient r. What is the regression equation? y=+x (Round to one decimal place as needed.) What is the best predicted weight of a bear with a chest size of 63 inches? The best predicted weight for a bear with a chest size of 63 inches is pounds. (Round to one decimal place as needed.) Is the result close to the actual weight of 522 pounds? O A. This result is not very close to the actual weight of the bear. O B. This result is exactly the same as the actual weight of the bear. O C. This result is close to the actual weight of the…Roller coasters get their speed as a result of dropping down a steep incline. The table below gives height of a drop and the speed achieved for different roller coasters around the world. Drop in Meters (m 25 30 28 58 55 40 90 100 Speed of the coaster (k.p.h) 80 90 78 93 90 93 105 128 Have your calculator find the linear regression for the roller coaster data above. Write the linear model, clearly define the variables and state correlation coefficient and clearly interpret the slope. Use the model to find the drop height that will result in a coaster speed of 150 kph. Have your calculator find the exponential regression for this data. Write the exponential model, clearly define the variables and state correlation coefficient. Use the model to find the drop height that will result in a…Calculate the regression coefficient and obtain the lines of regression for the following data: USE MICROSOFT EXCEL
- Find the regression equation Student 1 4 5 6 7 8 9 10 Placement Exam 83 89 92 77 81 81 85 85 87 84 GWA 83 85 90 80 83 85 87 84 90 85 Select the correct response: y=35.39-0.5901x y=35.39x+0.5901 y=0.5901x-35.39 y=35.39+0.5901x 3. 2.We run the following regression using a sample of 148 women living in the western states. The variable wage is the hourly wage ( in USD) and educ is the number of years of education completed. wage = -3.07 + 1.52 educ Which statement best described the estimated slope coefficient? Group of answer choices A 1.52 extra years of education is predicted to increase wage by 1 USD An extra year of education is predicted to increase hourly wage by 1.52 USD A 1 percent increase in education is predicted to increase wage by 1.52 percent A 1.52 percent increase in education is predicted to increase wage by 1Develop a scatterplot and explore the correlation between customer age and net sales by each type of customer (regular/promotion). Use the horizontal axis for the customer age to graph. Find the linear regression line that models the data by each type of customer. Round the rate of changes (slopes) to two decimal places and interpret them in terms of the relation between the change in age and the change in net sales. What can you conclude? Hint: Rate of Change = Vertical Change / Horizontal Change = Change in y / Change in x
- Run a regression analysis on the following data set, where y is the final grade in a math class and x is the average number of hours the student spent working on math each week. hours/week Grade y 8. 72.2 8. 58.2 10 60 11 71.4 14 92.6 14 76.6 14 75.6 14 83.6 15 81 20 100 State the regression equation with constants accurate to 2 decimal places. ý = What is the predicted value for the final grade when a student spends an average of 15 hours each week on math? Round to 2 decimal places. Question Help: D Video 1 Video 2 Submit QuestionRun the Linear Regression Analysis for the following: → Submit Excel FileThe average driving distance (yards) and driving accuracy (percent of drives that land in thefairway) for 8 golfers are recorded in the table below.Rank Driving Distance (yards) Driving Accuracy (%)1 316.3 41.72 304.9 48.83 310.8 42.34 312.5 41.25 294.5 54.76 290.7 54.47 296.9 54.28 295.6 53.9a- Write the equation of a straight-line model relating driving accuracy (y) to drivingdistance (x). Select one of the following:i. Y = β1x + εii. Y = β1x2+ β0iii. Y = β1xiv. Y = β0 + β1x + εb- Fit the model and give the least square prediction equation c- Interpret the estimated y-intercept of the line. Choose the correct answer below:i. Since a drive with 0% accuracy is outside the range of the sample data, the y-intercepthas no practical interpretation.ii. Since a drive with distance 0 yards is outside the range of the sample data, the y-intercept has no practical interpretation.iii. For…The data show the chest size and weight of several bears. Find the regression equation, letting chest size be the independent (x) variable. Then find the best predicted weight of a bear with a chest size of 58 inches. Is the result close to the actual weight of 662 pounds? Use a significance level of 0.05. Chest size (Inches) 46 57 53 41 40 40 Weight (Pounds) 384 580 542 358 306 320