01:44:36The arm span and foot length were both measured (in centimeters) for each of 20 students in a biology class. Thecomputer output displays the regression analysis.PredictorCoef SE Coeft-ratioConstant-7.6112.5672.9650.046Arm span0.1860.0355.3770.000S = 1.61R-Sq = 63.0%R-Sq (Adj) = 64.9%Which of the following is the best interpretation of the standard deviation of the residuals?O The typical arm span is 161 centimeters.O The typical foot length is 16.1 centimeters.The typical distance between the observed and predicted arm spans is 1.61 centimeters.O The typical distance between the observed and predicted foot lengths is 1.61 centimetersSubmitNextSave and ExitMark this and return

For an estimated regression model of y on x such as

Answered: The arm span and foot length were both…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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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-a+bx a=37,525 b=-1.215 =0,790321 T==0,889 Use this to predict the number of situps a person who watches 6.5 hours of TV can do (to one decimal place) 19.7 Question Help: o Message instructor Submit Question MacBook Pro esc 288 F4 F3 F7 #3 & 2 6 7 Q T Y A S G C V B * LO
Find the regression equation, letting the diameter be the predictor (x) variable. Find the best predicted circumference of a beachball with a diameter of 44.4 cm. How does the result compare to the actual circumference of 139.5 cm? Use a significance level of 0.05. Baseball Basketball Golf Soccer Tennis Ping-Pong Volleyball Diameter 7.3 23.8 4.2 22.3 7.1 4.0 21.2 Circumference 22.9 74.8 13.2 70.1 22.3 12.6 66.6 LOADING... Click the icon to view the critical values of the Pearson correlation coefficient r.
Listed below are foot lengths (mm) and heights (mm) of males. Find the regression equation, letting foot length be the predictor (x) variable. Find the best predicted height of a male with a foot length of 272.7 mm. How does the result compare to the actual height of 1776 mm? Foot Length 282.3 277.8 252.8 258.7 279.0 258.4 274.1 261.7 Height 1785.0 1771.0 1675.7 1645.7 1859.3 1710.2 1789.2 1737.0 The regression equation is ŷ = + (x. (Round the y-intercept to the nearest integer as needed. Round the slope to two decimal places as needed.) The best predicted height of a male with a foot length of 272.7 mm is (Round to the nearest integer as needed.) How does the result compare to the actual height of 1776 mm? O A. The result is close to the actual height of 1776 mm. O B. The result is exactly the same as the actual height of 1776 mm. O C. The result is very different from the actual height of 1776 mm. O D. The result does not make sense given the context of the data. C mm.
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=a+bx b=-0.997 a=20.107 r2=0.950625 r=-0.975 Use this to predict the number of situps a person who watches 0.5 hours of TV can do.Round to one decimal place.
An automobile rental company wants to predict the yearly maintenance expense (Y) for an automobile using the number of miles driven during the year () and the age of the car (, in years) at the beginning of the year. The company has gathered the data on 10 automobiles and run a regression analysis with the results shown below:. Summary measures Multiple R 0.9689 R-Square 0.9387 Adj R-Square 0.9212 StErr of Estimate 72.218 Regression coefficients Coefficient Std Err t-value p-value Constant 33.796 48.181 0.7014 0.5057 Miles Driven 0.0549 0.0191 2.8666 0.0241 Age of car 21.467 20.573 1.0434 0.3314 Use the information above to estimate the annual maintenance expense for a 10 years old car with 60,000 miles.
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=-1.34 b=36.954 r2=0.966289 r=-0.983 Use this to predict the number of situps a person who watches 5 hours of TV can do (to one decimal place)
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=a+bx b=-0.993 a=29.135 r2=0.463761 r=-0.681 Use this to predict the number of situps a person who watches 6 hours of TV can do. Round to one decimal place.
A regression analysis was performed to determine if there is a relationship between hours of TV watched per day () and number of sit ups a person can do (y). The results of the regression were: y=ax+b a=-1.153 b=22.899 r2=0.649636 r=-0.806 Use this to predict the number of sit ups a person who watches 8 hours of TV can do, and please round your answer to a whole number.
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 = a x + b a = -1.043 b = 29.088 %3D %3! r2 = 0.390625 r = -0.625 Use this to predict the number of situps a person who watches 6.5 hours of TV can do. situps : %3D [one decimal accuracy1
A regression was run to determine if there is a relationship between hours of study per week (xx) and the test scores (yy).The results of the regression were:y=ax+b a=6.686 b=35.26 r2=0.6084 r=0.78 Use this to predict the final exam score of a student who studies 8.5 hours per week, and please round your answer to a whole number.
A regression was run to determine if there is a relationship between hours of TV watched per day (xx) and number of situps a person can do (yy).The results of the regression were:y=ax+b a=-0.629 b=39.045 r2=0.806404 r=-0.898 Use this to predict the number of situps a person who watches 6.5 hour(s) of TV can do, and please round your answer to a whole number.
A regression was run to determine if there is a relationship between hours of study per week (xx) and the final exam scores (yy).The results of the regression were: y=ax+b a=5.614 b=32.79 r2=0.808201 r=0.899 Use this to predict the final exam score of a student who studies 1 hours per week, and please round your answer to a whole number.

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