Listed below are altitudes (thousands of feet) and outside air temperatures (°F) recorded during a flight. Find the (a)explained variation, (b) unexplained variation, and (c) indicated prediction interval. There is sufficient evidence to support aclaim of a linear correlation, so it is reasonable to use the regression equation when making predictions. For theprediction interval, use a 95% confidence level with the altitude of 6327 ft (or 6.327 thousand feet).AltitudeTemperaturea. Find the explained variation.25584010370.30(Round to two decimal places as needed.)b. Find the unexplained variation.53.70845(Round to five decimal places as needed.)c. Find the indicated prediction interval.°F

Answered: Image /qna-images/answer/2daa8274-d91e-43a6-98c9-35c90e9a2b42.jpg

Answered: Listed below are altitudes (thousands…

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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Listed below are amounts of court income and salaries paid to the town justices for a certain town. All amounts are in thousands of dollars. Find the (a) explained variation, (b) unexplained variation, and (c) indicated prediction interval. There is sufficient evidence to support a claim of a linear correlation, so it is reasonable to use the regression equation when making predictions. For the prediction interval, use a 99% confidence level with a court income of $800,000. $68 $31 Court Income $415 $1592 $1150 $268 $252 $117 $163 $28 Justice Salary $47 $95 $52 $40 $55 $26 $26 $15 a. Find the explained variation. (Round to three decimal places as needed.)
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The table below lists measured amounts of redshift and the distances (billions of light-years) to randomly selected astronomical objects. Find the (a) explained variation, (b) unexplained variation, and (c) indicated prediction interval. There is sufficient evidence to support a claim of a linear correlation, so it is reasonable to use the regression equation when making predictions. For the prediction interval, use a 90% confidence leve with a redshift of 0.0126. Redshift Distance 0.0231 0.31 a. Find the explained variation. 0.0536 0.77 (Round to six decimal places as needed.) b. Find the unexplained variation. (Round to six decimal places as needed.) c. Find the indicated prediction interval. 0.0716 0.0395 1.01 0.53 billion light-years
The accompanying table lists overhead widths (cm) of seals measured from photographs and the weights (kg) of the seals. Find the (a) explained variation, (b) unexplained variation, and (c) prediction interval for an overhead width of 8.9 cm using a 99% confidence level. There is sufficient evidence to support a claim of a linear correlation, so it is reasonable to use the regression equation when making predictions. Click the icon to view the seal data. a. The explained variation is (Round to the nearest integer as needed.) b. The unexplained variation is. (Round to the nearest integer as needed.) c. The 99% prediction interval for an overhead width of 8.9 cm is kg
A researcher has developed a regression model from fourteen pairs of data points. He wants to test to determine if the slope is significantly different from zero. He uses a two-tailed test and a = 0.01. The critical tablet value is 2.718 3.012 2.650 O 3.055 O 2.168
Listed below are systolic blood pressure measurements (in mm Hg) obtained from the same woman. Find the regression equation, letting the right arm blood pressure be the predictor (x) variable Find the best predicted systolic blood pressure in the left arm given that the systolic blood pressure in the right arm is 95 mm Hg. Use a significance level of 0.05 Right Arm 103 102 94 79 80 D Left Arm 176 170 144 146 145 : Click the icon to view the critical values of the Pearson correlation coefficient r The regression equation is y= +x (Round to one decimal place as needed.) ater/ Enter your ahswer in the edit fields and then click Check Answer
Listed below are altitudes (thousands of feet) and outside air temperatures (°F) recorded during a flight. Find the (a) explained variation, (b) unexplained variation, and (c) indicated prediction interval. There is sufficient evidence to support a claim of a linear correlation, so it is reasonable to use the regression equation when making predictions. For the prediction interval, use a 95% confidence level with the altitude of 6327 ft (or 6.327 thousand feet). Altitude Temperature a. Find the explained variation. (Round to two decimal places as needed.) 2 55 8 40 13 25 20 - 3 28 - 26 31 - 41 34 - 53
A regression analysis was performed to predict weight (y, in kg) using height (x, in cm) among 150 children. The coefficient of determination was . Which of the following is a valid interpretation? a. For each 1-cm increase in height, weight tends to increase by about 0.32 kg b. There is no association between weight and height c. Height accounts for about 32% of the total variability in weight d. The correlation between weight and height is about 0.32
The table below lists measured amounts of redshift and the distances (billions of light-years) to randomly selected astronomical objects. There is sufficient evidence to support a claim of a linear correlation, so it is reasonable to use the regression equation when making predictions. For the prediction interval, use a 90% confidence level with a redshift of 0.0126. Find the explained variation. Redshift Distance OA, 0.157904 OB. 0.444904 OC. 0.000582 OD. 0.219582 0.0231 0.34 0.0536 0.76 0.0715 0.99 0.0391 0.56 0.0438 0.63 0.0109 0.15
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.
Listed below are systolic blood pressure measurements (in mm Hg) obtained from the same woman. Find the regression equation, letting the right arm blood pressure be the predictor (x) variable. Find the best predicted systolic blood pressure in the left arm given that the systolic blood pressure in the right arm is 90 mm Hg. Use a significance level of 0.05. Right Arm 101 100 92 75 75 O Left Arm 174 167 181 149 147 Click the icon to view the critical values of the Pearson correlation coefficient r The regression equation is y = + x. (Round to one decimal place as needed.) Given that the systolic blood pressure in the right arm is 90 mm Hg, the best predicted systolic blood pressure in the left arm is mm Hg. (Round to one decimal place as needed.)
Listed below are systolic blood pressure measurements (in mm Hg) obtained from the same woman. Find the regression equation, letting the right arm blood pressure be the predictor (x) variable. Find the best predicted systolic blood pressure in the left arm given that the systolic blood pressure in the right arm is 80 mm Hg. Use a significance level of 0.05. Right Arm 100 99 93 77 77 Q Left Arm 174 168 148 148 146 Click the icon to view the critical values of the Pearson correlation coefficient r The regression equation is ŷ=+x. (Round to one decimal place as needed.) mm Hg. Given that the systolic blood pressure in the right arm is 80 mm Hg, the best predicted systolic blood pressure in the left arm is (Round to one decimal place as needed.) Data table Critical Values of the Pearson Correlation Coefficient r α = 0.05 α = 0.01 0.950 0.990 0.959 0.878 0.811 0.917 0.754 0.875 0.707 0.834 0.666 0.798 0.632 0.765 0.602 0.735 0.576 0.708 0.553 0.684 0.532 0.661 0.514 0.641 0.497 0.623 0.482…

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