Ke thisUse the given data set to complete parts (a) through (c) below. (Use a = 0.05)98.76Xy109.13Click here to view a table of critical values for the correlation coefficient.OAAy10-84D6-4-2+40+a. Construct a scatterplot. Choose the correct graph below.0yo888.1512 16Q15 2OG138.73View an example Get more help.4017B.4-Ay10-8-6-4-2-0-0000004812 164+Q5119.25fgI◄◄99+Hf10CHED148.09a▶II108-6-42-0-b. Find the linear correlation coefficient, r, then determine whether there is sufficient evidence to support the claim of a linear correlation between the two variables.The linear correlation coefficient is r=(Round to three decimal places as needed.)0fil00000466.128DI12 16Qf12343.11insprt sc129.13Ay10-8-64-2+0+toodelete8 12 16Clear all77.25Qhome54.74Check answer7:37 PM4/27/2023end19Opg uppo

The data shows x and y variable with 11 observations.

Answered: b. Find the linear correlation…

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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Which of the following correlation coefficient would reflect a conclusion made by a researcher that two variables are not related? r = –.30 r = +.02 r = +.36 r = –.98
find the linear correlation coefficent and the critical value(s)
Match the following scatterplots to their correct correlation coefficients (r). 1. r= -0.852. r= -0.403. r= 04. r= 0.505. r= 0.906. r= 0.99
For a data set of chest sizes (distance around chest in inches) and weights (pounds) of eight anesthetized bears that were measured, the linear correlation coefficient is r = 0.996. Use the table available below to find the critical values of Ⓡ r. Number of Pairs of Data n 4 5 6 7 8 9 10 11 12 OA.-0.707 OB. +0.707 OC.-0.707, +0.707 OD. +0.754 Critical Value of r 0.950 0.878 0.811 0.754 0.707 0.666 0.632 0.602 0.576
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This is a continuation of the last question I posted: f. The value of linear correlation coefficient r is i. 0.948 ii. 0.974 iii. -0.948 iv. -0.974 g.Based on your answer in part (f), the linear dependance between weekly exercise time and change in cholesterol level is strong negative i. True ii. False H. When weekly exercise time is 460 minutes (approximately 66 minutes per day), the 95 percent confidence interval for change in cholesterol level is (197.3, 225.4). This means that: i. If one person exercises 460 minutes per week, his/hers change in cholesterol level will be between 197.3 and 225.4 with 95 percent certainty.ii. If several people exercise 460 minutes per week, their average change in choles- terol level will be between 197.3 and 225.4 with 95 percent certainty. I. When weekly exercise time is 460 minutes (approximately 66 minutes per day), the 95 percent prediction interval for change in cholesterol level is (155.1, 267.6). This means that i. If one person…
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Listed below are systolic blood pressure measurements ( in mm Hg ) from the same woman. Right arm 102 101 94 79 79 Left arm 175 169 182 146 144 a) Find the linear coefficient correlation r. What is the critical value in this case ?? ? b) Is there sufficient evidence to conclude that there is a linear correlation between right and left arm systolic blood pressure measurements? c) Find the equation of the regression line between right and left arm systolic. d) What proportion of left arm systolic can be explained by the right arm systolic ? e) Find the best predicted systolic blood pressure in the left arm given that the systolic blood pressure in the right arm is 100 mm Hg.

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b. Find the linear correlation coefficient, r, then determine whether there is sufficient evidence to support the claim of a linear correlation between the two variables. The linear correlation coefficient is r= (Round to three decimal places as needed.).