a) Are the body mass leanness and heat-loss ratecorrelated? Explain your reasoning. b) Calculate the standard error of the correlation coefficient; what does this value represent?Using the standard error how do you report (write) your results in a scientific paper? 3. To investigate whether subcutaneous fat provides insulation in humans, Sloan and Leatinge (1973) measured therate of heat loss by boys swimming for up to 40 minutes in water at 20.3 °C and expending energy at about 4.8kcal/min. Heat loss was measured by the change in body temperature, recorded using a thermometer under the tongue,divided by time spent swimming, in minutes. The authors measured an index of body "leanness" on 12 boys as thereciprocal of the skin-fold thickness adjusted for total skin surface area (in meters squared) and body mass (in kg).X =3.96667Y = 0.04833E (X, -X) = 41.14667E(Y, -Y)' =0.01696E (X, -X)(Y, – Y)=0.78053

Given X¯=3.96667, Y¯=0.04833 ∑Xi-X¯2=41.14667 ∑Yi-Y¯2=0.01696…

Answered: a) Are the body mass leanness and…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

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a) Are the body mass leanness and heat-loss ratecorrelated? Explain your reasoning.

b) Calculate the standard error of the correlation coefficient; what does this value represent?Using the standard error how do you report (write) your results in a scientific paper?

3. To investigate whether subcutaneous fat provides insulation in humans, Sloan and Leatinge (1973) measured the
rate of heat loss by boys swimming for up to 40 minutes in water at 20.3 °C and expending energy at about 4.8
kcal/min. Heat loss was measured by the change in body temperature, recorded using a thermometer under the tongue,
divided by time spent swimming, in minutes. The authors measured an index of body "leanness" on 12 boys as the
reciprocal of the skin-fold thickness adjusted for total skin surface area (in meters squared) and body mass (in kg).
X =3.96667
Y = 0.04833
E (X, -X) = 41.14667
E(Y, -Y)' =0.01696
E (X, -X)(Y, – Y)=0.78053

Expert Solution

Step 1

Given

$\bar{X} = 3.96667, \bar{Y} = 0.04833$

$\sum {(X_{i} - \bar{X})}^{2} = 41.14667$ $\sum {(Y_{i} - \bar{Y})}^{2} = 0.01696$ $\sum (X_{i} - \bar{X}) (Y_{i} - \bar{Y}) = 0.78053$

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