Question

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Cardiarespiratry fitness is wicely recognized as a major camponent of overall physical well-being. Direct measurement consuming and expensive. It is therefore desirable to have a prediction equation for VO2max in terms of easily obtained quantities. Consider the variables f maximal oxygen uptake (VOmax) is the single best measure of such fitness, but direct measurement is time- vo2max (L/min) X1weight (kg) x age (yr) x time necessary to walk 1 mile (min) X4 heart rate at the end of the walk (beats/m in ) Consider the model of the entire population Here is one possible model for male students: Y5.0D.012x1 0.05x - 0.D8x - 0.4 0.01x (a) Interpret Holding all ether variables constant, a 1 kg increase in weight will result a D.012 L/min increase in VOzmax Holding all ether variables constant, a 0.012 kg increase in weight will result in a 1 L/min increase in VOzmax. Holding all other variables constant, a 1 kg increase in weight will result a 0.012 Lmin decrease in VO2max. Holding all other variables constant, a 0.012 kg increase in weight will result in a 1 L'min decrease in VOmax. Interpret Holding all other variables constant, a 1 min increase walk time will result in a 0.08 Lmin decrease in VOzmax Hold ng all other variables constant, a 0.08 min increase in walk time will result in a 1 Lmin increase in VO)max. Hold ng all ether variables constant, walk tme wll result in a 0.08 Lmin increase in VO2max. 1 min increase Hold ng all other variables constant. o 08 min increase in walk time "i result in a L'min decrease in VO,max 134 b/m (b What is the expecbed value 79 kg, age is 24 yr, walk time is 12 min, and beart rate VOmax, when weight Lfmin (c) what is the probability vo2max is between 1.928 and 2.968, when welght is 79 kg, age is 24 yr, walk time is 12 min, and heart rate is 134 b/m? Hint: The response varles normally with a mean of and a standard deviation of a.

Answer 1

In a regression equation, the coefficient of a predictor can be interpreted in such a way that if predictor is X_i and its coefficient is a_i , then holding all other variables constant, a 1 unit increase in X_i is going to cause a_i units change in Y(response variable).

(a) Interpret  ,

31

Holding all other variables constant, a 1 kg increase in  weight will result in a 0.012 L/min increase in VO₂max.

Interpret  $\beta_3$,

Holding all other variables constant, a 1 min increase in walk time will result in a 0.08 L/min decrease in VO₂max.

(b) x₁ = 79 , x₂ = 24 , x₃ = 12 , x₄ = 134

Expected value of Y in this case is

y5.00.012 x 79 0.05 x 24 0.08 x 12 0.01 x 134

= 2.448

(c) When x₁ = 79 , x₂ = 24 , x₃ = 12 , x₄ = 134 , the Y is normally distributed

Y ~ N(
lY
= 2.448 , $\sigma$ = 0.4)

We need to find,

P[ 1.928 < Y < 2.968 ]

= P[ (1.928 -2.448)/4 < (Y - 2.448)/4 < (2.968 - 2.448)/4]

= P[ -0.13 < Z < 0.13] ,{where Z = (Y - 2.448)/4 ~ N(0,1) }

=

D(0.13) D(-0.13

= {since, Z is symmetrically distributed}

D(0.13) 1 (0.13)

=

2D(0.13) 1

= 2 x 0.55172 - 1 {value is obtained from a standard normal table}

= 0.10344