Reference no: EM131723574

Question:

Below is how I solved a problem to "determine the regression formula" to predict maximal oxygen consumption (VO2max) from scores on a 12 min. run. The information given and how I solved for a and b are shown below.

The next question says, "Using the prediction formulas developed in problem 3, what is the predicted VO2max for a participant who ran 2,954 m in 12 min?"

If X stays the same (12 min) but the mean of X changes from 2,853 m to 2,954 m, wouldn't I have to also recalculate the standard deviation and the correlation (r)?

If so, how do I do that when I do not have N or the original data?

Information given

X (12 min run)

M_x (mean of X) = 2,853 m

s_x (Standard Deviation of X) = 305 m

r (correlation) = .79

Y = V·O 2max

M_y (mean of Y) = 52.6 ml·kg·min^-1

s_y (Standard Deviation of Y) = 6.3 ml·kg·min^-1

Solving for b

b = r (s_r/ s_x)

r (correlation) = .79

s_y (Standard Deviation of Y) = 6.3 ml·kg·min^-1

s_x (Standard Deviation of X) = 305 m

b=0.79 * (6.3/305)

b = 0.0163

Solving for a

a = M_y - b (M_x)

M_y (mean of Y) = 52.6 ml·kg·min^-1

M_x (mean of X)= 2,853 m

a = 52.6 - (0.0163)* (2,853)

a = 52.6 - 46.5039

a = 6.0961

Solving for Y

Y = a + bX

a = 6.0961

b = 0.0163

X = 12 min.

Y = 6.0961 + 0.0163(12)

=6.2917 V·O 2max