Question

Air Force servicemen and servicewomen were tested for maximal oxygen use in a 12 min distance run versus a 1.5 mile run. The participant runs on a level terrain for the prescribed distance. The technicians records the participant’s time to the closet second, the heart rate for 15 s immediately after the participant crosses the finish line and then estimate the VO2 max consumption using a Wilmore and Bergfeld fitness table. Below is listed the data for both the 12 min and the 1.5 mile run with the corresponding VO2 maximum oxygen levels

Data 12 Min Distance (n=25)
x	y
Miles	VO2Max(ml/kg.min)
1.1	23.2
1.15	30
1.2	30
1.15	32.5
1.25	36.3
1.26	37
1.28	40
1.31	32.3
1.3	35
1.38	40
1.4	45.1
1.43	47.5
1.45	50
1.5	46.2
1.6	41
1.65	46.4
1.67	49
1.67	50
1.7	48
1.8	47.5
1.85	51.5
1.9	59.8
2	49.8
2.15	60
2.2	59

Data Two: 12 Min distance vs. VO2 Consumed

9. Write the equation of the least-squares line( line of fit). ________

10. What is the slope of the line of fit?_____________________

11. What is the marginal change for the 1 min of distance to one ml/kg.min. of oxygen consumption?

12. If the 12 min distance is 2.3 miles, what is the amount of oxygen consumed?

Answer 1

> x<-c(1.1,1.15,1.2,1.15,1.25,1.26,1.28,1.31,1.3,1.38,1.4,1.43,1.45,1.5,1.6,1.65,1.67,1.67,1.7,1.8,1.85,1.9,2,2.15,2.2)
> y<-c(23.2,30,30,32.5,36.3,37,40,32.3,35,40,45.1,47.5,50,46.2,41,46.4,49,50,48,47.5,51.5,59.8,49.8,60,59)
> mod1<-lm(y~x)
> summary(mod1)

Call:
lm(formula = y ~ x)

Residuals:
Min 1Q Median 3Q Max
-8.2540 -2.9448 -0.2994 2.7462 8.8444

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.9632 4.3977 0.219 0.829
x 27.7189 2.8099 9.865 9.88e-10 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 4.36 on 23 degrees of freedom
Multiple R-squared: 0.8088,   Adjusted R-squared: 0.8005
F-statistic: 97.31 on 1 and 23 DF, p-value: 9.878e-10

$So,\ eqn\ of\ line : \ \ y = 27.7189x + 0.9632 \\ Slope : 27.7189 \\ Since, \ slope = 27.7189, \ marginal\ change\ for\ 1\ min\ of\ distance\ to\\ one\ ml/kg-min\ of\ oxygen\ is\ 27.7189 \\ Now, \ x=2.3, \ \implies\ y=27.7189*2.3 + 0.9632 = 64.7167$