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

1. Make a scatter diagram for these pairs of data.   Sketch below.  

2. According to the scatter plot, as the distance increases, what happens to the oxygen consumed? _________

3. Does this indicate a positive or negative correlation? ________________

4. Complete the steps to find the correlation coefficient, the line of fit and the values for the linear regression equation.

Answer 1

1)

select data -> insert -> scatterplot

scatterplot y = 27.719x + 0.9632 R2 = 0.8088 70 60 50 40 VO2 Consumed 30 20 10 0 0 0.5 1 1.5 2 2.5 distance

2)
The scatterplot shows an upward trend. thus, as the distance increases, oxygen consumed also increases.

3)
The scatterplot shows an upward trend. thus there is a positive correlation

4)

x	y	x^2	y^2	xy
1.1	23.2	1	538	26
1.15	30	1	900	35
1.2	30	1	900	36
1.15	32.5	1	1056	37
1.25	36.3	2	1318	45
1.26	37	2	1369	47
1.28	40	2	1600	51
1.31	32.3	2	1043	42
1.3	35	2	1225	46
1.38	40	2	1600	55
1.4	45.1	2	2034	63
1.43	47.5	2	2256	68
1.45	50	2	2500	73
1.5	46.2	2	2134	69
1.6	41	3	1681	66
1.65	46.4	3	2153	77
1.67	49	3	2401	82
1.67	50	3	2500	84
1.7	48	3	2304	82
1.8	47.5	3	2256	86
1.85	51.5	3	2652	95
1.9	59.8	4	3576	114
2	49.8	4	2480	100
2.15	60	5	3600	129
2.2	59	5	3481	130

sum x	sum y	sum x^2	sum y^2	sum xy	n
38.35	1087.1	61	49559	1734	25

slope, b1=   {n*sum(xy) - sum(x)*sum(y)} / {n*sum(x^2) - [sum(x)]^2 }
b1 =    {25*1734.353 - 38.35*1087.1} / {25*61.2367 - [38.35]^2 }
b1 =    27.7189
  
  
intercept, bo=   {sum(y)*sum(x^2) - sum(x)*sum(xy)} / {n*sum(X^2) - [sum(X)]^2}
bo =    {1087.1*61.2367 - 38.35*1734.353} / {25*61.2367 - [38.35]^2}
bo =    1.0
  
  
regression equation:  
y = bo + b1*x  
y = 0.9632 + 27.7189*x  
  
corr coef, r =    (n*sum(XY) - sum(X)*sum(Y)) / ( sqrt(n*sum(x^2)-[sum(x)]^2) * sqrt(n*sum(y^2)-[sum(y)]^2) )
corr coef, r =    (25*1734.353-38.35*1087.1) / ( sqrt(25*61.2367-[38.35]^2) * sqrt(25*49558.71-[1087.1]^2) )
corr coef, r =    0.899351