A sports medicine major wanted to conduct an experiment to determine if there is a correlation between the members of the soccer team's leg strength and the time it takes for them to sprint 40 yards. He sets up the following test and records the data:
Every day for a week, he counts how many times each player can leg press 350 pounds. The following week, he has each player sprint 40 yards every day. The table shows the average number of leg-press repetitions and the average 40-yard dash time (in seconds) for seven randomly selected players. What is the equation of the line of best fit? How many seconds should he expect a player to take to run 40 yards if that player can do 22 leg-press repetitions? Round any values to the nearest tenth, if necessary.
Leg Press (reps) 12
32
11 23
28 15
40-yard Dash (s) 8.6 14.6 7.1 8.3 11.9 13.4 9.5
To determine the equation of the line of best fit, we can use linear regression analysis. Using the given data, we calculate the equation as follows:
Sum of x values (leg press repetitions): 12 + 32 + 11 + 23 + 28 + 15 = 121
Sum of y values (40-yard dash times): 8.6 + 14.6 + 7.1 + 8.3 + 11.9 + 13.4 + 9.5 = 73.4
Sum of x^2 values: 12^2 + 32^2 + 11^2 + 23^2 + 28^2 + 15^2 = 4,240
Sum of xy values: (12 * 8.6) + (32 * 14.6) + (11 * 7.1) + (23 * 8.3) + (28 * 11.9) + (15 * 13.4) = 1,692
In terms of the equation y = mx + b, where y represents the 40-yard dash time and x represents the leg press repetitions, we have the following equations:
m (slope) = (nΣxy - ΣxΣy) / (nΣx^2 - (Σx)^2)
b (y-intercept) = (Σy - mΣx) / n
Using the given data, we can calculate the values:
m = ((6 * 1,692) - (121 * 73.4)) / ((6 * 4,240) - (121)^2) ≈ -0.048
b = (73.4 - (-0.048 * 121)) / 6 ≈ 10.076
Therefore, the equation of the line of best fit is:
y ≈ -0.048x + 10.076
To estimate the time it should take for a player to run 40 yards if they can do 22 leg press repetitions, we plug in x = 22 into the equation:
y ≈ -0.048(22) + 10.076 ≈ 8.6
Therefore, the expected time for a player who can do 22 leg press repetitions is approximately 8.6 seconds.