A group of students want to determine if a person's height is linearly related to the distance they are able to jump.
To determine the relationship between a person's height and the distance they are able to jump, the group of students measured the height, in inches, of each person in their class and then measured the distance, in feet, they were able to jump from a marked starting point.
Each student was given three tries at the jump and their longest jump distance was recorded. The data the students collected is shown below.
Height (in.) Jump Distance
(m)
59
5.4
60
5.2
65
6.5
74
6.6
72
6.9
66
6.6
63
6.0
70
6.8
61
5.5
62
5.9
64
6.1
65
6.0
67
6.7
60
5.7
68
6.8
67
6.5
Use a form of technology to compute the correlation coefficient, r,
for the linear fit between the person's height and the distance they were able to jump, where rxy=∑i=1n(xi−x¯¯¯)(yi−y¯¯¯)∑i=1n(xi−x¯¯¯)2∑i=1n(yi−y¯¯¯)2⎷ and n is the number of students and x represents the person's height and y
represents the distance they were able to jump.
Enter the correlation coefficient. Round your answer to the nearest hundredth.
So what’s the answer
whats the answer??
To compute the correlation coefficient, we need to calculate the following values:
1. Find the mean (average) of the height (x) values:
x̄ = (59 + 60 + 65 + 74 + 72 + 66 + 63 + 70 + 61 + 62 + 64 + 65 + 67 + 60 + 68 + 67) / 16 = 65.9375
2. Find the mean (average) of the jump distance (y) values:
ȳ = (5.4 + 5.2 + 6.5 + 6.6 + 6.9 + 6.6 + 6.0 + 6.8 + 5.5 + 5.9 + 6.1 + 6.0 + 6.7 + 5.7 + 6.8 + 6.5) / 16 = 6.21875
3. Calculate the sum of the squared differences from the mean for both height (x) and jump distance (y):
∑(xi - x̄)² = (59 - 65.9375)² + (60 - 65.9375)² + (65 - 65.9375)² + (74 - 65.9375)² + (72 - 65.9375)² + (66 - 65.9375)² + (63 - 65.9375)² + (70 - 65.9375)² + (61 - 65.9375)² + (62 - 65.9375)² + (64 - 65.9375)² + (65 - 65.9375)² + (67 - 65.9375)² + (60 - 65.9375)² + (68 - 65.9375)² + (67 - 65.9375)²
∑(xi - x̄)² ≈ 228.75
∑(yi - ȳ)² = (5.4 - 6.21875)² + (5.2 - 6.21875)² + (6.5 - 6.21875)² + (6.6 - 6.21875)² + (6.9 - 6.21875)² + (6.6 - 6.21875)² + (6.0 - 6.21875)² + (6.8 - 6.21875)² + (5.5 - 6.21875)² + (5.9 - 6.21875)² + (6.1 - 6.21875)² + (6.0 - 6.21875)² + (6.7 - 6.21875)² + (5.7 - 6.21875)² + (6.8 - 6.21875)² + (6.5 - 6.21875)²
∑(yi - ȳ)² ≈ 2.01875
4. Calculate the sum of the product of the differences from the mean for height (x) and jump distance (y):
∑(xi - x̄)(yi - ȳ) = (59 - 65.9375)(5.4 - 6.21875) + (60 - 65.9375)(5.2 - 6.21875) + (65 - 65.9375)(6.5 - 6.21875) + (74 - 65.9375)(6.6 - 6.21875) + (72 - 65.9375)(6.9 - 6.21875) + (66 - 65.9375)(6.6 - 6.21875) + (63 - 65.9375)(6.0 - 6.21875) + (70 - 65.9375)(6.8 - 6.21875) + (61 - 65.9375)(5.5 - 6.21875) + (62 - 65.9375)(5.9 - 6.21875) + (64 - 65.9375)(6.1 - 6.21875) + (65 - 65.9375)(6.0 - 6.21875) + (67 - 65.9375)(6.7 - 6.21875) + (60 - 65.9375)(5.7 - 6.21875) + (68 - 65.9375)(6.8 - 6.21875) + (67 - 65.9375)(6.5 - 6.21875)
∑(xi - x̄)(yi - ȳ) ≈ -1.103125
5. Calculate the correlation coefficient:
r = (∑(xi - x̄)(yi - ȳ)) / √(∑(xi - x̄)² * ∑(yi - ȳ)²)
r = -1.103125 / √(228.75 * 2.01875) ≈ -0.2644
Therefore, the correlation coefficient (r) for the linear fit between a person's height and the distance they were able to jump is approximately -0.26.
To find the correlation coefficient, we will use the formula:
rxy = ∑(xi - x̄)(yi - ȳ) / √[∑(xi - x̄)² * ∑(yi - ȳ)²]
where:
- rxy is the correlation coefficient
- xi represents the person's height
- x̄ represents the mean of the person's height
- yi represents the distance they were able to jump
- ȳ represents the mean of the distance they were able to jump
First, let's calculate the means of the height (xi) and the jump distance (yi):
x̄ = (59 + 60 + 65 + 74 + 72 + 66 + 63 + 70 + 61 + 62 + 64 + 65 + 67 + 60 + 68 + 67) / 16 = 64.25 (rounded to two decimal places)
ȳ = (5.4 + 5.2 + 6.5 + 6.6 + 6.9 + 6.6 + 6.0 + 6.8 + 5.5 + 5.9 + 6.1 + 6.0 + 6.7 + 5.7 + 6.8 + 6.5) / 16 = 6.29 (rounded to two decimal places)
Next, let's calculate the numerator of the correlation coefficient formula:
∑(xi - x̄)(yi - ȳ) = (59 - 64.25)(5.4 - 6.29) + (60 - 64.25)(5.2 - 6.29) + ...
= (-5.25)(-0.89) + (-4.25)(-1.09) + ...
Calculating the sum for all the data points, we get:
∑(xi - x̄)(yi - ȳ) = -4.6725 + 4.6225 + ...
Now, let's calculate the denominator:
∑(xi - x̄)² = (59 - 64.25)² + (60 - 64.25)² + ... = (-5.25)² + (-4.25)² + ...
∑(yi - ȳ)² = (5.4 - 6.29)² + (5.2 - 6.29)² + ... = (-0.89)² + (-1.09)² + ...
Finally, let's substitute the values into the correlation coefficient formula and calculate the result:
rxy = ∑(xi - x̄)(yi - ȳ) / √[∑(xi - x̄)² * ∑(yi - ȳ)²]
After calculating the numerator and denominator, divide the numerator by the denominator, and take the square root of the result. Round the final answer to the nearest hundredth.
Once you have performed the necessary calculations, you can enter the correlation coefficient.