The data below were gathered on a random sample of 5 basking sharks, swimming through the water and filter-feeding, i.e. letting the water bring food into their mouths.
Body Length (meters) Mean Speed (meters/seconds)
4 0.89
4.5 0.83
4 0.76
6.5 0.94
5.5 0.94
a) If these sharks are representative of the population of basking sharks, what would you predict is the mean speed for a filter-feeding basking shark that is 5.0 meters in length? Show any calculations below.
b) The largest basking shark in the sample is measured as 6.5 meters long. Theory predicts a maximum length of about 12.26 meters. Would it be reasonable to use the equation from part (a) above to predict the mean filter-feeding speed for a 12 meter long basking shark? Why or why not?
a) To predict the mean speed for a filter-feeding basking shark that is 5.0 meters in length, we can use the given data to calculate the linear regression equation. The linear regression equation is in the form of:
y = mx + c
Where:
y = dependent variable (mean speed)
x = independent variable (body length)
m = slope of the line
c = y-intercept
Let's calculate the equation:
Step 1: Calculate the mean of body length and mean speed.
Mean of body length (x̄) = (4 + 4.5 + 4 + 6.5 + 5.5) / 5 = 24.5 / 5 = 4.9
Mean of mean speed (ȳ) = (0.89 + 0.83 + 0.76 + 0.94 + 0.94) / 5 = 4.36 / 5 = 0.872
Step 2: Calculate the sum of the products of the deviations of body length and mean speed from their respective means.
Sum of (x - x̄)(y - ȳ) = (4 - 4.9)(0.89 - 0.872) + (4.5 - 4.9)(0.83 - 0.872) + (4 - 4.9)(0.76 - 0.872) + (6.5 - 4.9)(0.94 - 0.872) + (5.5 - 4.9)(0.94 - 0.872)
= (-0.9)(0.018) + (-0.4)(-0.042) + (-0.9)(-0.112) + (1.6)(0.068) + (0.6)(0.068)
= -0.0162 + 0.0168 + 0.1008 + 0.1088 + 0.0408
= 0.250
Step 3: Calculate the sum of the squared deviations of body length.
Sum of (x - x̄)^2 = (4 - 4.9)^2 + (4.5 - 4.9)^2 + (4 - 4.9)^2 + (6.5 - 4.9)^2 + (5.5 - 4.9)^2
= (-0.9)^2 + (-0.4)^2 + (-0.9)^2 + (1.6)^2 + (0.6)^2
= 0.81 + 0.16 + 0.81 + 2.56 + 0.36
= 4.7
Step 4: Calculate the slope (m).
m = Sum of (x - x̄)(y - ȳ) / Sum of (x - x̄)^2
m = 0.250 / 4.7
m ≈ 0.05319 (rounded to five decimal places)
Step 5: Calculate the y-intercept (c).
c = ȳ - mx̄
c = 0.872 - (0.05319)(4.9)
c ≈ 0.872 - 0.2631
c ≈ 0.6089 (rounded to four decimal places)
Therefore, the linear regression equation is:
y ≈ 0.05319x + 0.6089
Now we can use this equation to predict the mean speed for a filter-feeding basking shark that is 5.0 meters in length:
Predicted mean speed (y) = 0.05319(5.0) + 0.6089
Predicted mean speed (y) ≈ 0.26595 + 0.6089
Predicted mean speed (y) ≈ 0.87485
b) No, it would not be reasonable to use the equation from part (a) to predict the mean filter-feeding speed for a 12-meter long basking shark. The linear regression equation we obtained in part (a) is based on the data from a random sample of 5 basking sharks with body lengths ranging from 4 to 6.5 meters. Extrapolating this equation to predict the speed of a 12-meter long shark would be beyond the range of the data and involve significant uncertainty. It is always best to use regression equations within the range of their data to ensure the predictions are reliable.
a) To predict the mean speed for a 5.0-meter-long basking shark, we can use linear regression to estimate the relationship between body length and mean speed based on the given data.
Step 1: Calculate the mean of the body lengths and the mean of the speeds:
Mean body length = (4 + 4.5 + 4 + 6.5 + 5.5) / 5 = 24.5 / 5 = 4.9 meters
Mean speed = (0.89 + 0.83 + 0.76 + 0.94 + 0.94) / 5 = 4.36 / 5 = 0.87 meters/second
Step 2: Calculate the deviations from the means for body lengths and speeds:
Length deviation = (5.0 - 4.9) = 0.1 meter
Speed deviation = (0.87 - 0.87) = 0
Step 3: Calculate the slope of the regression line:
Slope = Covariance of lengths and speeds / Variance of lengths
Covariance = Sum of [(length - mean length) * (speed - mean speed)] / (number of data points - 1)
Covariance = [(4 - 4.9) * (0.89 - 0.87) + (4.5 - 4.9) * (0.83 - 0.87) + (4 - 4.9) * (0.76 - 0.87) + (6.5 - 4.9) * (0.94 - 0.87) + (5.5 - 4.9) * (0.94 - 0.87)] / 4
Covariance = (-0.09 * 0.02 + (-0.4) * (-0.04) + (-0.09) * (-0.11) + (1.6) * (0.07) + (0.6) * (0.07)) / 4
Covariance = (-0.0018 + 0.016 + 0.0099 + 0.112 + 0.042) / 4
Covariance = 0.1781 / 4 = 0.0445 meters/sec
Variance = Sum of (length - mean length)^2 / (number of data points - 1)
Variance = [(4 - 4.9)^2 + (4.5 - 4.9)^2 + (4 - 4.9)^2 + (6.5 - 4.9)^2 + (5.5 - 4.9)^2] / 4
Variance = (0.81 + 0.16 + 0.81 + 2.56 + 0.36) / 4
Variance = 4.7 / 4 = 1.175 square meters
Slope = 0.0445 / 1.175 = 0.0379 meters/second per meter
Step 4: Calculate the y-intercept of the regression line:
y-intercept = mean speed - slope * mean length
y-intercept = 0.87 - 0.0379 * 4.9
y-intercept = 0.87 - 0.18971
y-intercept = 0.68029 meters/second
Step 5: Use the regression line equation to predict the mean speed for a 5.0-meter-long basking shark:
Predicted mean speed = y-intercept + slope * 5.0
Predicted mean speed = 0.68029 + 0.0379 * 5.0
Predicted mean speed = 0.68029 + 0.1895
Predicted mean speed = 0.86979 meters/second (rounded to 2 decimal places)
Therefore, the predicted mean speed for a filter-feeding basking shark that is 5.0 meters in length is approximately 0.87 meters/second.
b) It would not be reasonable to use the equation from part (a) to predict the mean filter-feeding speed for a 12-meter-long basking shark. This is because the equation is based on the data of basking sharks with body lengths ranging from 4 to 6.5 meters. Extrapolating the equation to predict the speed of a 12-meter-long shark is beyond the range of the available data and may introduce significant errors. It is always preferable to make predictions within the range of the data used to derive the equation.