The toco toucan, the largest member of the toucan family, possesses the largest beak relative to body size of all birds. This exaggerated feature has received various interpretations, such as being a refined adaptation for feeding. However, the large surface area may also be an important mechanism for radiating heat (and hence cooling the bird) as outdoor temperature increases. Here are data for beak heat loss, as a percent of total body heat loss, at various temperatures in degrees Celsius:
Temperature( o C)
15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Percent heat loss from beak 34 33 33 36 37 45 55 54 41 53 43 53 60 61 62 65
The equation of the least-squares regression line for predicting beak heat loss, as a percent of total body heat loss from all sources, from temperature is
y^=____ +____ x
(Use decimal notation. Give your answer to four decimal places.)
Use the equation to predict (±0.01)beak heat loss, as a percent of total body heat loss from all sources, at a temperature of 25 degrees Celsius.
____%
What percent (±0.01) of the variation in beak heat loss is explained by the straight-line relationship with temperature?
_____%
Find the correlation r
r(± 0.001) between beak heat loss and temperature:
r=_____
y^=44.9072 + 0.9072x
44.90%
87.90%
r=0.945
To find the equation of the least-squares regression line, we can use the given data. The equation is in the form y^ = mx + b, where y^ represents the predicted beak heat loss, x represents the temperature, m represents the slope, and b represents the y-intercept.
1. First, calculate the mean of the temperature values:
Mean of temperature = (15 + 16 + 17 + ... + 29 + 30) / 16 = 22.5
2. Calculate the mean of the beak heat loss values:
Mean of beak heat loss = (34 + 33 + 33 + ... + 62 + 65) / 16 = 45.1875
3. Calculate the differences from the mean for both temperature and beak heat loss:
Temperature deviations: (15 - 22.5), (16 - 22.5), (17 - 22.5), ... , (29 - 22.5), (30 - 22.5)
Beak heat loss deviations: (34 - 45.1875), (33 - 45.1875), (33 - 45.1875), ... , (62 - 45.1875), (65 - 45.1875)
4. Calculate the product of the temperature deviations and beak heat loss deviations:
Product = (15 - 22.5)(34 - 45.1875) + (16 - 22.5)(33 - 45.1875) + (17 - 22.5)(33 - 45.1875) + ... + (29 - 22.5)(62 - 45.1875) + (30 - 22.5)(65 - 45.1875)
5. Calculate the squared deviations for temperature:
Temperature squared deviations: (15 - 22.5)^2, (16 - 22.5)^2, (17 - 22.5)^2, ... , (29 - 22.5)^2, (30 - 22.5)^2
6. Calculate the squared deviations for beak heat loss:
Beak heat loss squared deviations: (34 - 45.1875)^2, (33 - 45.1875)^2, (33 - 45.1875)^2, ... , (62 - 45.1875)^2, (65 - 45.1875)^2
7. Calculate the sum of the squared deviations for temperature and beak heat loss:
Sum of temperature squared deviations = (15 - 22.5)^2 + (16 - 22.5)^2 + (17 - 22.5)^2 + ... + (29 - 22.5)^2 + (30 - 22.5)^2
Sum of beak heat loss squared deviations = (34 - 45.1875)^2 + (33 - 45.1875)^2 + (33 - 45.1875)^2 + ... + (62 - 45.1875)^2 + (65 - 45.1875)^2
8. Calculate the sum of the cross-products:
Sum of cross-products = (15 - 22.5)(34 - 45.1875) + (16 - 22.5)(33 - 45.1875) + (17 - 22.5)(33 - 45.1875) + ... + (29 - 22.5)(62 - 45.1875) + (30 - 22.5)(65 - 45.1875)
9. Calculate the slope (m):
m = Sum of cross-products / Sum of temperature squared deviations
10. Calculate the y-intercept (b):
b = Mean of beak heat loss - m * Mean of temperature
11. Substitute the slope (m) and y-intercept (b) values into the equation y^ = mx + b in decimal notation.
y^ = __________ + __________ x
To predict the beak heat loss at a temperature of 25 degrees Celsius, substitute the value of x = 25 into the equation you derived. Round the result to two decimal places for the final answer.
The percent of the variation in beak heat loss explained by the straight-line relationship with temperature is determined by the correlation coefficient (r).
r = __________ (± 0.001)