Question 5
2 answers:
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Answer:
Step-by-step explanation:
Hello!
The variables of interest are:
Y: % of new sparrow hacks.
X: % of sparrow hacks that return after migration.
a. see attachment
b. Using a statistics software I've estimated the regression line:
^Y= 30.29 -0.26X
The second attachment shows scatterplot with the estimated regression line.
c.
The estimated slope is 30.29
30.29% is the estimated average percentage of new sparrows hawk in the colony when the percentage of returned sparrow hawks is zero.
-0.29 is the modification of the estimated average percentage of new sparrow hawks in the colony when the percentage of returned sparrow hawks increases by 1%.
d. To know how does the model fits the regression you have to calculate the determination coefficient.
R²= 0.43
The coefficient of determination gives you an idea of how much of the variability of the dependent variable (Y) is due to the explanatory variables. Its range is from 0 to 1, where zero means that the regression line doesn't fit the data and 1 means that there is a perfect fit.
Now for the estimated model, only 43% of its variability is given by the explanatory variable, this is too low so the data doesn't fit the model.
e. The residues represent the distance between each observation Yi and the regression line, symbolically:
ei= Yi-^Yi
The % of new sparrow hawks that is further away from the regression line is Y=8, the residue corresponding to this value is ei= -5.45
f. You need to predict the percentage of new birds in the colony if the return percentage is 70%, symbolically:
Y/X=70
To do so you need to replace the given value of X is the estimated regression model:
^Y= 30.29 -0.26X
^Y= 30.29 -0.26*10
^Y= 12.09%
I hope this helps!
