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Answer :
Final answer:
The residual is calculated by finding the difference between the predicted age from the regression model and the actual age. For the lion with an 11% black nose, the residual is approximately 0.151, indicating that the model overestimated the age.
Explanation:
The question is based on residuals in regression analysis. Here, we're given the linear regression model y hat = 0.8790 + 10.6471x, where x is the proportion of the lion's nose that's black, and y hat is the predicted age of a lion. We also know the actual age of a lion with 11% black on its nose - 1.9 years.
Calculating the residual involves computing the difference between the predicted age and the real age. First, we use the regression equation to predict the age of the lion. Plugging in x = 0.11 (since 11% is equivalent to 0.11 in decimal), we'll get y hat = 0.8790 + 10.6471*0.11 = 2.051082. Then, we subtract the lion's actual age from this prediction: 2.051082 - 1.9 = 0.151082.
So, the residual (the difference between the observed and predicted values) for the age of the lion with an 11% black nose is approximately 0.151. It means the regression equation slightly overestimated the lion's age.
Learn more about Residuals in Regression here:
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Answer:
The residual age of a lion whose nose is 11% black and is 1.9 years old is -0.15.
Step-by-step explanation:
In regression, the difference between the observed value of the dependent variable (y) and the predicted value ([tex]\hat y[/tex]) is known as the residual (e).
[tex]e=y-\hat y[/tex]
The least square regression line is used to predict the value of the response or dependent variable (y) from the known value of the explanatory or independent variable (x).
The general form of a least square regression line is:
[tex]\hat y=\alpha +\beta x[/tex]
The equation of the least squares regression line to predict the relationship between age (in years) and proportion of blackness in the lion’s nose is:
[tex]\hat y=0.8790+10.6471 x[/tex]
Compute the predicted value of y for x = 0.11 as follows:
[tex]\hat y=0.8790+10.6471 x[/tex]
[tex]=0.8790+(10.6471\times 0.11)\\=0.8790+1.171181\\=2.050181\\\approx 2.05[/tex]
The predicted value of y is, [tex]\hat y=2.05[/tex].
The observed value of the age of lion whose nose is 11% black is, y = 1.90.
Compute the residual age of this lion as follows:
[tex]e=y-\hat y[/tex]
[tex]=1.90-2.05\\=-0.15[/tex]
Thus, the residual age of a lion whose nose is 11% black and is 1.9 years old is -0.15.
