Answer:
[tex]e_i = Y_i -\hat y_i , i =1,2,...,n[/tex]
If we find the residuals for this case we got:
[tex]e_1 = -2.7 -(-2.84)= 0.14[/tex]
[tex]e_2 = -0.9 -(-0.81)= -0.09[/tex]
[tex]e_3 = 1.1 -(1.22)= -0.12[/tex]
[tex]e_4 = 3.2 -(3.25)= -0.05[/tex]
[tex]e_5 = 5.4 -(5.28)= 0.12[/tex]
And we can construct the residual plot as we can ee in the figure attached and after see the pattern of this graph we can conclude that the pattern for this case not seems to be linear.
And on this case the best option would be:
No, the points are in a curved pattern.
Step-by-step explanation:
The residual values are defined as the observed values minus the predicted values with the following formula:
[tex]e_i = Y_i -\hat y_i , i =1,2,...,n[/tex]
If we find the residuals for this case we got:
[tex]e_1 = -2.7 -(-2.84)= 0.14[/tex]
[tex]e_2 = -0.9 -(-0.81)= -0.09[/tex]
[tex]e_3 = 1.1 -(1.22)= -0.12[/tex]
[tex]e_4 = 3.2 -(3.25)= -0.05[/tex]
[tex]e_5 = 5.4 -(5.28)= 0.12[/tex]
And we can construct the residual plot as we can ee in the figure attached and after see the pattern of this graph we can conclude that the pattern for this case not seems to be linear.
And on this case the best option would be:
No, the points are in a curved pattern.
