Respuesta :
Answer:
Predicted values:
[tex]\hat y_1 = 49.3 (1.47)^1 =72.471[/tex]
[tex]\hat y_2 = 49.3 (1.47)^2 =106.5324[/tex]
[tex]\hat y_3 = 49.3 (1.47)^3 =156.6026[/tex]
[tex]\hat y_4 = 49.3 (1.47)^4 =230.2058[/tex]
[tex]\hat y_5 = 49.3 (1.47)^5 =338.4025[/tex]
[tex]\hat y_6 = 49.3 (1.47)^6 =497.4517[/tex]
[tex]\hat y_7 = 49.3 (1.47)^7 =731.254[/tex]
[tex]\hat y_8 = 49.3 (1.47)^8 =1074.943[/tex]
Residuals:
[tex]e_1 = 65-72.471=-7.471[/tex]
[tex]e_2 = 90-106.5324=-16.5324[/tex]
[tex]e_3 = 162-156.6026 = 5.3974[/tex]
[tex]e_4 = 224-230.2058 = -6.2058[/tex]
[tex]e_5 = 337-338.4025 = -1.4025[/tex]
[tex]e_6 = 466-497.4617 = -31.4517[/tex]
[tex]e_7 = 780-731.254 = 48.7459[/tex]
[tex]e_8 = 1087-1074.493 = 12.0566[/tex]
Step-by-step explanation:
For this case we assume the following exponential function:
[tex]\hat y_i = 49.3 (1.47)^x [/tex]
Where x represent the hours and y the predicted values for each hour. We can find the estimated values like this:
[tex]\hat y_1 = 49.3 (1.47)^1 =72.471[/tex]
[tex]\hat y_2 = 49.3 (1.47)^2 =106.5324[/tex]
[tex]\hat y_3 = 49.3 (1.47)^3 =156.6026[/tex]
[tex]\hat y_4 = 49.3 (1.47)^4 =230.2058[/tex]
[tex]\hat y_5 = 49.3 (1.47)^5 =338.4025[/tex]
[tex]\hat y_6 = 49.3 (1.47)^6 =497.4517[/tex]
[tex]\hat y_7 = 49.3 (1.47)^7 =731.254[/tex]
[tex]\hat y_8 = 49.3 (1.47)^8 =1074.943[/tex]
Now we can find the residuals with this formula:
[tex] e_i = Y_i -\hat y_i [/tex]
And replacing we got:
[tex]e_1 = 65-72.471=-7.471[/tex]
[tex]e_2 = 90-106.5324=-16.5324[/tex]
[tex]e_3 = 162-156.6026 = 5.3974[/tex]
[tex]e_4 = 224-230.2058 = -6.2058[/tex]
[tex]e_5 = 337-338.4025 = -1.4025[/tex]
[tex]e_6 = 466-497.4617 = -31.4517[/tex]
[tex]e_7 = 780-731.254 = 48.7459[/tex]
[tex]e_8 = 1087-1074.493 = 12.0566[/tex]
