Question

Complete the following table for residuals for the exponential function: f(x) = 49.3(1.47)

Answer 1

Answer:

Predicted values:

$\hat y_1 = 49.3 (1.47)^1 =72.471$

$\hat y_2 = 49.3 (1.47)^2 =106.5324$

$\hat y_3 = 49.3 (1.47)^3 =156.6026$

$\hat y_4 = 49.3 (1.47)^4 =230.2058$

$\hat y_5 = 49.3 (1.47)^5 =338.4025$

$\hat y_6 = 49.3 (1.47)^6 =497.4517$

$\hat y_7 = 49.3 (1.47)^7 =731.254$

$\hat y_8 = 49.3 (1.47)^8 =1074.943$

Residuals:

$e_1 = 65-72.471=-7.471$

$e_2 = 90-106.5324=-16.5324$

$e_3 = 162-156.6026 = 5.3974$

$e_4 = 224-230.2058 = -6.2058$

$e_5 = 337-338.4025 = -1.4025$

$e_6 = 466-497.4617 = -31.4517$

$e_7 = 780-731.254 = 48.7459$

$e_8 = 1087-1074.493 = 12.0566$

Step-by-step explanation:

For this case we assume the following exponential function:

$\hat y_i = 49.3 (1.47)^x$

Where x represent the hours and y the predicted values for each hour. We can find the estimated values like this:

$\hat y_1 = 49.3 (1.47)^1 =72.471$

$\hat y_2 = 49.3 (1.47)^2 =106.5324$

$\hat y_3 = 49.3 (1.47)^3 =156.6026$

$\hat y_4 = 49.3 (1.47)^4 =230.2058$

$\hat y_5 = 49.3 (1.47)^5 =338.4025$

$\hat y_6 = 49.3 (1.47)^6 =497.4517$

$\hat y_7 = 49.3 (1.47)^7 =731.254$

$\hat y_8 = 49.3 (1.47)^8 =1074.943$

Now we can find the residuals with this formula:

$e_i = Y_i -\hat y_i$

And replacing we got:

$e_1 = 65-72.471=-7.471$

$e_2 = 90-106.5324=-16.5324$

$e_3 = 162-156.6026 = 5.3974$

$e_4 = 224-230.2058 = -6.2058$

$e_5 = 337-338.4025 = -1.4025$

$e_6 = 466-497.4617 = -31.4517$

$e_7 = 780-731.254 = 48.7459$

$e_8 = 1087-1074.493 = 12.0566$