Please answer all 3 questions
1 answer:
The residuals of the linear regression equation are 0.329, -0.346, 0.064, -0.326, 0.049, -0.286, -0.161, 0.214, 0.429 and 0.059
<h3>How to determine the residuals?</h3>The regression equation is given as:
y = 0.005x + 3.111
Next, we calculate the predicted values (y) at the corresponding x values.
So, we have:
y = 0.005 * 4 + 3.111 = 3.131
y = 0.005 * 3 + 3.111 = 3.126
y = 0.005 * 5 + 3.111 = 3.136
y = 0.005 * 3 + 3.111 = 3.126
y = 0.005 * 8 + 3.111 = 3.151
y = 0.005 * 15 + 3.111 = 3.186
y = 0.005 * 10 + 3.111 = 3.161
y = 0.005 * 15 + 3.111 = 3.186
y = 0.005 * 6 + 3.111 = 3.141
y = 0.005 * 6 + 3.111 = 3.141
The residuals are then calculated using:
Residual = Actual value - Predicted value
So, we have:
y = 3.46 - 3.131 = 0.329
y = 2.78 - 3.126 = -0.346
y = 3.2 - 3.136 = 0.064
y = 2.8 - 3.126 = -0.326
y = 3.2 - 3.151 = 0.049
y = 2.9 - 3.186 = -0.286
y = 3 - 3.161 = --0.161
y = 3.4 - 3.186 = 0.214
y = 3.57 - 3.141 = 0.429
y = 3.2 - 3.141 = 0.059
Hence, the residuals of the linear regression equation are 0.329, -0.346, 0.064, -0.326, 0.049, -0.286, -0.161, 0.214, 0.429 and 0.059
See attachment for the residual plot.
The residual plot shows that the linear model from the regression calculator is a good model because the points are not on a straight line
Read more about residuals at:
#SPJ1
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