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Percentage errors If a line or curve equation is used to predict values then there will probably be an error as the points will not all lie on the line or curve The percentage error of a value predicted by a model from the true (observed) value gives you an idea of how good the model is.
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Percentage errors If a line or curve equation is used to predict values then there will probably be an error as the points will not all lie on the line or curve The percentage error of a value predicted by a model from the true (observed) value gives you an idea of how good the model is. Clearly the data points below lie close but not on the quadratic curve of best fit.
If the equation W = 2.2987t2 - 1.8229t + 5.5212 is used to predict the weight when t = 6 W = 2.2987t2 - 1.8229t + 5.5212 = 2.2987×62 - 1.8229×6 + 5.5212 = 77.34 Replace t by 6 The actual value is 72 from the table above so there is an error in the estimated value
Find the percentage error if the model predicts 25 and the actual value is 23.85. Take care to divide by the actual value. Percentage error to 3s.f. This means that if a particular equation is used to predict a value then there is a 4.8% error in the value obtained
Find the percentage error if the predicted value is 1.45 and the actual value is 1.50 Percentage error = A negative percentage error indicates the predicted value is less than the actual value.
Predicted Actual Predicted Actual Predicted Actual 1) 23 25 2) 25 23 3) 12.5 12.6 4) 12.6 12.5 5) 43.2 44.3 6) 1.43 1.4 Answers : Percentage Errors Answers are to 3sf when not exact. 1) – 8% 2) 8.70% 3) – 0.794% 4) 0.8% 5) – 2.48% 6) 2.14%