Question

A blood sample with a known glucose concentration of 102.0 mg/dL is used to test a new at home glucose monitor. The device is used to measure the glucose concentration in the blood sample five times. The measured glucose concentrations are 104.5 , 96.2 , 102.2 , 98.3 , and 101.8 mg/dL. Calculate the absolute error and relative error for each measurement made by the glucose monitor. A. 104.5 mg/dL absolute error = 2.5 mg / dL relative error = 0.025 B. 96.2 mg/dL absolute error = −5.8 mg / dL relative error = 0.057 C. 102.2 mg/dL absolute error = 0.2 mg / dL relative error = 0.020 D. 98.3 mg/dL absolute error = −3.7 mg / dL relative error = 0.036 E. 101.8 mg/dL absolute error = −0.2 mg / dL relative error =

Answer 1

Answer:

The Absolute Error is the difference between the actual and measured value.

$Absolute \:error = |Actual \:value - Measured \:value|$

The Relative Error is the Absolute Error divided by the actual measurement.

$Relative \:error = \frac{Absolute \:error}{Actual \:value}$

We know that the actual value is 102.0 mg/dL.

To find the absolute error and relative error for each measurement made by the glucose monitor you must use the above definitions.

a) For a concentration of 104.5 mg/dL the absolute error and relative error are

$Absolute \:error = \left|102-104.5\right|\\Absolute \:error =\left|-2.5\right|\\Absolute \:error =2.5$

$Relative \:error = \frac{2.5}{102.0}=0.0245$

b) For a concentration of 96.2 mg/dL the absolute error and relative error are

$Absolute \:error = \left|102.0-96.2\right|\\Absolute \:error =\left|5.8\right|\\Absolute \:error =5.8$

$Relative \:error = \frac{5.8}{102.0}=0.0569$

c) For a concentration of 102.2 mg/dL the absolute error and relative error are

$Absolute \:error = \left|102.0-102.2\right|\\Absolute \:error =\left|-0.2\right|\\Absolute \:error =0.2$

$Relative \:error = \frac{0.2}{102.0}=0.00196$

d) For a concentration of 98.3 mg/dL the absolute error and relative error are

$Absolute \:error = \left|102.0-98.3\right|\\Absolute \:error =\left|3.7\right|\\Absolute \:error =3.7$

$Relative \:error = \frac{3.7}{102.0}=0.0363$

e) For a concentration of 101.8 mg/dL the absolute error and relative error are

$Absolute \:error = \left|102.0-101.8\right|\\Absolute \:error =\left|0.2\right|\\Absolute \:error =0.2$

$Relative \:error = \frac{0.2}{102.0}=0.00196$