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3.1 Using and Expressing Measurements >
Accuracy vs. Precision
1
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
3.1 Using and Expressing Measurements > Accuracy, Precision,
and Error
Accuracy vs. Precision
Accuracy is a measure of how close a
measurement comes to the actual or true value.
Precision is a measure of how close a series of
measurements are to one another.
Good Accuracy,
Good
Accuracy
Good Precision
Good
Precision
2
Poor
PoorAccuracy,
Accuracy
Good
GoodPrecision
Precision
Poor
Accuracy,
Poor
Accuracy
Poor
Precision
Poor
Precision
Accuracy,
3.1 Using and Expressing Measurements > Precision, and
Error
Precision and Range
The range of group of measurements is the
amount of units that separate the largest
measured value from the smallest measured
value.
Range = largest value – smallest value
Precision = average ±
range
Average = Sum of measurements
# Measurements
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3.1 Using and Expressing Measurements > Sample Problem 3.2
Calculating Precision
Two students were ask to measure
the width of the door 3 times. Who is
more precise of the following data?
4
Brandon
Ruben
TRIAL #1
0.80 m
0.79 m
TRIAL #2
0.79 m
0.77 m
TRIAL #3
0.75 m
0.78 m
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
3.1 Using and Expressing Measurements > Sample Problem 3.2
1 Start with the equations.
Precision = average ± range
Range = largest value – smallest value
2 Find the average value for each student.
Average = Sum of measurements
# Measurements
Juan’s Average =
0.80m + 0.79m + 0.75m = 0.78m
3
Bryan’s Average =0.79m + 0.77m + 0.78m = 0.78m
3
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Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
3.1 Using and Expressing Measurements > Sample Problem 3.2
3 Calculate the range for each student.
Ahmad’s Range = 0.80 m – 0.75 m = 0.05 m
Jon’s Range = 0.79 m – 0.77 m = 0.02 m
4 Write the precision for each student
*Precision = average ± range
Juan’s Precision =
0.78 ± 0.05 m
Jon’s Precision = 0.78 ± 0.02 m
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Jon was more
precise because his
range is smaller.
Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
3.1 Using and Expressing Measurements > Accuracy, Precision,
and Error
Determining Error
• The difference between the experimental
value and the accepted value is called the
error.
Error = | experimental value – accepted value |
• The percent error of a measurement is
the absolute value of the error divided by
the accepted value, multiplied by 100%.
error
Percent error =
error
Percent error = accepted value
accepted value
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100%
100%
3.1 Using and Expressing Measurements > Sample Problem 3.2
Calculating Percent Error
The boiling point of pure water is
measured to be 99.1°C. Calculate
the percent error.
Think about it: Using
the absolute value of
the error means that
percent error will
always be a positive
value.
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3.1 Using and Expressing Measurements > Sample Problem 3.2
1 Analyze List the knowns and unknown.
UNKNOWN
Percent error =
?
2
KNOWNS
Experimental value = 99.1°C
Accepted value = 100.0°C
Calculate Solve for the unknown.
Start with the equations.
|error |
____________
Percent error =
accepted value
x 100%
Error = experimental value – accepted value
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3.1 Using and Expressing Measurements > Sample Problem 3.2
2
Calculate Solve for the unknown.
Substitute the equation for error, and then plug
in the known values.
|experimental
value – accepted value|
_______________________________
X 100%
Percent error =
accepted value
=
|99.1°C – 100.0°C|
X 100%
100.0°C
0.9°C
_______
=
X 100 % = 0.9%
100.0°C
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Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
3.1 Using and Expressing Measurements >
Practice Problems Pay attention to SigFigs!
• A student measures the depth of a swimming pool to
be 2.04 meters at its deepest end. The accepted
value is 2.00 m. What is the student’s percent error?
• In an experiment, a student estimates the
acceleration due to gravity is 9.31 m/s2. The actual
value is 9.8 m/s2. What is the student’s percent error?
• While doing a lab, a student found the density of a
piece of pure aluminum to be 2.85 g/cm3. The
accepted value for the density of aluminum is 2.70
g/cm3. What was the student's percent error?
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