Accuracy & Precision
Date: ________
(you must have a calculator for today’s lesson)
Accuracy
• Closeness of a measurement to an accepted
value
• Accepted value comes from reference
materials
• The quantitative expression of accuracy is
called relative error (in chemistry this may
have been called percentage error) and is a %
Relative Error
Er = | O – A | x 100
A
Er = relative Error
O = your observed or calculated value
A = the accepted value from reference materials
|brackets| = absolute value, so no matter which
is the larger number O or A, take the
difference and make the result positive.
Let’s practice
• In a density experiment, you determine the
following for a sample of aluminum;
• mass = 15.0 g ; volume = 4.91 cm3
• Reference materials, such as a chemistry text
or a trusted science website, will contain an
accepted value for the density of aluminum.
• A = 2.70 g/cm3
• What is the relative Error of your observed
density value?
First you must calculate the density
density = mass/volume
d = 15.0 g / 4.91 cm3
d = 3.05 g / cm3
Now you can calculate the relative Error
Er = | O – A | x 100
A
Er = | 2.70-3.05 | g/cm3
x 100
2.70 g/cm3
Er = 12.96% but we are limited to 3 significant
digits so the answer is 13.0%
You try! Work with your table partner
• A sample of lead…..Pb…… what does a good
reference source list as the accepted value of
density?
• http://periodictable.com/Properties/A/Density.
html
• Your observed values: m= 125g ; V = 11.3 cm3
• What is the relative Error for your data?
Answer
• d = 11.1 g/cm3
• Er = 2.11%
• If you did not successfully get both of these
answers, see me in aclab at 10:15 asap.
Precision
• Precision is the closeness of measurements to
each other. Some experimental data may not
have a accepted value, so you can only
express the precision of the data rather than
it’s accuracy.
• So what do we compare our data to if there is
no accepted value?
You compare you data to
the mean (or average) of your data
• The quantitative expression of precision is
Relative Deviation
Dr = | O – M| x 100
M
You should notice that this equation is almost
identical to the one for Relative Error, what is
the difference?
•
•
•
•
Dr = relative Deviation
O = your observed or calculated value
M = mean or average of your data
The structure of the equation is the same!
Let’s practice
• 3 students collect data on an unknown sample
and determine the following density values.
• Student 1: d= 5.92 g/cm3
• Student 2: d= 6.29 g/cm3
• Student 3: d= 6.01 g/cm3
• What is the relative Deviation of Student 1
density result compared to the entire data set
of the three density values?
First calculate the mean
• Add the density values and divide by the
number of samples
• Add the 3 density values then divide by 3:
5.92+6.29+6.01 g/cm3
3
M = 6.07 g/cm3
Now calculate the relative Deviation
Dr = | O – M| x 100
M
Dr = | 5.92 – 6.07 | g/cm3
6.07 g/cm3
Dr = 2.47%
x 100
General view of precision
Evaluate the range of data
Which of the following data sets is more
precise?
Set 1:
21.4 cm, 26.7 cm, 18.9 cm
Set 2:
17.2 cm, 24.6 cm, 25.3 cm
Set 1:
21.4 cm, 26.7 cm, 18.9 cm
Set 2:
17.2 cm, 24.6 cm, 25.3 cm
• Range: Set 1: 26.7 cm – 18.9 cm = 7.8 cm
• Range: Set 2: 25.3 cm – 17.2 cm = 8.1 cm
• Therefore Set 1 is more precise since it has a
smaller range of data.
You try
Work with your table partner
• Data set for measurement of velocity
• Student 1: 2.44 m/s
• Student 2: 2.89 m/s
• Student 3: 3.15 m/s
• Student 4: 3.07 m/s
Find the relative Deviation for Student 3 data.\
Hint: start with the mean and you are limited to
3 significant figures.
Answer
• Mean = 2.89 m/s
• Dr = 9.00%
• If you were not successful, see me in aclab at
10:15 asap.