Presentation on theme: "Accuracy & Precision Date: ________ (you must have a calculator for today’s lesson)"— Presentation transcript:
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 E r = | O – A | x 100 A E r = 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 cm 3 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 cm 3 d = 3.05 g / cm 3 Now you can calculate the relative Error E r = | O – A | x 100 A Er = | 2.70-3.05 | g/cm 3 x 100 2.70 g/cm 3 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 http://periodictable.com/Properties/A/Density. html Your observed values: m= 125g ; V = 11.3 cm 3 What is the relative Error for your data?
Answer d = 11.1 g/cm3 E r = 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 D r = | 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/cm 3 Student 2: d= 6.29 g/cm 3 Student 3: d= 6.01 g/cm 3 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/cm 3 3 M = 6.07 g/cm 3
Now calculate the relative Deviation D r = | O – M| x 100 M D r = | 5.92 – 6.07 | g/cm 3 x 100 6.07 g/cm 3 D r = 2.47%
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.