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Accuracy and Precision

Accuracy and Precision. Accuracy. Accuracy – how closely a measurement agrees with an accepted value. For example: The accepted density of zinc is 7.14 g/cm 3 Student A measures the density as 5.19 g/cm 3 Student B measures the density as 7.01 g/cm 3

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Accuracy and Precision

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  1. Accuracy and Precision

  2. Accuracy • Accuracy – how closely a measurement agrees with an accepted value. • For example: • The accepted density of zinc is 7.14 g/cm3 • Student A measures the density as 5.19 g/cm3 • Student B measures the density as 7.01 g/cm3 • Student C measures the density as 8.85 g/cm3 • Which student is most accurate?

  3. Error • All measurements have some error. • Scientists attempt to reduce error by taking the same measurement many times. • Assuming no bias in the instruments. • Bias – A systematic (built-in) error that makes all measurements wrong by a certain amount. • Examples: • A scale that reads “1 kg” when there is nothing on it. • You always measure your height while wearing thick-soled shoes. • A stopwatch takes half a second to stop after being clicked.

  4. Error • Error = experimental value – accepted value • Example: The accepted value for the specific heat of water is 4.184 J/gºC. Mark measures the specific heat of water as 4.250 J/gºC. What is Mark’s error? • Error = exp.value – acc.value • Error = 4.250 J/gºC – 4.184 J/gºC • Error = 0.066 J/gºC

  5. Percent Error │error │ • %Error = x 100% • Example: The accepted value for the molar mass of methane is 16.042 g/mol. Jenny measures the molar mass as 14.994 g/mol. What is Jenny’s percent error? • First, find the error: • Error = exp.value – acc.value = 14.994 g/mol – 16.042 g/mol • Error = -1.048 g/mol • Percent error = │error │/ acc.value x 100% • Percent error = (1.048 g/mol) / (16.042 g/mol) x 100% • Percent error = 0.06533 x 100% • Percent error = 6.533% acc.value

  6. Precision • Precision – describes the closeness of a set of measurements taking under the same conditions. • Good precision does not mean that measurements are accurate.

  7. Accuracy and Precision Decent accuracy, but poor precision: the average of the shots is on the bullseye, but they are widely spread out. If this were a science experiment, the methodology or equipment would need to be improved. Good precision, but poor accuracy. The shots are tightly clustered, but they aren’t near the bullseye. In an experiment this represents a bias. Good accuracy and good precision. If this were a science experiment, we would consider this data to be valid.

  8. Accuracy and Precision • Another way to think of accuracy and precision: • Accuracy means telling the truth… • Precision means telling the same story over and over. • They aren’t always the same thing.

  9. Accuracy and Precision • Four teams (A, B, C, and D) set out to measure the radius of the Earth. Each team splits into four groups (1, 2, 3, and 4) who compile their data separately, then they get back together and compare measurements. Their data are presented below: Group 1 Group 2 Group 3 Group 4 Averages Team A 6330.2 km 6880.3 km 6940.3 km 6752.0 km 6613.2 km Team B 6105.2 km 6130.7 km 6112.2 km 6099.5 km 6111.9 km Team C 6015.1 km 5810.0 km 6741.1 km 6912.9 km 6369.8 km Team D 6038.7 km 6380.0 km 6366.3 km 6400.1 km 6296.3 km

  10. Which team’s data were most precise? • Team B’s data was most precise, because their measurements were very consistent. • Which team’s data were most accurate? • We can’t say yet, because we don’t know the accepted value for the radius of Earth. • The accepted value is 6378.1 km. • % Error of Team A = 3.686% • % Error of Team B = 4.174% • % Error of Team C = 0.130% • % Error of Team D = 1.283% • Team C was the most accurate team, even though their data weren’t the most precise. Group 1 Group 2 Group 3 Group 4 Averages Team A 6330.2 km 6880.3 km 6940.3 km 6752.0 km 6613.2 km Team B 6105.2 km 6130.7 km 6112.2 km 6099.5 km 6111.9 km Team C 6015.1 km 5810.0 km 6741.1 km 6912.9 km 6369.8 km Team D 6038.7 km 6380.0 km 6366.3 km 6400.1 km 6296.3 km

  11. Example Problem • Working in the laboratory, a student finds the density of a piece of pure aluminum to be 2.85 g/cm3.  The accepted value for the density of aluminum is 2.699 g/cm3.  What is the student's percent error?

  12. Another Example Problem • A student takes an object with an accepted mass of 200.00 grams and masses it on his own balance.  He records the mass of the object as 196.5 g.   What is his percent error?

  13. Significant Figures

  14. Why do we have them? When we measure things, we want to measure to the place we are sure of and guess one more space.

  15. So, they show the uncertainty in our measurements

  16. A thought problem • Suppose you had to find the density of a rock. Density = mass / volume • You measure the rock’s mass as 45.59 g • You measure the rock’s volume as 9.3 cm3 • When you type 45.59 / 9.3 into the calculator, you get 4.902150538 g/cm3 • Should you really write all those digits in your answer, OR • Is the precision of your answer limited by your measurements?

  17. A thought problem • The calculator’s answer is misleading. • You don’t really know the rock’s density with that much precision. • It’s scientifically dishonest to claim that you do. • Your answer must be rounded to the most precise (but still justifiable) value. • How do scientists round numbers to avoid giving misleading answers?

  18. A thought problem • Scientists use the concept of significant figures to give reasonable answers. • If a scientist divided 45.59 g by 9.3 cm3, he or she would report the answer as 4.9 g/cm3. • In class we will NOT use all the rules of significant figures. • We will use the least number of digits after the decimal.

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