Accuracy & Precision Measurements & Error

How would you describe the ACCURACY & the PRECISION of the shooter? Four shots are fired at the target. The shooter was aiming for the bulls eye. How would you describe the ACCURACY & the PRECISION of the shooter?

Accuracy - refers to the agreement between experimental data and a known value. A highly accurate measuring device will provide measurements very close to the standard, true or known values. Precision - is the degree to which several measurements provide answers very close to each other, the degree of exactness to which a measurement can be reproduced. It is an indicator of the scatter in the data. The less the scatter, the higher the precision.

Bulls Eye Accuracy Bulls Eye Precision Accuracy - you can think of it in terms of a bulls eye in which the target is hit close to the center, yet the marks in the target aren't necessarily close to each other. Bulls Eye Precision If you hit a bulls eye precisely, then you are able to hit the same spot (or very close) on the target each time, even though that spot may be distant from the center.

Science Accuracy & Precision If we use a measuring device to measure gravity and we get a value of 9.799 𝑚 𝑠 2 then the device is very ACCURATE. If we carry out an experiment to determine gravity and we have 4 trials: 9.65 𝑚 𝑠 2 , 9.64 𝑚 𝑠 2 , 9.66 𝑚 𝑠 2 , & 9.65 𝑚 𝑠 2 then our values are very precise but not necessarily accurate.

A device can be both ACCURATE & PRECISE, it just has to meet the two criteria: Reproducible – every time you use it for the same measurement you get the same result. Agreement – the measurements are all very close to the standard value.

%Error %Error is a formula we can apply to determine the accuracy of a measurement. Usually if we are within 10% are measurement is fairly accurate. %𝐸= 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 −𝐸𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 100

Example: The standard value for gravity is 9 Example: The standard value for gravity is 9.8 𝑚 𝑠 2 and we have conducted an experiment and found gravity to be 9.95 𝑚 𝑠 2 , what is our %Error? %𝐸= 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 −𝐸𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 100 %𝐸= 9.8 𝑚 𝑠 2 − 9.95 𝑚 𝑠 2 9.8 𝑚 𝑠 2 100=1.5% Our answer is positive because of the absolute value brackets and we rounded to two significant figures.