Percentage Error Questions
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a percentage of one (or both) values Use Percentage Change when comparing an Old Value to a New Value Use Percentage Error when comparing an Approximate Value to an Exact Value Use Percentage Difference when both percentage error formula values mean the same kind of thing (one value is not obviously older
Percent Error Chemistry
or better than the other). (Refer to those links for more details) How to Calculate Step 1: Subtract one value from percent error calculator the other Step 2: Then divide by ... what? Percentage Change: Divide by the Old Value Percentage Error: Divide by the Exact Value Percentage Difference: Divide by the Average of The Two Values Step 3:
Can Percent Error Be Negative
Is the answer negative? Percentage Change: a positive value is an increase, a negative value is a decrease. Percentage Error: ignore a minus sign (just leave it off), unless you want to know if the error is under or over the exact value Percentage Difference: ignore a minus sign, because neither value is more important, so being "above" or "below" does not make sense. Step 4: Convert this into a negative percent error percentage (multiply by 100 and add a % sign) The Formulas (Note: the "|" symbols mean absolute value, so negatives become positive.) Percent Change = New Value - Old Value × 100% |Old Value| Example: There were 200 customers yesterday, and 240 today: 240 - 200 × 100% = (40/200) × 100% = 20% |200| A 20% increase. Percent Error = |Approximate Value - Exact Value| × 100% |Exact Value| Example: I thought 70 people would turn up to the concert, but in fact 80 did! |70 - 80| × 100% = (10/80) × 100% = 12.5% |80| I was in error by 12.5% (Without using the absolute value, the error is -12.5%, meaning I under-estimated the value) Percentage Difference = | First Value - Second Value | × 100% (First Value + Second Value)/2 Example: "Best Shoes" gets 200 customers, and "Cheap Shoes" gets 240 customers: | 240 - 200 | × 100% = |40/220| × 100% = 18.18...% (200+240)/2 Percentage Difference Percentage Error Percentage Change Percentage Index Search :: Index :: About :: Contact :: Contribute :: Cite This Page :: Privacy Copyright © 2014 MathsIsFun.com
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Images Coordinates in MaxIm Fits Header Graphing in Maxim Image Calibration in Maxim Importing Images into MaxIm Importing what is a good percent error Images into Rspec Measuring Magnitude in Maxim Observing with Rigel Photometry in Maxim Producing Color Images Stacking Images Using SpectraSuite Software Using Tablet Applications Using the Rise and Set Calculator http://www.mathsisfun.com/data/percentage-difference-vs-error.html on Rigel Wavelength Calibration in Rspec Glossary Kepler's Third Law Significant Figures Percent Error Formula Small-Angle Formula Stellar Parallax Finder Chart Iowa Robotic Telescope Sidebar[Skip] Glossary Index Kepler's Third LawSignificant FiguresPercent Error FormulaSmall-Angle FormulaStellar ParallaxFinder Chart Percent Error Formula When you calculate results that are aiming for known values, the percent error formula is useful tool for determining the http://astro.physics.uiowa.edu/ITU/glossary/percent-error-formula/ precision of your calculations. The formula is given by: The experimental value is your calculated value, and the theoretical value is your known value. A percentage very close to zero means you are very close to your targeted value, which is good. It is always necessary to understand the cause of the error, such as whether it is due to the imprecision of your equipment, your own estimations, or a mistake in your experiment.Example: The 17th century Danish astronomer, Ole Rømer, observed that the periods of the satellites of Jupiter would appear to fluctuate depending on the distance of Jupiter from Earth. The further away Jupiter was, the longer the satellites would take to appear from behind the planet. In 1676, he determined that this phenomenon was due to the fact that the speed of light was finite, and subsequently estimated its velocity to be approximately 220,000 km/s. The current accepted value of the speed of light is almost 299,800 km/s. What was the percent error of Rømer's estimate?Solution:experimental value = 220,000 km/s = 2.2 x 108 m/stheore
found by measurement and the "true value' of the quantity. eg an object that has a mass of 120 g may be shown to weigh 130 g on an imperfect weighing machine. True weight: 120 g Measured weight: 130 g Error: +10 g Measurement errors arise because http://www.staff.vu.edu.au/mcaonline/units/percent/pererr.html of inevitable imperfections in the measuring instrument and limitations of the human eye. Errors come in all sizes, and sometimes you need to decide if the error in your measurement is so big that it makes the measurement useless. (see examples below) Errors can be positive or negative. An electric current might be measured as Examples The effective size of the error depends on the actual size of the error the size of percent error the measurement itself Example 1 Measuring a Line Actual length of line: 11 cm Length of line when measured: 12 cm Error is (Measured Length - Actual Length) Error is (12 cm - 11 cm) = 1 cm. The error expressed as a fraction of the actual size is Example 2 Measuring the height of a person Actual height is 1.72 cm = 1270 mm If the error in measurement is only 1 mm, percentage error questions then expressing this as a fraction of the actual size Have a Go Problem 1 Voltage is measured with a multimeter. A particular multimeter is being tested. True voltage of the multimeter: 224 V. Measured voltage: 220 V. Calculate the actual error and the percentage error. You will notice that in this example the error is a negative value Problem 2 Another multimeter is being tested. True voltage of the multimeter: 150 V Measured voltage: 153 V Calculate the actual error and the percentage error. In this case the error has a positive value. Practice Questions Question 1 Answer 1.3 hectares Question 2 answer + or - 0.2% Question 3 answer + or - 0.2 cm Question 4 answer + or - 32.2 sec Question 5 answer + or - 0.2% Question 6 answer + or - 1.2% Solution 1 Actual Error = Measured Voltage -.True Voltage = 220 - 224 V = (-) 4 V back to Have a Go Solution 2 Actual Error = Measured Voltage- True Voltage = 153-150 V = (+) 3 V The multimeter is slightly less accurate than the one in the previous problem (This had an accuracy of 1.8%) back to Have a Go [Home][General][Business][Engineering][VCE][Learning Units][Tool Box][Glossary]
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