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Unformatted text preview: LABORATORY #1 CHEMICAL PRINCIPLES II LAB Statistical Treatment of Experimental Data Most experiments involve multiple measurements or determinations of the desired result. Usually these repeated trials are performed by one person, but they could also represent a grouping of results from the same experiment performed by several lab workers. Once the results have been calculated (correctly, we hope!) and tabulated, they must be reported in a manner which represents their reliability to the reader or prospective user of the information. The most common treatment of experimental data uses the science of statistics and what follows is only a brief summary of applications from that area. [Anyone majoring in scientific areas would benefit greatly from a course in basic statistics.] Two definitions are important to consider when using scientific data: precision and accuracy. These two words are often used interchangeably in everyday language, but there is a distinct difference between them. Precision is the closeness of measured results to each other. It is an indication of how reproducibly a result can be determined (i.e., if a procedure is done again and again, will the same answer be attained?). Accuracy is the comparison of the actual measured answer to a known true value. Consider the following examples of quiz grades: CASE I_ _CASE II_ _CASE III_ 97 % 58 % 97 % 58 % 53 % 99 % 21 % 55 % 98 % poor precision good precision good precision poor accuracy poor accuracy good accuracy Obviously, the desired goal would be to have good precision and good accuracy (as shown in CASE III) for scientific measurements and results. Let us consider the following results from an experiment to determine the atomic weight (mass) of the element molybdenum (Mo): 97.58, 92.17, 95.36, 90.11, 94.94, and 96.83 g/mole. The first calculation would involve finding the average (or mean). This is defined as the sum of the individual results ( ΣX i ) divided by the number of measurements or results (n): TRIAL (n) X i 1 97.58 2 92.17 3 95.36 4 90.11 5 94.94 6 96.83 SUM 566.99 X = 94.50 TRIAL (i) X i X  X i 1 2 3 4 5 6 97.58 92.17 95.36 90.11 94.94 96.83 3.08 2.33 0.86 4.39 0.44 2.33 SUM 566.99 13.43 X = 94.50 AD=2.24 X= ∑ X i n Let us begin by setting up a data table. So the average X (pronounced "X bar") will be (566.99 g/mole)/6 = 94.50 g/mole. Next we will calculate the deviation which is the absolute difference between the average ( X ) and the individual result (X i ), or I ( X  X i ) I. By adding deviations and dividing by the number of results, we obtain the average deviation (AD): (XX )  ∑ i average deviation (AD) = n For Trial #1, the deviation would be  94.50 97.58 I = 3.08 g/mole....
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This note was uploaded on 03/18/2010 for the course COS 101121501 taught by Professor Langner during the Fall '08 term at RIT.
 Fall '08
 LANGNER

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