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  <title>WEF Discussion Forums : Laboratory Management and Technical Issues : Solids Duplicate Determinations</title>
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  <description><![CDATA[<p>I'm new here and attempted to post this question last week, but it didn't display for some reason.</p>
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<p>I have a question regarding Standard Methods 2540 Solids section.  Specifically, TSS 2540D, end of section 3.c.:  "Duplicate determinations should agree within 5% of their average weight."  Can someone please clarify this statement?</p>
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<p>Does this mean to take the average of the two final weights and then compare each weight to the average?  If that comparison is within 5% of the average, then the values are acceptable?</p>
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<p>How are you applying this in your benchsheet?  We have an Excel spreadsheet in which our criteria is set via formulas.  I would like to understand that statement so I can make sure it's already included in our sheet criteria. If not, I'll need to figure out how to enter it.</p>
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<p>Thanks!</p>
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  <pubDate>Mon, 15 Oct 2012 19:18:23 GMT</pubDate>
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  <description><![CDATA[<p>We record relative percent difference of duplicates on the benchsheet. RPD is the <u>difference</u> between the two samples divided by the average. </p>
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<p>Excel formula  =ABS(A1-B1)/AVERAGE(A1,B1)<br /> </p>
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  <pubDate>Wed, 17 Oct 2012 11:55:12 GMT</pubDate>
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  <description><![CDATA[<p>You are correct in that it is a comparison between the two residue weights and their average.  The Excel forumla would be</p>
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<p>100*ABS(Cell1-Cell2)/(AVERAGE(Cell1:Cell2)*2)</p>
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<p>Cell1 and Cell2 are the cells with the residue weights (final - tare)</p>
<p>100 is to covert to percent and 2 is because the comparison is to each weight.</p>
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<p>Some may recognize the formula without the 2 factor as the one to compute % difference</p>
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<p>This suffers the same problem as all others where the determination is against a fixed value in that there are ranges of weights where the comparison does not make sense.  Below are some actual results and how they fared</p>
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<p>WT 1   WT 2   % differece to average</p>
<p>16.7   16.8          0.3 </p>
<p>  2.2     3.2        18.5</p>
<p>  3.1     3.3          3.1</p>
<p>  6.5     6.9          3.0</p>
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<p>As you can see the second pair compute to an 18.5% difference.  The sample is of a final effluent and 1000mL was sampled.  The only way to improve this would have been to use more sample and collect more residue.</p>
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<p>The good news is the wiggle word "should".  "Should" does not mean "Shall" so you can write up a work around.  In these cases, I suggest the "If..then" process where the 5% applies unless some limit is met then another criteria is used.  For example</p>
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<p>5% for weights over 5 mg, 1.0 mg for 5 mg or less</p>
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  <pubDate>Wed, 24 Oct 2012 13:57:44 GMT</pubDate>
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