Comparing your lab results with the others by one-way ANOVA

Transcription

1 Comparig your lab results with the others by oe-way ANOVA You may have developed a ew test method ad i your method validatio process you would like to check the method s ruggedess by coductig a simple cross-check program by comparig your ow test results agaist the results provided by the other collaboratig peer laboratories o a similar sample. What you may have doe is to sed them oe or more similar homogeeous samples, request them to follow your method closely ad report its repeatability data after their lab aalyses. Whe all the data are collated, you ca carry out a aalysis of variace (ANOVA) to see if all the data are sigificatly comparable or ot. I fact, ANOVA is a extremely powerful statistical techique which ca be used to separate ad estimate the differet causes of variatio. I this case, the causes of variatio are the laboratory factor ad the radom error i measuremet of each participatig laboratory. Sice there is oly oe cotrolled factor, the statistical approach is called oe-way ANOVA. Let s see a example to demostrate how oe-way ANOVA ca be carried out. The followig table show the results of 4 similarly prepared samples received by 4 differet laboratories i a simple cross-check programme orgaised for the measuremet of zic cotet i water i mg/l by the FAAS method: Trial No. Lab 1 Lab Lab 3 Lab Mea, mg/l Overall Mea : 10.6 mg/l The above table shows that the mea values for the 4 samples (from a same source) reported by the participatig laboratories were apparetly differet. But, were they really so? We kow that because of radom error, eve if the true value which we are tryig to measure, say 100 mg Z/L is uchaged, the sample mea may vary from oe laboratory to aother. 1

2 ANOVA therefore, tests whether the differece betwee the samples reported by the 4 laboratories is too great to be explaied by the radom error aloe. First of all, let s do a hypothesis testig as below: H o : all the samples tested by the laboratories were reportig comparable populatio mea results with a variace σ o, i.e. µ 1 = µ = µ 3 = µ 4 H 1 : all mea results were ot comparable, i.e. µ 1 µ µ 3 µ 4 O the basis of this hypothesis, the data variace σ o ca be estimated i ways: * oe ivolvig the variatio withi each laboratory, ad, * the other, the variatio betwee the laboratories. (i) Withi lab variatio Remember that variace is the square of stadard deviatio. Hece, upo calculatio, we have: Variace of Lab 1 = 4.00 Variace of Lab = 5.33 Variace of Lab 3 = 4.41 Variace of Lab 4 =.33 Averagig these values gives: Withi-sample estimate of σ o = ( )/4 = 4.0 This estimate has 8 degrees of freedom, v [each sample estimate had (3-1) or degrees of freedom ad there were 4 laboratories]. (ii) Betwee-lab variatio Sice the samples aalyzed by the 4 laboratories were all draw from a populatio (oe source) which has populatio variace σ o, their average of all the samples aalyzed should come from this populatio with variace σ o ( where square of σ o is the variace of the populatio). (Recall the well kow Cetral Limit Theorem equatio : σ µ = x± z ) Thus, if the ull hypothesis H o is true, the variace of the meas of the samples aalyzed by these laboratories gives a estimate of σ o /.

3 I this example, we have : Lab Lab Lab Lab Mea of Labs, mg/l 10.6 Number of labs () 4 Degree of freedom 3 Std deviatio (SD) 4.74 Lab mea variace = (SD) squared.314 That meas : σ o / 3 =.314 (where = 3 measuremets for each lab) Therefore, the betwee-sample (laboratory) estimate of σ o (mea square) =.314 x 3 = Note : this estimate of σ o does ot deped o the variability withi each laboratory, because it is calculated from the lab meas. I this case, the umber of degrees of freedom v is (4 1) or 3 ad the mea square is ad so the sum of squared terms is 3 x = 00.8 Now, let s summarize our calculatios so far: Withi-lab mea square = 4.0 with df v = 8 Betwee-lab mea square = with df v = 3 If our H o is correct, these estimates of σ o should ot differ sigificatly, or else, for H 1, the betwee-lab estimate of σ o will be greater tha the withi-lab estimate because of betweelab variatio. To test if it is sigificatly greater, a oe-tailed F-test is used: F 3,8 = / 4.0 = 16.7 From the F-table, we fid that the critical value at F 3,8 is 4.07 with 95% cofidece. Sice the calculated value of F is greater tha this critical value, the ull hypothesis Ho is rejected, i.e. the laboratory meas do differ sigificatly. I other words, the proposed aalytical method may ot be rugged eough ad does ot seem to be reproducible by all the participatig laboratories. 3

4 If we were to use the Data Aalysis Tool of Microsoft Excel spreadsheet uder the ANOVA Sigle Factor to aalyse the above data, we should get the followig results which are exactly the same as described above based o the first priciple: ANOVA: Sigle Factor SUMMARY Groups Cout Sum Average Variace Lab Lab Lab Lab ANOVA Source of Variatio SS df MS F P-value F crit Betwee Labs Withi Labs Total Aother questio: which lab or labs were the root cause for o-comparable results? Were they all sigificatly differet amogst themselves? To aswer this, we ca rearrage the meas of the 4 labs i icreasig order ad compare the differece betwee adjacet values with a quatity called the least sigificat differece (LSD), which has the followig expressio: LSD t s = h( 1) where s is the withi-lab estimate of σ o h = umber of factor (labs i this case) = umber of repeats i each factor ad h(-1) is the umber of degrees of freedom of this estimate. 4

5 I the above example, the sample meas arraged i icreasig order of size are: Lab 3 Lab 1 Lab Lab The degree of freedom v = 4(3-1) = 8 ad hece t v= 8 =.306. The least sigificat differece therefore is.306 x 4.0 x (/3) with 95% cofidece, givig Now, differece of Lab 3 ad Lab 1 = = 3.6 differece of Lab 1 ad Lab = =.3 differece of Lab ad Lab 4 = = 5.4 Therefore whe we compare this LSD value with the differeces betwee the meas, we ote that : - the mea of Lab 4 differs sigificatly from those of Lab 1, ad 3; - the mea values of Lab 1, ad 3 do ot differ sigificatly from each other. 5