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Development and optimization of microbeam x-ray fluorescence analysis of Sr in speleothems Jessica A. BucklesHarold D. RoweSupplementary MaterialSr Counts-to-Concentration CalibrationTable S1. Calibration standard data for Sr Counts-to-Concentration for Speleothems. Note that the numbering of the standard label is not sequential.Standard 1Table S2. ?-XRF data for Standard 1 at 30s, 60s, 90s, and 120s. Ca and Sr K12 counts measurements were taken 5 times at each setting. Averages of the five were then used to calculate the average counts (avg), standard deviation (std), and coefficient of variation (cv).Standard 2Table S3. ?-XRF data for Standard 2 at 30s, 60s, 90s, and 120s. Ca and Sr K12 counts measurements were taken 5 times at each setting. Averages of the five were then used to calculate the average counts (avg), standard deviation (std), and coefficient of variation (cv).Standard 3Table S4. ?-XRF data for Standard 3 at 30s, 60s, 90s, and 120s. Ca and Sr K12 counts measurements were taken 5 times at each setting. Averages of the five were then used to calculate the average counts (avg), standard deviation (std), and coefficient of variation (cv).Standard 4Table S5. ?-XRF data for Standard 4 at 30s, 60s, 90s, and 120s. Ca and Sr K12 counts measurements were taken 5 times at each setting. Averages of the five were then used to calculate the average counts (avg), standard deviation (std), and coefficient of variation (cv).Standard 5Table S5. ?-XRF data for Standard 5 at 30s, 60s, 90s, and 120s. Ca and Sr K12 counts measurements were taken 5 times at each setting. Averages of the five were then used to calculate the average counts (avg), standard deviation (std), and coefficient of variation (cv).Table S6. Compilation of Sr ppm of known calibration standards and reported ?-XRF counts. Slope and intercept (int) calculated for the four time settings. Figure S1. Calibration curve for Sr 30s count time. Linear regression plotted for the standards of known concentrations.Figure S2. Calibration curve for Sr 60s count time. Linear regression plotted for the standards of known concentrations.Figure S3. Calibration curve for Sr 90s count time. Linear regression plotted for the standards of known concentrations.Figure S4. Calibration curve for Sr 120s count time. Linear regression plotted for the standards of known concentrations.Counting Statistics and Optimization of Analysis Point estimate of the repeatability of an X-Ray Fluorescence instrumentAnalysis is done in R. R commands are show in the format below, where ‘>’ is the R command prompt:> command(Commands shown are for the Apple OS X platform. They may vary slightly for other platforms, particularly in the case of manipulating graphics files. In particular, the “quartz.save()” function is peculiar to the Mac platform. On other platforms, the “png()” function should provide the same functionality.)(Note also that it will not be possible to exactly duplicate the simulations performed here, since the set of random values generated will be unique. With the large sample sizes, however, results should be quite similar.)In the absence of prepared standards, a single point on a field sample was analyzed 30 times using the Bruker Artax ?-XRF instrument producing measurements of Ca and Sr as counts. The repeatability of the instrument is estimated for these two elements and the Sr/Ca ratio.The data is exported from Excel as a comma separated variables file (xrfrr.csv) to allow it to be imported into R for analysis.CaSr123137648223347582323987566423187709523767850623457908724617622823857753923587610102385777511236278421224597654132498782714249478181523557673162421772117242176941823097648192364783020237475922124947854222326772523237978172424297761252315778126244877822722907883282344767729239979033023277757Table S7. Count values for 30 identical analyses using the Bruker Artax ?-XRF instrument producing measurements of Ca and Sr as counts.The data is first imported into R as a data frame named ‘data’. The counts data for Ca and Sr are then copied into two vector variables named ‘Ca’ and ‘Sr’.> data = read.csv( “xrfrr.csv” )> Ca = data$Ca> Sr = data$SrAnalysis of Ca counts dataThe mean and standard deviation of Ca counts are found> mean( Ca )[1] 2382.7> sd( Ca )[1] 59.13873A histogram of Ca counts is produced and saved to a file for inclusion in a report> hist( Ca, main=”Histogram of Ca counts”, breaks=seq( 2250, 2500, by=25 ) )> quartz.save(“histCa.png” )Figure S5. Histogram of Ca counts values for the 30 scans.The counts data for the ?-XRF is expected to reflect a Poisson distribution. The parameter of a Poisson distribution is lambda, the mean of the distribution. The variance of the Poisson distribution is also lambda.The histogram is redrawn as a probability density, and a Poisson distribution with lambda = 2383 is overlaid.> hist( Ca, main=”Density of Ca counts vs Poisson”, breaks=seq( 2250, 2500, by=25 ),prob=TRUE )> lines( 2250:2500, dpois( 2250:2500, 2383 ), type=”h”, col=”yellow” )> abline( v=2383, col=”red” )> quartz.save(“densCa.png” )Figure S6. Density of Ca counts values for the 30 scans versus a superimposed Poisson distribution.Finally, a Kolmogorov-Smirnov test is applied to test the hypothesis that the sample data is drawn from a Poisson distribution with lambda = 2383.>ks.test( Ca, ppois, lambda=2383 )One-sample Kolmogorov-Smirnov testData: CaD = 0.1132, p-value = .8365Alternative hypothesis: two-sidedWarning message:In ks.test(Ca, ppois, lambda = 2383):Cannot compute correct p-values with tiesThe returned p-value of 0.8365 indicates that there is no reason to conclude that the data do not follow a Poisson distribution. The warning about ties applies only to continuous distributions and can be ignored as the Poisson is a discrete distribution.(In a statistical test, the smaller a p-value, the stronger the evidence that the “null hypothesis”, in this case that the data follows a Poisson distribution, is false. A p-value of 0.05 indicates 95% certainty that the hypothesis is false.)Note that the average of the Ca counts is only an estimate of the mean of the distribution from which the sample is drawn. We may calculate a confidence interval for the true mean using the knowledge that for large n, the distribution of the mean of n values drawn from any distribution, normal or not, approaches a normal distribution with mean equal to the mean of the distribution and variance equal to the variance of the distribution divided by n.Thus with sample mean of 2383 and sample standard deviation of 59.14, the 95% confidence interval for the mean is from 2361 to 2404. (The 95% confidence interval lies between 1.96 sigma below and 1.96 sigma above the mean).Since the variance of a Poisson distribution is equal to the mean, the sample mean of 2383 would suggest a sample standard deviation of 48.82. The actual sample standard deviation was 59.14.In the absence of a theoretical distribution for the sampling distribution of the variance from a Poisson distribution, a simulation was conducted.A set of 1000 samples, each of size 30, were drawn from a Poisson distribution of mean 2383. The standard deviations of the samples were calculated and a histogram plotted.> x = rpois( 30000, 2383 )> dim( x ) = c( 1000, 30 )> sdx = apply( x, 1, sd )> hist( sdx, “Histogram of sd, 1000 samples, mean=2383, n=30” )> abline( v=sqrt(2383), col=”red” )> quartz.save( “sd-2383-30.png” )Figure S7. Histogram for the standard deviations of 1000 samples, each of size 30, that were drawn from a Poisson distribution of mean 2383 (Ca counts mean).The red line is the expected sd of sqrt(2383). No values below 30 or above 75, but values between 50 and 60 were quite common. The cumulative distribution function for the generated standard deviations was determined, and the 0.025 and 0.975 points determined to find a 95% probability window.> foo = ecdf( sdx )> plot( foo, main=”ECDF of sd, 1000 samples, mean=2383, n=30” )> abline( h=0.025, col=”red” )> abline( h=0.975, col=”red” )> foo025 = function( x ){ foo( x ) - .025 }> uniroot( foo025, c(30, 40 ) )$root[1] 36.79317> abline( v=36.79317, col=”red” )> foo975 = function( x ){ foo( x ) - .975 }> uniroot( foo975, c(50,70) )$root[1] 61.19742> abline( v=61.19742, col=”red” )Figure S8. Cumulative distribution function of the standard deviations of the simulated 1000 samples, each of size 30, that were drawn from a Poisson distribution of sample mean 2383 (Ca counts mean) with sample standard deviation of 48.8.In the simulation, 95% of the standard deviations fell between 36.79 and 61.20. A standard deviation of 59.14 is inside the 95% window. We cannot reject the hypothesis that the sample is drawn from a Poisson distribution with mean 2383.In order to test for time-dependent errors, a lag test for autocorrelation is performed. The lag test is a test for correlation between the each data value and the data value which is offset in time by a specified offset in time. For example, in a lag test of lag 1, each point is compared to the successive point.> w1 = Ca[ 1:29 ]> w2 = Ca[ 2:30 ]> cor( w1, w2 )[1] -0.03524947> plot( w1, w2, main=”Scatter Plot of Ca[i] vs Ca[i+1]”)Figure S9. Lag test for autocorrelation, where each point (Ca count) has been compared to the successive point.Neither the correlation coefficient of -0.3525 nor the scatter plot suggests the data is correlated. There is no evidence of time dependent behavior in the data. Finally, as a final test that the data is drawn from a single Poisson distribution, a C control chart is produced.> cl = mean( Ca )> cl[1] 2382.7> ucl = cl + 3 * sqrt( cl )> ucl[1] 2529.139> lcl = cl – 3 * sqrt( cl )> lcl[1] 2236.261> plot( Ca, type=”b”, ylim=c(2200,2600),main=”C Chart for Ca” )> abline( h=cl, col=”red” )> abline( h=ucl, col=”red” )> abline( h=lcl, col=”red” )> quartz.save( “CCa.png” )Figure S10. C control chart for Ca counts.Analysis of this chart is as follows: If any of the following occur, that is an indication that the data is not from a Poisson distribution:A point above the upper red line (Upper Control Limit)A point below the lower red line (Lower Control Limit)Six or more consecutive points above the central lineSix or more consecutive points below the central lineSix or more consecutive points monotonically increasing or decreasingNone of these indicators are present.Analysis of Sr counts data> mean( Sr )[1] 7742.067> sd( Sr )[1] 99.70264A histogram of Sr counts is produced and saved to a file for inclusion in a report> hist( Sr, main=”Histogram of Sr Counts” )> quartz.save( “histSr.png” )Figure S11. Histogram of Sr counts values for the 30 scans.The Sr counts data is also expected to reflect a Poisson distribution.The histogram is redrawn as a probability density, and a Poisson distribution with lambda = 7742 is overlaid.> hist( Sr, ylim=c(0,.005), main=”Density of Sr counts vs Poisson”, prob=TRUE )> lines( 7550:7950, dpois( 7550:7950, 7742 ), type=”h”, col=”yellow” )> abline( v=7742, col=”red” )> quartz.save(“densCa.png” )Figure S12. Density of Sr counts values for the 30 scans versus a superimposed Poisson distribution.Finally, a Kolmogorov-Smirnov test is applied to test the hypothesis that the sample data is drawn from a Poisson distribution with lambda = 2383.>ks.test( Sr, ppois, lambda=7742 )One-sample Kolmogorov-Smirnov testData: SrD = 0.138, p-value = .6169Alternative hypothesis: two-sidedWarning message:In ks.test(Sr, ppois, lambda = 7742):Cannot compute correct p-values with tiesIn this case also the returned p-value of 0.6169 indicates that there is no reason to conclude that the data does not follow a Poisson distribution. The warning about ties applies only to continuous distributions and can be ignored as the Poisson is a discrete distribution.We calculate a confidence interval for the true mean of Sr. With sample mean of 7742 and sample standard deviation of 99.7, the 95% confidence interval for the mean is from 7706 to 7777.In order to test the sample standard deviation of 99.70 against the theoretical standard deviation of 87.99 (the square root of 7742), a simulation similar to that described for Ca above was conducted.(The R code has been omitted, but is equivalent to that shown earlier.)Figure S13. Cumulative distribution function of the standard deviations of the simulated 1000 samples, each of size 30, that were drawn from a Poisson distribution of sample mean 7742 (Sr counts mean) with a sample standard deviation of 99.7.The 95% window for the cumulative distribution of 1000 simulated samples was from 64.58 to 109.71. The sample standard deviation of 99.70 falls within that window. We cannot reject the hypothesis that the sample was drawn from a Poisson distribution of mean 7742.In order to test for time-dependent errors, a lag test for autocorrelation is performed. > w1 = Sr[ 1:29 ]> w2 = Sr[ 2:30 ]> cor( w1, w2 )[1] -0.06252135> plot( w1, w2, main=”Scatter Plot of Sr[i] vs Sr[i+1]”)Figure S14. Lag test for autocorrelation, where each point (Sr count) has been compared to the successive point.Neither the correlation coefficient of -0.0625 nor the scatter plot suggests the data is correlated. There is no evidence of time dependent behavior in the data. Finally, as a final test that the data is drawn from a single Poisson distribution, a C control chart is produced.> cl = mean( Sr )> cl[1] 7742.067> ucl = cl + 3 * sqrt( cl )> ucl[1] 8006.034> lcl = cl – 3 * sqrt( cl )> lcl[1] 7478.1> plot( Sr, type=”b”, ylim=c(7400,8100),main=”C Chart for Sr” )> abline( h=cl, col=”red” )> abline( h=ucl, col=”red” )> abline( h=lcl, col=”red” )> quartz.save( “CSr.png” )Figure S15. C control chart for Sr counts data.Once again, none of the indicators which would raise a question are present.Analysis of the Sr/Ca ratioThe Sr/Ca ratio is not a measurement, but a transformation on two random variables. Thus it should introduce no new source of variation.This can be described through a Taylor series approximation for the variance of the ratio of two random variables described in terms of the expected value (mean) and variance of the two variables. This is given asWhere E(x) is the mean of x, Var(x) is the variance of x, and Cov(x,y) is the covariance of x and y. (Standard deviation is the square root of variance.)Since the numerator and denominator are independent random variables, the covariance (Cov) term becomes zero.Since the variance of a Poisson random variable is equal to its mean, the equation simplifies to:First, we want to confirm that the Ca and Sr counts data are independent. We create a scatter plot to get a visual idea.> plot( Sr, Ca, main=”Scatter Plot of Sr counts vs Ca counts”)> quartz.save(“SrvsCa.png”)Figure S16. Scatterplot comparison of Sr counts and Ca counts values.And we calculate the correlation to confirm numerically.> cor( Sr, Ca )[1] 0.1247869Since the correlation is low, and the scatter plot shows no sign of a relationship, it seems reasonable to assume independence.To test the reasonableness of the proposed formula for variance, we turn once again to a simulation.> s = rpois( 10000, 7742 )> c = rpois( 10000, 2383 )> r = s / c> mean( r )[1] 3.250387> sd( r )[1] 0.07684884> sqrt( ( 7742 / 2383 )^2 * ( 1/7742 + 1/2383 ) )[1] 0.07610935The standard deviation estimated from the mean count values closely matches the actual calculated standard deviation.At this point, we have no information about the shape of the distribution of the ratio. We know only that it is a discrete distribution (defined only at specific points). It is therefore definitely NOT a normal distribution.But we can look at a histogram for some clue.> hist( r, main=”Histogram of simulated ratio”)> quartz.save( “histsimr.png” )Figure S17. Histogram of simulated Sr/Ca ratio values.This “looks” normally distributed, so we plot as a density and superimpose a fitted normal distribution with mean of 3.25 and standard deviation of 0.076845.> hist( r, main=”Histogram of simulated ratio”, prob=T)> curve( dnorm(x, mean=3.25, sd= 0.07685, add=T )> quartz.save( “histsimr2.png” )Figure S18. Histogram of simulated Sr/Ca ratio values with a superimposed fitted normal distribution with mean of 3.25 and standard deviation of 0.076845.The fitted distribution is very close to the simulated data.As a further check, a Kolmogorov-Smirnov test is applied.> ks.test( r, pnorm, mean=3.25, sd=0.07685 )The returned p=value of 0.745 says we cannot reject the hypothesis of approximate normality (although the presence of ties in the data confirms that it is not a continuous distribution).Finally, we apply the Kolmogorov-Smirnov test to the actual measured ratio data.> ks.test( R, pnorm, mean=3.25, sd=0.07685 )The returned p-value of 0.6903 says we cannot reject the hypothesis that the distribution of the Sr/Ca ratios is approximately normal with standard deviation as predicted by the formula above.Thus the tolerance of the ratio measurement, as a fraction of the ratio measurement is a function of numerator and denominator counts given byThe shape of the surface represented by the formula is not easy to visualize. We construct a surface plot of tolerance as a function of numerator and denominator counts as follows.> rr = function( num, den ) { (den/num) * 3 * sqrt( (num/den)^2 * ( 1/num + 1/den ) }> num = seq( 10, 500, by=10 )> den = seq( 10, 500, by=10 )> z = outer( num, den, rr )> persp( num, den, z, theta=120, phi=20,ticktype=”detailed”, main=”Three sigma tolerance of ratio vs Num, Den” )Figure S19. Surface plot of tolerance as a function of numerator and denominator counts.SummaryFor measurements involving a single count, there is no evidence that the measurement uncertainty is not simply the variability of a Poisson distribution with mean equal to counts. Since the variance of a Poisson distribution is equal to its mean, the three sigma limits of a measurement would be given by where is the measured value in counts. Given as a fraction of the measurement, this is Over the range of 10 to 10,000 counts, this is given by> x=seq(10,10000,by=10)> y= 3/sqrt(x)> plot( x, y, log=”x”, type=”l”, main=”Measurement uncertainty vs Counts” )> quartz.save( “EvsC.png” )Figure S20. Measurement uncertainty, as three sigma limits of a measurement (y-axis), versus counts values (x-axis).In the absence of known deterioration of measurements at higher counts, this argues for longer sampling times to increase accuracy.Since the distribution is Poisson, rather than normal, the three sigma limits do not represent the same percentage of the data as if it were normal, however for all reasonable values of counts, more than 99.7% of the data will fall within the three sigma limits.For measurements involving a ratio of two counts, the standard deviation of the ratio is approximated byThe ratio is approximately normally distributed, implying that 99.7% of the values will fall within +/- 3 standard deviations of the mean. ................
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