General rules for error in spectrum chemical analysis and data processing in laboratories
Some standard content:
Guiding Technical Documents of the Ministry of Electronics Industry of the People's Republic of China SJ/Z3206.14-89
General Rules for Errors in Spectrochemical Analysis and Experimental Data Processing This standard applies to the classification and representation methods of errors in spectral chemical analysis, as well as the statistical processing methods of experimental data. It is used to test the reliability of measurement data, so as to make a reasonable evaluation of the analysis method. This standard is also applicable to the processing of element concentrations and other measurement data related to element concentrations in spectral chemical analysis. 1 Preamble
1.1 The terms used in this general rule can be found in GB9259-88 (Terms and Terms for Emission Spectroscopy Analysis). 2 Errors and their classification
2.1 Error (E) The difference between the measured value (X) and the true value (μ). That is,
E=X-μ
2.2 Classification of errors
2.2.1 Systematic error The deviation of all or most of the measured values from the true value in one direction, which has repeatability and numerical constancy. The cause of its occurrence can be found and corrected or eliminated. It determines the accuracy of the measurement result.
2,2.2 Random error Measurement error caused by accidental changes in some factors in measurement. Relative to the true value, the error size and positive and negative directions are uncertain. From the overall observation, the probability of small errors is greater than that of large errors. It is difficult to avoid, but has statistical regularity (see 5). Random error determines the precision of the measurement result. 2.2.3 Negligent error Error caused by operational or calculation errors, so it can be avoided. It is not considered when estimating the overall error.
3 Deviation
Absolute deviation (d) The difference between the measured value (X,) and the average value (x) of a series of measured values. 3.1
(3)
Where is the arithmetic mean, n is the number of measurements. When it increases infinitely and there is no systematic error, the value converges to the true value μ. At this time, the deviation can be treated as an error. 3.2 Standard deviation The sum of squared deviations divided by the square root of the number of measurements. 3.2.1 When the number of measurements n is infinite (n>20), the standard deviation is represented by the symbol 6. >
(X,-X)2/n
Approved by the Ministry of Electronics Industry of the People's Republic of China on February 10, 1989·(4)
Implemented on March 1, 1989
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3.2.2 When the number of measurements is limited (n<20), the standard deviation is represented by the symbol S, S=
(X;-X)2)(n-1)
Where (-1)=f, called the degree of freedom. It means that there is only #-1 variable error in the measurement. The calculation procedure for this is shown in Appendix C2.1.
The percentage of the standard deviation of a single measurement to the mean value. 3.2.3 Relative standard deviation (RSD)
S×100
For the procedure to implement this measurement, see Appendix C2.1.
3.2.4 Standard deviation of the mean (S yuan) The standard deviation of a single measurement divided by the square root of the number of measurements. S
Sa=±Vn
(Xi-x)/n(n-1)
Calculation of total standard deviation When there are K groups (K>1) of similar repeated measurements, the total standard deviation is the sum of the individual standard deviations.
((n-1)S,+(n-1)s,*+.+(n-1)Sr)Se\V
(ni-1)+(nz-1)+...+(nt-1)3.3 Variance (S*) The square of the standard deviation. s1
n-1i=1
(XX)
For the implementation of this calculation procedure, see Appendix C2.1.
(9)
·(10)
3.4Range (R) The difference between the highest measurement value (X) and the lowest measurement value (X,) in n measurement data under the same conditions.
The R value describes the spread of a series of measurement values. For the implementation of this calculation procedure, see Appendix C2.1.
4Probability distribution of random errors
·(11)
4,1Normal distribution The distribution law followed by random errors in large-scale measurements, and the frequency of measurement values with zero error is the highest. Relative to the mean, the frequency of measurement values with equal errors and opposite signs is the same. As the error increases, the frequency decreases exponentially.
4.1.1 Error interval probability If the error is expressed in units of standard deviation and 1=distribution law, the probability of the error appearing in any interval can be known. 2
Fly, according to the normal
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When u=±1, the interval probability of ×=μ±16 is 68.3%, when u=±2, the interval probability of X=μ+26 is 95.5%, when u=±3, the interval probability of ×=μ±36 is 99.7%. 4.2T distribution In a finite number of measurements, the random error does not follow the distribution law. Because there is an error between the mean and the true value, the distribution law of the random error needs to be calculated according to the new variable (t) value. The meaning of the t value is the value of the error of the mean expressed in units of the standard deviation of the mean. ()
μ=#±ts/Vn
where t is related to the degree of freedom and the confidence level. Vn
(12)
(13)
4.3 Confidence Level The probability that the mean value (X) falls within the range of ±t·s/Vn is expressed as P%. The probability outside this range is α%. The relationship between P and α is P=1-α. The confidence level of P=95% is usually used in analytical chemistry.
The procedure for implementing this calculation method is shown in Appendix C3.1. 4.4 Confidence Limit (CL) The true value falls within a certain limit around the measured value at a specified confidence level. For example, in (13), t·s/n is the confidence limit of the true value (μ). Therefore, the size of the confidence limit depends on the measurement precision, the number of measurements and the confidence level. The procedure for implementing this calculation method is shown in Appendix C3.1. 5 Data Evaluation
5.1 The test for outliers is used to determine suspicious values that are deviated greatly due to errors. If it is determined to be the result of obvious negligence after inspection, it should be discarded. The following test method can be used to determine the discard. 5.1.1 Q test method Arrange the measured data in order of size, X., X., X..X., where X or X. is a suspicious outlier. Calculate the difference between the outlier and the adjacent value and divide it by the range. The quotient obtained is the Q test value. Q=(X,-X)/(XX,)..
Q.=(XX-1)/(XX,)
If the calculated Q value is greater than the Q value listed in Table 1, the value should be discarded as an outlier. Table 1 Q value table
Number of measurements
(14)
(15)
, 0.90, 0.95, 0.99 confidence levels. SJ/Z3206.14-89
For the procedure to implement this calculation method, see Appendix C1. 5.1.2 G test method Arrange the measured data in order of size: X, Xa, Xa, ….X., where X, or X. is a detectable outlier, calculate the difference between the outlier and the mean, divide it by the standard deviation, and the quotient is the G test value.
Gi=(XX.)/SG.=(xX)/S.
If the calculated G value is greater than the G value listed in Table 2, the outlier should be discarded. Table 2 G value table
Number of measurements
*0.95, 0.99 are confidence levels.
(16)
5,2 Significance test Use statistical methods to test whether there is a statistically significant difference in the problem being dealt with.
5.2.1 t test method Two groups of measurements from the same population should have the same standard deviation (which can be determined by the F test). Otherwise, the difference in their mean values is not only due to random errors, but also to systematic errors. The t test method is mainly used to determine the systematic error in the mean value. 5.2.1.1 Comparison with standard samples Compare the mean value of n measurements with the nominal value of the standard sample (regarded as the true value μ) to determine whether the two are consistent. Use formula (12) to determine. If the calculated t value is greater than the tabulated t value (see Appendix A), it is considered that there is a significant difference between the measured mean value and the nominal value, that is, there is a systematic error. 5.2.1.2 Comparison between two average values If two laboratories or two different methods are used to measure the same sample n, and n. times, and the corresponding average values Xi and X, standard deviations S, and S2 are obtained, the total standard deviation should be: SJ/Z 3206.14-89
/(n1-1)St*+(n2-1)S2\/(ni+n,-2)Stotal=y
Then t=|Xt-Xa/.Vgrna/(n,+n,)
(17)
If the calculated t value is greater than the tabulated t value (see Appendix A), it is considered that there is a significant difference between the two mean values at the selected confidence level.
See Appendix C4 for the procedure to implement this calculation method. 5.2.2 F test method Variance and standard deviation are important indicators reflecting measurement precision. Compare the variances of two groups of data to compare whether there is a significant difference in their precision. Suppose the variances of the two groups of measurement data are S, and S,2 respectively, and let S,2>S. , then the statistic of the F test is: F=S,/S,2
.......(18)
If the calculated F value is greater than the tabulated F value (see Appendix B), it indicates that there is a significant difference in the precision of the two groups of measurement values at the selected confidence level.
For the procedure to implement this calculation method, see Appendix C5. 6 Analysis of variance
Analysis of variance is to decompose the total variance of the experiment into the variance components of each factor, so that the influence of each factor can be distinguished, and then use the significance test method to judge the relative size of the influence of each factor on the measured data. The analysis of variance is based on the rule of summation of variances, and the total variance should be the sum of the variances of each factor. Each factor is represented by the following symbols: L-laboratory, [-number of laboratories, M-analysis method, m-number of methods, S-sample, K-number of samples, A-analyst, g-number of analysts, n-number of analysis repetitions. 6,1 Single-factor analysis of variance refers to an experiment in which only one certain factor may affect the measured data. For example, the same sample is sent to two laboratories for analysis, and each laboratory repeats the analysis n times. It is required to judge whether there is a significant difference between the average values of one laboratory.
Here, the sum of squares of the total deviation comes from the sum of squares of deviations between laboratories and the sum of squares of deviations within each laboratory.
6.1.1 Calculation steps
6.1.1.1 Calculate the total measurement value (TL) of each factor, that is, the total measurement value of each laboratory. Tt-2X
Where. L=1, 2, l, X is each individual measurement value, i=1, 2, nl. 6.1.1.2 Calculate the total measurement value (G) Add up all the measurement values of all factors, 1
G=ZTL=ZXi
L=li=l
6.1.1.3 Calculate the correction factor (CF) The total square of the measurement value divided by the total number of measurements. CF=G/ln
6.1.1.4 Calculate the total square sum (Qtotal)
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The sum of the squares of all the measurement values minus the correction factor. Qe=Ex.-G*/1n=zX.-CF
6.1.1.5 Calculate the sum of squares between factors (Q闻) correction factor.
Q间=—≥T,*-G+*/ln
Calculate the sum of squares within factors (Q内)
Q内=Q总-Q间=X*
Add the total sum of squares of each laboratory and divide it by the number of measurements, then subtract the total sum of squares minus the sum of squares between laboratories.
Calculate the average square between factors (ngL) and divide the sum of squares between laboratories by the degrees of freedom. M6L.=
(1-ZTi-G*/ln
Calculate the mean square within the factor (6.) The sum of squares within the factor is divided by the degrees of freedom. 6.
1(n-1)
Establish variance analysis table 3,
sum of squares
degrees of freedom
(ZX'nET)
One-way variance analysis table
average square
l(n-1)
)/(1-1)
1.2T)/(n-1)
Ihn)/(l~1)
The procedure for implementing this calculation method is shown in Appendix C6.1. Variance composition
6. a+n 62
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6.1.1.10 Calculate the F value and perform a significance test. The ratio of the inter-laboratory variance component to the intra-laboratory component. F=(60*+n6,*)/60z
If the calculated F value is greater than the tabulated F value (see Appendix B) at the selected confidence level, it means that there are significant differences between factors (laboratories).
6.2 Three-factor variance analysis In analytical chemistry, the influence of three factors on measurement data is often avoided. For example, samples with K different contents are distributed to 9 analysts, and each uses m different methods for analysis. Each method is repeated times, and the interaction effects between factors are considered. It is required to judge whether there are significant differences between factors and between the interaction effects of factors.
The variance here comes from: a. Between samples, b. Between methods, c. Between analysts, d. Interaction effects between factors, e. Experimental errors within each factor.
6.2.1 Calculation steps
6.2.1,1 Calculate the sum of the measured values of each factor (T) Sum of values:
n·m·k
S=-l...K.
Al......q.
6.2.1.2 Calculate the sum of all measured values (G): k·m·n·g
6.2.1.3 Calculate the correction factor (CF)
CF-G\/kmnq
6.2.1.4 Calculate the total sum of squares (Qtotal)
The total sum of measured values of each sample, each method, and each analyst is respectively The sum of squares is the sum of squares of all measured values for the total number of measurements minus the correction factor: Qtotal=Zx.-G*/k·min.g
6.2.1.5 Calculate the sum of squares between factors. Add the total squares of each factor and divide it by the product of the level of other factors except the level of this factor and , and then subtract the correction factor. The sum of squares between samples Qs-
ZTs\-CF
The sum of squares between analysts QA=K·m·n
The sum of squares between methods Q=
ZTA-CF
ZTm-CF
6.2.1.6 Calculate the total sum of squares of each interaction effect SJ/Z 8206.14-88
a. The total sum of squares of the interaction effect between the sample and the analyst (expressed as S·A), assuming that the total measured values in the "cell" are C, then
1-ZC*-CF-(Qa+Qs)
Note, "cell" refers to the data measured under the same conditions. Generally, they are placed in the same cell when listing data. For details, see the calculation application example 7 (below)
b The total sum of squares of the interaction effect between the sample and the analytical method (S·M), assuming that the total measured values in the "cell" are d, then
d-CF-(Qs+Qm)
C. The total sum of squares of the interaction effect between the analyst and the analytical method (A·M), assuming that the total measured values in the "cell" are eShun Xie
1,-Ze*-CF-(QA+Qu)
d. The total sum of squares of the sample, analysis method and analyst interaction effects (S·M·A). Assuming that the total number of measurements within the "cell" is , then
1Zf*-CF-(Qg+Qn+QA+Qs.m+Qs.A+Qm.A)Qs.MA=
6.2.1.7 Calculate the sum of squares of all measurements of the total intra-factor error (Q intra), minus the sum of squares of the total measurement values within the "cell" of the three-factor interaction effect divided by the number of measurements. Q inside -2x*-——2f*
Establish variance analysis table 4
Variance source
Mutual S·M
Effect A·M
Response sM·A
Sum of squares Q
ET\-CF
m·nq
ET*-CF
k·m·n
Degrees of freedom f
ZC*-CF-(Q+Q)J(k-1)(Q-1)
Zd*-CF-(Qs+Q)
Ze\-CF-(QA+Qu)
1 Zf*-CF-(Qs+Qn+Q)
+Q5.M+QS.A++QA.n
x\-G\/k..nqn
(k-1)(m-1)
(q-1)(m-1)
(g-1)(m-1)
kq.m(n-1)
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Table 4 Three-factor variance Analysis table
Average square
(ki)(q-1)
(k-1)(m-1)
(q-1)(m-1)
(g-1)(m-1)(k-1)
Zx*-G\/km-qn
kqm(n-1)
C+nosa+qno sm+mnosa+qmnggwwW.bzxz.Net
g,++ngsAm+knoAm+mngsm+kmnga
gi+ng\Am+kngAm+qngsu+kqnom
O.*+OSAx+m.ngSA
g,*+ngSAn+qngom
o,\+ngAm+kngA
ga+ngiAn
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6.2,19 Let us calculate the F value. Perform a significance test and make a judgment. The procedure for implementing this calculation method is shown in Appendix C7.1. 7 Calculation and Application
7.1 Outlier Test
Example 1: A group of b measurements are arranged in order of size as 0.02, 0.12, 0.16, 0.18, 0.18, 0.20. The first data is a suspicious value, and it is determined whether it is an outlier. The calculation is as follows:
a. Arrange the data in order of 0.02, 0.12, 0.16, 0.18, 0.18, 0.20. b. Calculate according to formula (14): Q, = 0,12-0.02/(0.20-0.02) = 0.56c Judgment: When n = 6, look up Table 1Q. .9 = 0.56, which is equal to the calculated value, and the data should be retained. The procedure for implementing this calculation method is shown in Appendix C1.2. 7.2 Precision calculation
Example 2: A set of 10 measurement data is shown in Table 5 below: Table 5 Measurement data deviation statistics
Standard deviation (S): From formula (5), S = V0.3306/9 = 0.192. Variance (S\): From formula (10), S = 0.3306/9 = 0.0367. Relative standard deviation (RSD): From (6), we get RSD = 0.192 × 100/4.81 = 3.35%.
Range (R): From (11), we get R = 5.20-4.50 = 0.70. (Xx)2
For the procedure to implement this calculation, see Appendix C2.2. 7.3 Confidence limits (CL) of the mean value
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Example: A group of 3 measurements with a low of 1.11, 112, 1,13, 115, 1.16. Calculate the limit of the mean value when the confidence level is 95%.
a, calculate the mean value × = 1.13, standard deviation S = 0.02. b. From Appendix A, we can find out that t(4, 0.5) = 2.78C. From (13), we can get CL = 1.13 ± 2.78 × 0.02/V5 = 1,13 ± 0.027. d. The average value is between 1.10 and 1.16.
The procedure for implementing this calculation method is shown in Appendix C3.2. 7.4 Comparison of average values
7.4.1 Comparison with standard samples
Example 4: The content of element in the standard sample is 0.123. The results of the four determinations using the new method are 0.112, 0.115, 0.118, and 0.11. Is there a systematic error between the average result and the content of the standard sample? a. The average value × = 0.116, and the standard deviation S = 0.0032. b. From Appendix A, we can find out t(s, 0.5) = 3.18c. From (12), we can calculate t = [0.116-0.123/V4/0.0032 = 4.38d. The calculated t value is equal to the t value in the table. There is a systematic error. It can also be seen intuitively that the average value of the measured result is 0.007 lower than the standard value, and there is a negative systematic error. The procedure for implementing this calculation method is shown in Appendix C4.2. 7.4.2 Comparison of two half means
Example 5: Two analysts use the same method to analyze the same batch of samples, 11=7, 12=9 times respectively, and the average values X, -92.0B, X, = 93.08; their variances S, = 0.6505, S2 = 0.6354. Determine whether there is a systematic error in the two analysis results.
a The consistency of the precision of the two groups of measurement results was checked by the F test method, F=0.6505/0.6354=1.02
Check the F distribution (see Appendix B) to get F.9588)=3.53, which is greater than the calculated value, proving the consistency of precision. b Total standard deviation (Stotal): From (9), Stotal=V(7-1)×0.6505+(9-1)×0.6354/(7+9-2)=0.80c. From (17), the t value is calculated
t=[(93.08-02.08)/0.803×V7×9/(7+9)=2.48d. Check the t distribution (see Appendix A). t(14,0.0s)=2.14e, the calculated t value is greater than the t value, and there is a systematic error between the two analysis results. f. When the direct confidence level is 5% and f=14, the confidence interval is: X1-X=±2.15×0.800
02=±0.37
Therefore, the interval of the systematic error is (1.00~0.87, 1.00-0.87) or (0.13, 1,87). 11
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The procedure for implementing this calculation method is shown in Appendix C5.2. 7.5 Analysis of variance
7.5.1 Single-factor analysis of variance
Example 6: The same sample was sent to three laboratories for analysis, and the data are shown in Table 6 below. Determine whether there are any significant differences between the laboratories, and calculate the confidence limits of the average values of each experimental analysis. Table 6
Analysis data of three laboratories
This example is a single-factor equal-repeated analysis of variance. The total variance comes from the variance between laboratories and the variance within laboratories.
For the convenience of calculation, all measured values are subtracted by 0.77 and then multiplied by 100, that is, X=(X-0.77)×100. The simplified data is shown in Table 7 below:
Table 7 Data after conversion
Calculation steps are as follows:
a The sum of the measured values of each laboratory: TA=11, T=-51, Tc--11.12123/V4/0.0032=4.38d, the calculated t value is less than the listed t value. There is a systematic error. It can also be seen intuitively that the average value of the measured result is 0.007 lower than the standard value, and there is a negative systematic error. The procedure for implementing this calculation method is shown in Appendix C4.2. 7.4.2 Comparison of two half means
Example 5: Two analysts use the same method to analyze the same batch of samples, 11=7, 12=9 times respectively, and the average values X, -92.0B, X, = 93.08; their variances S, = 0.6505, S2 = 0.6354, determine whether there is a systematic error in the two analysis results.
a The consistency of the precision of the two groups of measurement results was checked by the F test method, F=0.6505/0.6354=1.02
Check the F distribution (see Appendix B) to get F.9588)=3.53, which is greater than the calculated value, proving the consistency of precision. b Total standard deviation (Stotal): From (9), Stotal=V(7-1)×0.6505+(9-1)×0.6354/(7+9-2)=0.80c. From (17), the t value is calculated
t=[(93.08-02.08)/0.803×V7×9/(7+9)=2.48d. Check the t distribution (see Appendix A). t(14,0.0s)=2.14e, the calculated t value is greater than the t value, and there is a systematic error between the two analysis results. f. When the direct confidence level is 5% and f=14, the confidence interval is: X1-X=±2.15×0.800
02=±0.37
Therefore, the interval of the systematic error is (1.00~0.87, 1.00-0.87) or (0.13, 1,87). 11
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The procedure for implementing this calculation method is shown in Appendix C5.2. 7.5 Analysis of variance
7.5.1 Single-factor analysis of variance
Example 6: The same sample was sent to three laboratories for analysis, and the data are shown in Table 6 below. Determine whether there are any significant differences between the laboratories, and calculate the confidence limits of the average values of each experimental analysis. Table 6
Analysis data of three laboratories
This example is a single-factor equal-repeated analysis of variance. The total variance comes from the variance between laboratories and the variance within laboratories.
For the convenience of calculation, all measured values are subtracted by 0.77 and then multiplied by 100, that is, X=(X-0.77)×100. The simplified data is shown in Table 7 below:
Table 7 Data after conversion
Calculation steps are as follows:
a The sum of the measured values of each laboratory: TA=11, T=-51, Tc--11.12123/V4/0.0032=4.38d, the calculated t value is less than the listed t value. There is a systematic error. It can also be seen intuitively that the average value of the measured result is 0.007 lower than the standard value, and there is a negative systematic error. The procedure for implementing this calculation method is shown in Appendix C4.2. 7.4.2 Comparison of two half means
Example 5: Two analysts use the same method to analyze the same batch of samples, 11=7, 12=9 times respectively, and the average values X, -92.0B, X, = 93.08; their variances S, = 0.6505, S2 = 0.6354, determine whether there is a systematic error in the two analysis results.
a The consistency of the precision of the two groups of measurement results was checked by the F test method, F=0.6505/0.6354=1.02
Check the F distribution (see Appendix B) to get F.9588)=3.53, which is greater than the calculated value, proving the consistency of precision. b Total standard deviation (Stotal): From (9), Stotal=V(7-1)×0.6505+(9-1)×0.6354/(7+9-2)=0.80c. From (17), the t value is calculated
t=[(93.08-02.08)/0.803×V7×9/(7+9)=2.48d. Check the t distribution (see Appendix A). t(14,0.0s)=2.14e, the calculated t value is greater than the t value, and there is a systematic error between the two analysis results. f. When the direct confidence level is 5% and f=14, the confidence interval is: X1-X=±2.15×0.800
02=±0.37
Therefore, the interval of the systematic error is (1.00~0.87, 1.00-0.87) or (0.13, 1,87). 11
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The procedure for implementing this calculation method is shown in Appendix C5.2. 7.5 Analysis of variance
7.5.1 Single-factor analysis of variance
Example 6: The same sample was sent to three laboratories for analysis, and the data are shown in Table 6 below. Determine whether there are any significant differences between the laboratories, and calculate the confidence limits of the average values of each experimental analysis. Table 6
Analysis data of three laboratories
This example is a single-factor equal-repeated analysis of variance. The total variance comes from the variance between laboratories and the variance within laboratories.
For the convenience of calculation, all measured values are subtracted by 0.77 and then multiplied by 100, that is, X=(X-0.77)×100. The simplified data is shown in Table 7 below:
Table 7 Data after conversion
Calculation steps are as follows:
a The sum of the measured values of each laboratory: TA=11, T=-51, Tc--11.12
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