Statistical interpretation of data - Multiple comparison for test results
Some standard content:
ICS03.120.30
National Standard of the People's Republic of China
GB/T10092—2009
Replaces GB/T10092-1988
Statistical interpretation of dataMultiple comparison for test results2009-10-15Published
General Administration of Quality Supervision, Inspection and Quarantine of the People's Republic of ChinaStandardization Administration of the People's Republic of China
Digital anti-counterfeiting
2009-12-01Implementation
Normative references
Terms and definitions
Test leadership team and its responsibilities
Collation, calculation and verification of test results
5.1Collation of test results
Calculate the mean of each treatment and the sample variance of the treatment5.2
5.3Test for homogeneity of variance
Recalculate the mean of the treatment and calculate the estimate of the common variance5.4
Multiple comparison procedures
6.1 Comparison between treatments and reference treatment6.2 Pairwise comparison of treatments ( (T method)
6.3 Comparison of treatment means between child groups (S method) 7 Application examples
Appendix A (Normative Appendix) Selection of different multiple comparison methods and determination of the number of experimental repetitions n Appendix B (Informative Appendix) Statistical tables for multiple comparisons GB/T10092—2009
“Statistical processing and interpretation of data” includes the following national standards: GB/T3359
GB/T3361
GB/T4087
GB/T4088
GB/T4089
GB/T4882
GB/T4883
GB/T4885
GB/T4889
GB/T4890
GB/T80 55
GB/T8056
GB/T6380
GB/T10092
GB/T10094
Statistical processing and interpretation of data
Statistical processing and interpretation of data
Statistical processing and interpretation of data
Statistical processing and interpretation of data
Statistical processing and interpretation of data
Statistical processing and interpretation of data
Statistical processing and interpretation of data
Determination of statistical tolerance interval
GB/T10092—2009
Comparison of two means in the case of paired observationsOne-sided lower confidence limit of reliability of binomial distribution
Estimation and test of binomial distribution parameters
Estimation and test of Poisson distribution parameters Test
Normality test
Judgment and treatment of outliers in normal samples
Lower confidence limit of reliability of complete sample of normal distributionStatistical processing and interpretation of data
Statistical processing and interpretation of data
Statistical processing and interpretation of data
Statistical processing and interpretation of data
Statistical processing and interpretation of data
Statistical processing and interpretation of data
Estimation and test of mean and variance of normal distributionPower of test for mean and variance of normal distributionParameter estimation of F distribution (Pearson type III distribution)Judgment and treatment of outliers in samples from exponential distributionJudgment and treatment of outliers in samples from type 1 extreme value distributionStatistical processing and interpretation of dataMultiple comparisons of test resultsConfidence limits of quantiles and coefficient of variation of normal distributionThis standard replaces GB/T10092-1988 "Multiple Comparisons of Test Results". Compared with GB/T10092-1988, the main changes of this standard are as follows: the standard format has been revised according to the requirements of GB/T1.1-2000 "Guidelines for Standardization Part 1: Structure and Writing Rules of Standards";
added terms and definitions;
changed "consistency test of variance" to "homogeneity test of variance"; modified the note after 4.3.2 of "Consistency test of variance"; unified "comparison of treatment results" to "comparison of treatments"; corrected some erroneous formulas;
deleted GB/T10092-1 988. Appendix A of this standard is a normative appendix, and Appendix B is an informative appendix. This standard is proposed and managed by the National Technical Committee for Standardization of Statistical Methods (SAC/TC21). Drafting units of this standard: China National Institute of Standardization, Institute of Mathematics and Systems Science of the Chinese Academy of Sciences, Shenzhen Institute of Metrology and Testing, Beijing Institute of Technology.
Main drafters of this standard: Ding Wenxing, Xiang Kefeng, Zhang Peng, Xie Tianfa, Yu Zhenfan, Chen Min, Wu Guofu, etc. The previous versions of the standards replaced by this standard are: GB/T100921988.
SE SONI
1 Scope
Statistical processing and interpretation of data
Multiple comparisons of test resultsbzxz.net
GB/T10092—2009
This standard specifies the basic principles and methods for multiple comparison tests and statistical analysis of the same single indicator of multiple treatments, in order to obtain comparative conclusions.
This standard applies to any comparison of the same single indicator (mean) in production and scientific experiments. For example, comparing the quality indicators of several products, the results of several process conditions or several test methods. Note: This standard assumes that the test results of the same treatment are from the same normal population, and the variances of the different treatments involved in the comparison are basically the same. 2 Normative referenced documents
The clauses in the following documents become clauses of this standard through reference in this standard. For any dated referenced documents, all subsequent amendments (excluding errata) or revisions are not applicable to this standard. However, parties to an agreement based on this standard are encouraged to study whether the latest versions of these documents can be used. For any undated referenced documents, the latest versions are applicable to this standard. Standard. GB/T3358.1 Statistical Vocabulary and Symbols Part 1: General Statistical Terms and Terms Used in Probability (GB/T3358.1-2009, ISO3534-1:2006, IDT)
GB/T3358.2 Statistical Vocabulary and Symbols Part 2: Applied Statistics (GB/T3358.22009, ISO3534-2:2006 IDT)
GB/T3358.3 Statistical Vocabulary and Symbols Part 3: Experimental Design (GB/T3358 .3—2009, ISO3534-3:1999, IDT)
GB/T6379.2 Accuracy (trueness and precision) of measurement methods and results Part 2: Basic methods for determining the repeatability and reproducibility of standard measurement methods (GB/T6379.22004, ISO5725-2:1994, IDT)3 Terms and definitions
GB/T3358.1. The terms and definitions established in GB/T3358.2 and GB/T3358.3 and the following terms and definitions apply to this standard. 3.1
Test resultstestresults
The characteristic value obtained by the specified test method Note 1: The test method should indicate whether the observation is one or more, and whether the reported test result is the average of the observed values or its other function (such as the median or standard deviation). It may require correction according to the applicable standard, such as the correction of gas volume according to standard temperature and pressure. Therefore, a test result can be the result calculated from several observations. In the simplest case, the test result is the observation itself. Note 2: A test method is defined in ISO/IEC Guide 2 as "a specified technical procedure for completing a test". 3.2
Treatment
The various objects involved in the comparison are called treatments. 3.3
Reference treatmentreferencetreatment
A special treatment that serves as a benchmark in the comparison of treatments 3.4
Multiple comparisonsmultiplecomparison
A statistical test for comparing multiple treatments simultaneously for significant differences GB/T10092—2009
Testsampletestsample
A prepared sample that can be used for one or more tests or analyses, 3.6
Outlier
One or more observations in a sample that are far from other observations, suggesting that they may come from different populations. The experimental leadership group and its responsibilities
Multiple comparison experiments should be carried out in an organized and planned manner. The unit or department responsible for the experiment shall organize the experimental leadership group. At least one member of the group shall have knowledge of statistical data analysis and understand the application of multiple comparison methods. The leadership group shall discuss and determine
the specific indicators to be treated in this experiment,
the method to be used in this experiment, and the method to be used for comparison. Once determined, it cannot be changed at will.
How to arrange
How to ensure
Ensure that the repetitions of the same
treatment are carried out under basically the same conditions. Randomness of sampling
How to ensure the uniformity of samples in the preparation, distribution, transportation, storage and testing of samples? Conduct statistical analysis, discuss the reports on statistical analysis, and make comparative conclusions on the test
Collation of test results, calculation and inspection
Test results
Arrange the test results in the form of Table 1
Number of test repetitions
Reference treatment
In the table:
y is the observed value of the th treatment;
Observation value 3g
yiiyiz
Yaya,ye
yaryoz..yom
n is the number of test repetitions of the ith treatment. When designing an experiment, it is generally required that n be as identical as possible. T
5.2 Calculate the mean value of each treatment and the sample variance of each treatment. Calculate the mean value of each treatment according to formula (1): yi
Calculate the sample variance of each treatment according to formula (2): S
Arrange the calculation results in the form of Table 2.
Reference treatment
5.3 Variance homogeneity test
T,i=o,lk
Number of test repetitions
,i=0,1,..,k
Mean value of each treatment
GB/T10092—2009
（1）
（2）
Sample variance of each treatment S
5.3.1 In order to ensure that the variances of different treatments involved in the comparative test are basically equal, a variance homogeneity test must be performed. Before testing, it should be required that there are no outliers in the repeated test results of each treatment. For this purpose, it is necessary to use the Grubbs test and Dixon test specified in GB/T6379.2 to find outliers. 5.3.2 This standard stipulates that the Cochran test method is used to test the homogeneity of variance of each treatment. The value of the Cochran test statistic C is calculated according to formula (3).
Where:
The maximum value among all S.
For the critical value table of the Cochran test, please refer to Table B.1. In the table, n is the number of test repetitions. Note: If the number of test repetitions of some treatments is different, the number of repetitions of most treatments is taken. · (3)
If the value of the Cochran test statistic C is greater than the critical value of 0.05 or 0.01), after removing S, continue to test the largest variance among the remaining k-1 variances until the remaining variances meet the requirement of variance homogeneity. Treatments that are excluded because of excessive variance cannot be included in the comparison, and all their data should be eliminated. Note: The Cochran test specified in this standard is a one-sided variance homogeneity test. It only detects those with the largest variance that do not meet the homogeneity requirement. In fact, some treatments may have small variances that do not meet the homogeneity requirement. This standard does not consider the test of small variances. 5.4 Recalculate the mean of the treatments and calculate the estimated value of the common variance After the outlier test, variance homogeneity test and correction or elimination of data, the number of treatments, the number of experimental repetitions of each treatment, the mean value and the sample variance of the treatments may change. Therefore, the mean value of some treatments should be recalculated and the estimated value of the common variance should be calculated. The calculation results are organized in the form of Table 3. For simplicity, the values of symbols , n and S in Table 3 may change, but the symbols remain unchanged.
GB/T10092—2009
Reference treatment
In the table:
Number of test repetitions n
Average value of each treatment
Sample variance of each treatment S
The table is the number of treatments finally retained after the Cochran test; n: is the number of test repetitions of the ith treatment after eliminating outliers; the estimated value of common variance
is the average value of the ith treatment calculated after outlier test and correction or elimination of data, and S is the sample variance of the th treatment calculated after outlier test and correction or elimination of data; and. is the estimated value of common variance, calculated as follows: when there is no reference treatment result,
(n;—1)
where f=Nk is the degree of freedom:
when there is a reference treatment result,
where =N-(+1) is the degree of freedom.
6 Multiple comparison procedures
6.1 Comparison between k treatments and reference treatmentsf
(4)
-(5)
6.1.1 The case where the number of experimental repetitions of the k treatments involved in the comparison and the reference treatment is equal (n=n,=nz=.=n=n) (D method) Implementation steps:
a) According to the procedure in Chapter 5, obtain the mean value1 of the reference treatment and other k treatments in Table 3, and the estimated value of the common variance.
b) Given a significance level α (usually α=0.05 or 0.01), calculate the degree of freedom = N-(k+1), and according to α, look up Table B.2 to obtain the value of d.(k,于)
Note: In this case, the value in the upper right corner of d.(k,了) can be ignored. c) Calculate 1 using the following formula. The value of:
Lo=d.(h,f)
Conclusion: In 2,…,, all treatments falling within the interval of (1, + 1) are judged to have no significant difference with the reference treatment d
at level α; all treatments falling outside the interval of (. - l., y. + 1.) are judged to have significant difference with the reference treatment at level . 6.1.2 The case where the number of test repetitions of the reference treatment is greater than the number of test repetitions of the other k treatments, and the number of test repetitions of the other treatments is equal (ng>n=nz=n=n) (D method)
Implementation steps:
a) Same as 6.1.1a);
GB/T10092—2009
b) Given a significance level α (usually α=0.05 or 0.01), calculate the degree of difference = N-(k+1). According to α, d.(k,) is obtained by looking up Table B.2. The upper right corner small number g is used to calculate the corrected value d:(k,) when n:≥n. The formula is: d (k,于)=[1+g(1-n)/100a.(k,F) For example: k=2, ng=21, n=nz11.=40, a=0.05 According to Table B.2, we get: 2.291-1, that is, do.os(2,40)=2.29g1.4 Then [1+1.(12)/100|×2.29=2.31
des(2,40)=
）The value of 1 is calculated by the following formula:
l. = d: (k,f)
d)Same as 6.1.1d)
6.1.3The results of the reference treatment are known. Implementation steps:
According to the procedures in Chapter 5, the average values of the treatments in Table 3 and the estimated common variance are obtained. a)
（8）
Given the significance level, calculate the degrees of freedom = N- and α = 1-(1-a), = 1-α/2. From f, check the quantile table of t distribution in Table B,3 and get the value of t. (). The value of () calculated by the following formula:
()=t,()
（9）
Conclusion: Compare the mean values of the treatments,,, respectively, and the difference between the known values of the reference treatment. All the treatments of d)
i) are judged to have no significant difference with the reference treatment at level α, and all the treatments of () are judged to have significant difference with the reference treatment at level. 6.2k pairwise comparison of treatments (T method)
In some practical problems, there is no reference treatment as a comparison benchmark, and any two of the treatments need to be compared. There are a total of
groups for the two k(k-1)
comparisons, which can be divided into the following two situations:
=n;=n) situation
6.2.1 The number of experimental repetitions of the k treatments involved in the comparison is equal (n,=n=.. Implementation steps:
a) Same as 6.1.3a);
Given the significance level a, calculate the degree of self-determination = Nk, and look up the value of g(, f) from table B.4 by nk, f; b)
The value is calculated by the following formula.
, = g(kn)
（10）
d) Conclusion: For a specific treatment i, all treatments falling within the interval of (一＋1) are judged to have no significant difference with the i-th treatment at level α: all treatments falling outside the interval of (一＋1.) are judged to have significant difference with the i-th treatment at level α. 6.2.2 Implementation steps for the case where the number of experimental repetitions of each treatment involved in the comparison is not equal:
a) Same as 6.1.3a);
b) Same as 6.2.1b);
) is calculated by the following formula. ;
+l),=1,..j-1,..i+j.11)
t(ij)=g(k)
GB/T10092—2009
Where: n, n, are the number of experimental repetitions of any two treatments i, respectively. d) Conclusion: If l,≤l(i,), then it is judged that there is no significant difference between the ith treatment and the ith treatment at level α; if l,>l(i,j), then it is judged that there is a significant difference between the ith treatment and the ith treatment at level α. 6.3 Comparison of the means of several groups of treatments (S method) Some practical problems require comparison between the means of several groups of treatments, such as the comparison between the mean of a certain treatment and the total average, the comparison between the mean of the th treatment and the means of the 6 treatments, etc. These comparisons are collectively called arbitrary linear comparisons. This standard only specifies the following two special cases for any linear comparison: 6.3.1 Comparison of the mean of one treatment with the total mean of k treatments Implementation steps:
a) According to the procedures of 5.1 to 5.3, calculate the mean of the first treatment: and the total mean of k treatments and the estimated value of the common variance, where multiple =
Given a significance level α, from αk-1,f, look up Table B.5, and get the value of S(k,). c
Calculate. -
d) Calculate the value of L(i) by the following formula: .()=s.(kf)
（12）
e) Conclusion: Whenever ≤l(i), it is judged that the mean of the ith treatment is not significantly different from the total mean at level α, and whenever [一>(), it is judged that the mean of the ith treatment is significantly different from the total mean at level α. 6.3.2 Comparison between the mean of any a treatment among k treatments and the mean of any other b treatments Implementation steps:
a) According to the procedures of 5.1 to 5.3, calculate the mean of any a treatment among k treatments (assuming it is the first a treatment) and the mean and estimated value of common variance of any other b treatments (assuming it is the next 6 treatments). Same as 6.3.1b);
2a+b≤k
[()+() value;
Calculate the value of l.(a,b) by the following formula: d)
t.(ab)=S.(kf)
·（13）
Conclusion: If. If 一≤l.(a, b), then the mean of the α treatment and the mean of the other b treatments have no significant difference at level e
. If 一>l(a, b), then the mean of the a treatment and the mean of the other b treatments have significant difference at level α.
7 Application Example
The color characteristics of cotton are one of the main indicators for evaluating cotton grades. When the color characteristics of cotton are measured by a cotton colorimeter, the reflectivity R. and the yellow depth + 6 are measured at the same time. At present, my country's cotton standards are divided into 7 grades. From grade 1 to grade 7, the R. value gradually decreases, while the + 6 value gradually increases. People believe that there is little difference between grades 12.3. Now 12 sets of cotton samples of grades 1 to 7 are selected by 15 provinces and cities, and the results are measured by a colorimeter as shown in Table 4 and Table 4. As shown in Figure 5, this batch of data was used to compare whether there were significant differences between the grades, 6
GB/T10092—2009
Analysis: (1) This is a comparison problem of two indicators (R. and +6). The two indicators were calculated and compared separately to compare two single indicators. In this experiment, treatment means different grades of cotton. (2) According to experience, the values of R. and +6 obey the normal distribution, so there is no need to conduct a normality test. (3) The color characteristics of cotton grades 1 to 7 are quite different, so the variances of different grades (i.e. different treatments) may not be equal. The implementation steps of multiple comparisons are as follows:
The test results of 15 provinces (cities) are organized into the form of Tables 4 and 5. Calculate the mean value of each treatment and the variance of each treatment respectively. The sample variance is listed in the form of Table 6 and Table 7. Table 4
1 (level)
2 (level)
3 (level)
4 (level)
5 (level)
6 (level)
7 (level)
1 (level)
2 (level)
3 (level)
4 (level)
5 (level)
Number of test repetitions
Number of test repetitions
Test results
Reflectivity R.
Test results
Yellow depth +6
GB/T10092—2009
6 (level)
7 (level)
Test Repeat times
Test repeat times
Test repeat times
Table 5 (continued)
Test results
Yellow depth+6
Average value of each treatment
S yuan, =77.91
Average value of each treatment
Sample variance of each treatment S%
S%——5.24
Sample variance of each treatment S
The repeated test results of each treatment were tested. In Table 5, the data 12.0 was found to be a highly significant outlier (significant at the 1% level). The reason was not found, so it was eliminated and the number of test repeats, treatment half mean and sample variance of the treatment were modified accordingly in Table 9. No outliers were found in other treatments.
Test the homogeneity of variance of treatments. The variances of treatments 5, 6, and 7 in Table 6 and treatment 67 in Table 7 are all significantly larger. The treatments in Tables 6 and 7 are divided into two categories, one category is treatments 1 to 4, and the other category is treatments 5 to 7. The homogeneity of variances is tested for the first and second categories respectively. The results show that the variances of the first category are basically equal, and the variances of the second category are basically equal, but the variances of the first and second categories are not equal. The common variance estimates are calculated for the first and second categories of treatments respectively, and are listed in Tables 8 and 9. Table 8
Number of experimental repetitions
Average value of each treatment
3x,—77.26
Degree,—75.46
Variance of samples within each treatment S
S%, =3.88
Common variance estimate
Equivalent variance ...Get the value of S(k,). c
Calculate. -
d) Calculate the value of L(i) by the following formula: .()=s.(kf)
（12）
e) Conclusion: Whenever 一≤l(i), it is judged that the mean of the ith treatment is not significantly different from the total mean at level α, and whenever [一>(), it is judged that the mean of the ith treatment is significantly different from the total mean at level α. 6.3.2 Comparison between the mean of any a treatment among k treatments and the mean of any other b treatments Implementation steps:
a) According to procedures 5.1 to 5.3, calculate the mean of any a treatment among k treatments (assume to be the first a treatments). And the mean and common variance estimate of any other b treatments (assume to be the next 6 treatments) 2. Same as 6.3.1b);
2a+b≤k
[()+() value;
Calculate the value of l.(a,b) by the following formula: d)
t.(ab)=S.(kf)
·（13）
Conclusion: If. If 一≤l.(a, b), then the mean of the α treatment and the mean of the other b treatments have no significant difference at level e
. If 一>l(a, b), then the mean of the a treatment and the mean of the other b treatments have significant difference at level α.
7 Application Example
The color characteristics of cotton are one of the main indicators for evaluating cotton grades. When the color characteristics of cotton are measured by a cotton colorimeter, the reflectivity R. and the yellow depth + 6 are measured at the same time. At present, my country's cotton standards are divided into 7 grades. From grade 1 to grade 7, the R. value gradually decreases, while the + 6 value gradually increases. People believe that there is little difference between grades 12.3. Now 12 sets of cotton samples of grades 1 to 7 are selected by 15 provinces and cities, and the results are measured by a colorimeter as shown in Table 4 and Table 4. As shown in Figure 5, this batch of data was used to compare whether there were significant differences between the grades, 6
GB/T10092—2009
Analysis: (1) This is a comparison problem of two indicators (R. and +6). The two indicators were calculated and compared separately to compare two single indicators. In this experiment, treatment means different grades of cotton. (2) According to experience, the values of R. and +6 obey the normal distribution, so there is no need to conduct a normality test. (3) The color characteristics of cotton grades 1 to 7 are quite different, so the variances of different grades (i.e. different treatments) may not be equal. The implementation steps of multiple comparisons are as follows:
The test results of 15 provinces (cities) are organized into the form of Tables 4 and 5. Calculate the mean value of each treatment and the variance of each treatment respectively. The sample variance is listed in the form of Table 6 and Table 7. Table 4
1 (level)
2 (level)
3 (level)
4 (level)
5 (level)
6 (level)
7 (level)
1 (level)
2 (level)
3 (level)
4 (level)
5 (level)
Number of test repetitions
Number of test repetitions
Test results
Reflectivity R.
Test results
Yellow depth +6
GB/T10092—2009
6 (level)
7 (level)
Test Repeat times
Test repeat times
Test repeat times
Table 5 (continued)
Test results
Yellow depth+6
Average value of each treatment
S yuan, =77.91
Average value of each treatment
Sample variance of each treatment S%
S%——5.24
Sample variance of each treatment S
The repeated test results of each treatment were tested. In Table 5, the data 12.0 was found to be a highly significant outlier (significant at the 1% level). The reason was not found, so it was eliminated and the number of test repeats, treatment half mean and sample variance of the treatment were modified accordingly in Table 9. No outliers were found in other treatments.
Test the homogeneity of variance of treatments. The variances of treatments 5, 6, and 7 in Table 6 and treatment 67 in Table 7 are all significantly larger. The treatments in Tables 6 and 7 are divided into two categories, one category is treatments 1 to 4, and the other category is treatments 5 to 7. The homogeneity of variances is tested for the first and second categories respectively. The results show that the variances of the first category are basically equal, and the variances of the second category are basically equal, but the variances of the first and second categories are not equal. The common variance estimates are calculated for the first and second categories of treatments respectively, and are listed in Tables 8 and 9. Table 8
Number of experimental repetitions
Average value of each treatment
3x,—77.26
Degree,—75.46
Variance of samples within each treatment S
S%, =3.88
Common variance estimate
Equivalent variance ...Get the value of S(k,). c
Calculate. -
d) Calculate the value of L(i) by the following formula: .()=s.(kf)
（12）
e) Conclusion: Whenever 一≤l(i), it is judged that the mean of the ith treatment is not significantly different from the total mean at level α, and whenever [一>(), it is judged that the mean of the ith treatment is significantly different from the total mean at level α. 6.3.2 Comparison between the mean of any a treatment among k treatments and the mean of any other b treatments Implementation steps:
a) According to procedures 5.1 to 5.3, calculate the mean of any a treatment among k treatments (assume to be the first a treatments). And the mean and common variance estimate of any other b treatments (assume to be the next 6 treatments) 2. Same as 6.3.1b);
2a+b≤k
[()+() value;
Calculate the value of l.(a,b) by the following formula: d)
t.(ab)=S.(kf)
·（13）
Conclusion: If. If 一≤l.(a, b), then the mean of the α treatment and the mean of the other b treatments have no significant difference at level e
. If 一>l(a, b), then the mean of the a treatment and the mean of the other b treatments have significant difference at level α.
7 Application Example
The color characteristics of cotton are one of the main indicators for evaluating cotton grades. When the color characteristics of cotton are measured by a cotton colorimeter, the reflectivity R. and the yellow depth + 6 are measured at the same time. At present, my country's cotton standards are divided into 7 grades. From grade 1 to grade 7, the R. value gradually decreases, while the + 6 value gradually increases. People believe that there is little difference between grades 12.3. Now 12 sets of cotton samples of grades 1 to 7 are selected by 15 provinces and cities, and the results are measured by a colorimeter as shown in Table 4 and Table 4. As shown in Figure 5, this batch of data was used to compare whether there were significant differences between the grades, 6
GB/T10092—2009
Analysis: (1) This is a comparison problem of two indicators (R. and +6). The two indicators were calculated and compared separately to compare two single indicators. In this experiment, treatment means different grades of cotton. (2) According to experience, the values of R. and +6 obey the normal distribution, so there is no need to conduct a normality test. (3) The color characteristics of cotton grades 1 to 7 are quite different, so the variances of different grades (i.e. different treatments) may not be equal. The implementation steps of multiple comparisons are as follows:
The test results of 15 provinces (cities) are organized into the form of Tables 4 and 5. Calculate the mean value of each treatment and the variance of each treatment respectively. The sample variance is listed in the form of Table 6 and Table 7. Table 4
1 (level)
2 (level)
3 (level)
4 (level)
5 (level)
6 (level)
7 (level)
1 (level)
2 (level)
3 (level)
4 (level)
5 (level)
Number of test repetitions
Number of test repetitions
Test results
Reflectivity R.
Test results
Yellow depth +6
GB/T10092—2009
6 (level)
7 (level)
Test Repeat times
Test repeat times
Test repeat times
Table 5 (continued)
Test results
Yellow depth+6
Average value of each treatment
S yuan, =77.91
Average value of each treatment
Sample variance of each treatment S%
S%——5.24
Sample variance of each treatment S
The repeated test results of each treatment were tested. In Table 5, the data 12.0 was found to be a highly significant outlier (significant at the 1% level). The reason was not found, so it was eliminated and the number of test repeats, treatment half mean and sample variance of the treatment were modified accordingly in Table 9. No outliers were found in other treatments.
Test the homogeneity of variance of treatments. The variances of treatments 5, 6, and 7 in Table 6 and treatment 67 in Table 7 are all significantly larger. The treatments in Tables 6 and 7 are divided into two categories, one category is treatments 1 to 4, and the other category is treatments 5 to 7. The homogeneity of variances is tested for the first and second categories respectively. The results show that the variances of the first category are basically equal, and the variances of the second category are basically equal, but the variances of the first and second categories are not equal. The common variance estimates are calculated for the first and second categories of treatments respectively, and are listed in Tables 8 and 9. Table 8
Number of experimental repetitions
Average value of each treatment
3x,—77.26
Degree,—75.46
Variance of samples within each treatment S
S%, =3.88
Common variance estimate
Equivalent variance ...0 is a highly significant outlier (significant at the 1% level), and the reason was not found. It was eliminated and the number of test repetitions, treatment half means and sample variance of treatments were modified accordingly in Table 9. No outliers were found in other treatments.
Tested the homogeneity of variances of treatments. The variances of treatments 5, 6, and 7 in Table 6 and treatments 67 in Table 7 were significantly larger. The treatments in Tables 6 and 7 were divided into two categories, one for treatments 1 to 4 and the other for treatments 5 to 7. The homogeneity of variances was tested for the first and second categories respectively. The results showed that the variances of the first category were basically equal, and the variances of the second category were basically equal, but the variances of the first and second categories were not equal. The common variance estimates were calculated for the first and second types of treatments, respectively, and are listed in Tables 8 and 9. Table 8
Number of experimental repetitions
Average value of each treatment
3x,—77.26
Degree,—75.46
Sample variance within each treatment S
S%, =3.88
Common variance estimate
device-3. 720 is a highly significant outlier (significant at the 1% level), and the reason was not found. It was eliminated and the number of test repetitions, treatment half means and sample variance of treatments were modified accordingly in Table 9. No outliers were found in other treatments.
Tested the homogeneity of variances of treatments. The variances of treatments 5, 6, and 7 in Table 6 and treatments 67 in Table 7 were significantly larger. The treatments in Tables 6 and 7 were divided into two categories, one for treatments 1 to 4 and the other for treatments 5 to 7. The homogeneity of variances was tested for the first and second categories respectively. The results showed that the variances of the first category were basically equal, and the variances of the second category were basically equal, but the variances of the first and second categories were not equal. The common variance estimates were calculated for the first and second types of treatments, respectively, and are listed in Tables 8 and 9. Table 8
Number of experimental repetitions
Average value of each treatment
3x,—77.26
Degree,—75.46
Sample variance within each treatment S
S%, =3.88
Common variance estimate
device-3. 72
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