Statistical interpretation of data - Techniques of estimation and tests relating to means and variances
Some standard content:
ICS 03. 120. 30
National Standard of the People's Republic of China
GB/T 4889—2008
Replaces GB/T4889—1985
Statistical interpretation of data-Techniques ofestimation and tests relating to means and variances ofnormal distribution
(ISO2854.1976, MOD）
Released on July 28, 2008
General Administration of Quality Supervision, Inspection and Quarantine of the People's Republic of China Administration of Standardization of the People's Republic of China
Implementation on January 1, 2009
2 Normative references
3 Terms, definitions and symbols
3.1 Terms and definitions
3.2 Symbols
A General
5 Tests and estimation of the mean of a single normal population
5.1 Test of the mean of a single population
5. 2 Interval estimation of the mean of a single normal population 6 Comparison of the means of two normal populations
6.1 Test for comparison of two population means
6. 2 Interval estimation of the difference between two population means? 7.1 Tests and estimation of the variance or standard deviation of a single population 7.2 Interval estimation of the variance or standard deviation of a single population 8 Comparison of variances or standard deviations of two populations 8.1 Tests for comparison of variances or standard deviations of two populations 8.2 Interval estimation of the ratio of variances or standard deviations of two populations Appendix A (Normative Appendix) Table of statistical values Appendix B (Normative Appendix) Hypothesis testing using force values GB/T 4889—2008 GB/T 4889—2008 This standard modifies the international standard 1502851:1976. Compared with 1S02851:1976, the changes in the technical content of this standard mainly include:
---The methods and examples divided into two chapters in ISO are changed to be described in articles for easy understanding: In the description of many tables such as Table C, the original large number of repeated descriptions that are the same as the previous tables are omitted; the content of normality test is omitted: - The content of paired data comparison is omitted;
The structure is rearranged, and each method is followed by an illustrative example; - Normative Appendix B: Hypothesis testing using β values is added. This standard replaces GB/T48891985. The changes in the technical content of this standard compared with GB/T4889-1985 mainly include: the format of the standard has been revised according to the requirements of GB/T1.12000 Guidelines for Standardization Work Part 1: Structure and Writing Rules of Standards:
added terms, symbols and definitions;
changed sample size to sample quantity;
rearranged the structure, and each method was followed by an illustrative example; added normative Appendix B for hypothesis testing using force values, and both Appendix A and Appendix B of this standard are normative appendices. This standard was proposed and managed by the National Technical Committee for Standardization of Statistical Methods. The drafting units of this standard are East China Normal University, China Institute of Standardization and Chemical Technology, Peking University, and the Ministry of Railways. The main drafters of this standard are Pu Xiaolong, Li Yan, Wen Xing, Yu Zhenfan, Sun Ze, Qiu Hongjing, Gan Chunping, and Chen Yuzhong. The previous versions of the standards replaced by this standard are: GB/T4889-1985, 1 Scope
Statistical processing and interpretation of data
Estimation and test of mean and variance of normal distribution, the population involved in this standard is normally distributed, and this standard is applicable to the estimation or test of mean and variance of the population, 2 Normative references
GB/T 4889--2008
The clauses in the following documents become the clauses of this standard through reference in this standard. For the referenced documents with dates, all subsequent amendments (excluding errata) or revisions are not applicable to this standard. However, the parties to the agreement based on this standard are encouraged to study whether the latest versions of these documents can be used. For any referenced document without annotated date, the latest version shall apply to this standard. G13/T4882—2002 Statistical processing and interpretation of data Normality test IS03534-1=2006 Statistical vocabulary and symbols Part 1: General statistical terms and terms used in probability IS03534-2:2006 Statistical vocabulary and symbols Part 2: Applied statistics 3 Terms, definitions and symbols
The terms and definitions determined by ISO3534-1:2006 and ISO3534-2, 2006 and the following terms, definitions and symbols shall apply to this standard. For ease of reference, some terms are directly quoted from the above standards. 3.1 Terms and definitions
Significance level In hypothesis testing, the maximum probability of rejecting the null hypothesis (making a first-type error) when the null hypothesis is true. Note 1: For two-sided tests, the significance level is the probability of rejecting the null hypothesis when the null hypothesis is true. For single-sided tests, the significance level. is the maximum value of this probability.
Note 2: The value of should be selected according to the risk that the user is prepared to bear. In practice, it is often taken as 0.05 or 0.01. Since a hypothesis may be rejected when α=0.05 but not rejected when α=0.C1, it is appropriate to use the following statement: "The null hypothesis is rejected at the significance level α=0.05", or "The null hypothesis is not rejected at the significance level α=0.01". Go to 3: Pay special attention to the existence of the second type of error, which is committed when the null hypothesis is wrong but the null hypothesis is not rejected. For the technical spectrum of statistical tests, see ISO 3534-1:20cG.
Confidence level
The probability that the confidence interval of the estimated parameter contains the true value of the parameter. Note 1: The confidence level is usually 0.95, D.99, that is, a = 0.05 or -0.01. 3. 1.3
P quantile pquantile
The minimum value of the distribution function F() that is not less than (0 < 1). 1
GB/T 4889—2008
3.2 Symbol
1+22+***+2,
=(2 -)
s=(2-2)
4 General Provisions
Sample size
Population mean
Population variance
Population standard deviation
Simple random sample with sample size
Sample mean (
can also be expressed as, the same below)
Sample variance
Sample standard deviation
<Hypothesis test ≥ Significant sensitivity
Interval estimate ≥ Confidence level
Quantile of the standard normal distribution
Quantile of the distribution with degrees of freedom
Quantile of the \ distribution with degrees of freedom
Quantile of the F distribution with degrees of freedom and degrees of freedom
4.1 This standard assumes that samples are randomly drawn and independent. For infinite populations, independence is usually satisfied; for finite populations, when the population is large enough or the sampling ratio is small enough (for example, less than 1/10), independence can also be considered to be satisfied. 4.2 This standard assumes that the distribution of measured values is normal. If the actual distribution deviates little from the normal distribution and the approximation is not too small, it is reasonable to use the method specified in this standard, and its approximation is sufficient for most actual situations. At this time, the sample size in Tables A, B, C, and D is required to be no less than 5, and the sample size in other tables (A', B, C', D', E, F, G, and H) cannot be less than 20. 4.3 For details on the test method for the normality assumption, please refer to GB/T4882-2002. In practice, the normality of the population is usually assumed based on other information rather than the sample itself. When the normality assumption is rejected, non-parametric methods can be used to estimate or test the mean and variance, or they can be converted to an approximate normal distribution through appropriate transformations (such as log(a+a).1/z, Vr+a), and then the methods in this standard can be used. When using transformations, be cautious in choosing the transformation and interpreting the results. 4.4 If you only need to estimate the mean or variance of the variable X, regardless of whether the population follows a normal distribution, the sample mean s and the sample variance s are unbiased estimates of the population mean and population variance, respectively. 4.5 For each statistical analysis, detailed information about the data source and data collection method should be given. This information is helpful for the interpretation of the statistical analysis results. In particular, the measurement unit or the smallest unit of measurement should have practical significance. 4.6 If no experimental, technical or other reasons are found, even if the observation data is questionable, it cannot be eliminated or changed. Any observation data that is eliminated or changed and the reasons for it should be explained. 5 Test and estimation of the mean of a single normal population
5.1 Test of the mean of a single population
5.1.1 Test of the mean of a single population with known variance2
Technical characteristics of the population (see 4.5)
Technical characteristics of the sample (see 1, 5)
Divided or corrected observed value (see 4.6)
Statistical data
Sample size:
Observed sum:
Set value:
Known population variance:
Or standard deviation: -
Significance level: α=
Two-sided test:
Table A Comparison of population mean and set value (variance known) Calculation
[u-+Nna-
[u--nn o-
When |—「1/, reject the hypothesis (null hypothesis) that the population mean is equal to the set value. Single case test:
a) When --<—［u-/, reject the hypothesis (null hypothesis) that the population mean is not less than; b) When 一>「u-/n, reject the hypothesis (null hypothesis) that the population mean is not less than μ. Note 1: According to the definition of the α quantile of the standard normal distribution., we have u=一-., see the figure below: ()
:—/3
P/2/21
Double one case
P[/>-=
Single Xie case
Note 2: / is the standard deviation of the sample mean obtained by extracting observations. GB/T 48892008
Prs-a]-α
Note 3: For ease of use, the values of u1-/yuan and u1-a/2/ at α=0.05 and α=0.01 are given in Table A.1 of Appendix A. Note: The test can also be carried out using the value method, see Appendix B. Example A manufacturer claims that the average breaking strength of the cotton yarn it produces is 2.40. This can be tested. Assume that the passive conditions of the breaking strength of previous batches of cotton yarn are stable, and the standard deviation is <=0.3315. Assume that 10 samples are randomly selected and the breaking strength of the cotton yarn is measured. The data is as follows: 2.297
According to the method in Table A, the test process is as follows: 2.362
GB/T 48892008
Overall technical characteristics: A batch of cotton yarn was delivered to the manufacturer, with a total of 10,000 spools, packed in 100 boxes, each box with 100 spools. Technical characteristics of the sample: 10 boxes were randomly selected, and one spool was randomly selected from each box. 50 cm long cotton yarn was cut about 5 m from the end of the spool as the test material. In the actual experiment, only the middle 25 cm was taken. The unit of breaking strength is N. Observed values to be divided or corrected: None.
Statistical data
Sample size:
Observed values and:
Set value:
n—10
2x,—21. 761
Known standard deviation: #=0.3315
Significance level, α=0.05
Two-sided test:
From Appendix A Table A. 1
(u.ms//10)a=0.6198X0.3315-0.20551 yuan—μ [ - ↓ 2. 176 1-2. 40 =0. 223 9>0. 205 5Reject the population mean equal to 2. 40 at the 01. 05 significance level 5.1.2 Test table of single population mean with unknown variance A Comparison of population mean and set value (variance unknown) Technical characteristics of the population
Technical characteristics of the sample
Omitted or corrected observations
Statistical data
Sample plate:
Sum of observations:
Sum of squares of observations: =
Set value:
Degrees of freedom:
μ—n—1
Significance level: α
Two-sided test:
22(2-(2)/m
[sn]ti-.(v)-
[s/In]1-/2 (u)-
When—|>［-（，reject the hypothesis that the population mean is equal to the set value. One-sided test:
a) When -1「s/->, reject the hypothesis that the population mean is not less than; b) When -1>[s/t1-（>）, reject the hypothesis that the population mean is not greater than. -1
Note 1, by the definition of the α quantile of the distribution with degrees of freedom), we have. ()=—-（), see the figure below: (mt
fa)2(]=t1-α/2[5)
1-a/2 () [)
P[1()≤t /2()]-1-
Two-sided case
P[n)>hu()]-0
Single case
ta()--(u)
P[)-1-(]=α
Note 2: The estimate of the standard error of the sample mean obtained by drawing observations from the 5/yuan basis is the standard error of the sample mean. Note 3: For ease of use, the values of t1=/2()/yuan and 11-()//n at α=0.05 and a0.01 are given in Table A.3 of Appendix A. GB/T 4889-2008
Example Assume that the problem to be solved is the same as the example in 5.1.1, but this time the variance is unknown and must be estimated from the sample. This may be because there are no previous measurement data left, or it is believed that they are no longer applicable. Use the method in Table A' to solve 5.1.1 The data of the sample in the following example are analyzed: Overall technical characteristics: The manufacturer delivered a batch of cotton yarn, a total of 10,000 spools, which were packed in 100 boxes, each box containing 100 spools: Sample technical characteristics: 10 boxes were randomly selected, and one spool was randomly selected from each box. A 50 cm long cotton yarn was cut about 5 m from the end of the spool as the test material. In the actual experiment, only the middle 25 cm was taken. tm. The unit of the numerical strength is N. No observations to be excluded or corrected.
Statistical data
Sample:
Observation sum:
Z;=21.761
Observation sum of squares: Z-48.6105
Specified value:
Degrees of freedom
v-10--19
Significance level: 2-0.05
Two-sided test:
$18.610521,761/100. 139 6
s/ /10 0. 118 2
From Appendix A Table A.2:
(s//10)20.9(9)0.1182×2.2622=0.2673[=—f / -- |2, 176 1 -2, 40 / -0. 223 9-20. 267 3The hypothesis that the population mean is equal to 2.40 is not rejected at the 0.05 significance level. Note 1: Although the null hypothesis is not rejected, it cannot be confirmed that the manufacturer's claim is correct. Note 2: The sample standard deviation 5=0.3736 here is larger than the standard deviation (c=0.3315) used in the example in 5.1.1. Due to the different information, the conclusions obtained are different.
Note 3: When there is doubt as to whether the variance obtained from past experience is still applicable, it is more reliable to use an estimate of the variance. 5.2 Interval estimation of the mean of a single normal population
5.2.1 Interval estimation of the mean of a single population with known variance Table B Interval estimation of the population mean (with known variance) Technical characteristics of the population
Technical characteristics of the sample
Eliminated or corrected observations
Statistical data
Sample data:
Observations and:
Known population variance:. -
or standard:
Confidence level:
E-- n'a=
[up/2 N0-
Point estimate of the population mean: one changes to one
Two-sided confidence interval-[-2/o+[:-a/a
One-sided confidence interval: μ［u-
or- +N
The example still uses the data in the example in 5.1.1. In this example, we do not test whether the population mean is equal to a certain set value, but to find a confidence interval with a confidence level of 1-α for the mean μ. The method in Table B can be used. At this time, it is also sufficient to assume that the population variance has been obtained from historical data, which is =0. 331 5.
GB/T 4889—2008
Technical characteristics of the population:The manufacturer delivered a batch of ladder yarn, a total of 10,000 spools, packed in 100 pins, each pin has 10 spools. Technical requirements of the sample: 10 boxes were randomly selected, and one spool was randomly selected from each box. A 50cm long cotton yarn was cut about 5m from the end of the original spool as the test material. In the actual experiment, only the middle 25cm was taken. The unit of breaking strength is N. Observations to be eliminated or corrected: None.
Statistical data
Sample:
Observed value and:
Known standard deviation:
Confidence level:
2x,=21. 761
1—a0, 95
Point estimate of population mean μ: positive one to =2. 176 1 Two-sided confidence interval:
From Appendix A, Table A, 1.
(u. vs / V10)α= 0. 620 × 0. 331 5 - 0. 205 52. 176 1-0. 205 5;2. 176 1+0. 205 5 That is,
1.97062.3816
5.2.2 Interval estimation of the mean of a single population with unknown variance Table B
Technical characteristics of the population
Technical characteristics of the sample
Observed values that have been eliminated or corrected
Statistical data
Sample egg:
Observed value sum:
Observed value sum of squares: three-
Degrees of freedom
Confidence level:
Estimation of the population mean
Interval estimation of the population mean (unknown variance) Calculation
(-()/n
[snti--(-)=
[s/n_t-a/2 (u) -
Two-sided confidence interval—[s+1a()+[s//z()One-sided confidence interval: +[s/4n(v)
or μz}-[sn't-a(t)
The example problem is the same as the example in 5.2.1, except that the estimated value = s is used instead. And 1 (or t//n) is used instead of (or u//jin). Use the method in Table B' to find the confidence interval of u with a confidence level of 0.95. 6
GB/T 4889—2008
Technical characteristics of the population: The manufacturer delivered a batch of cotton yarn, a total of 10,000 spools, packed in 100 boxes, each box with 100 spools. Technical characteristics of the sample: 10 boxes were randomly selected, and one spool was randomly selected from each box. A 50cm long cotton yarn was cut about 5m from the end of the spool as the test material. In the actual experiment, only the middle 25cr was taken. The unit of breaking strength is N. Observations eliminated or corrected: None.
Statistical data
Sample size:
Sum of observations:
2;—21. 761
Sum of squares of observations: Yuan—48.6105
Degrees of freedom
Confidence level:
=10-1=9
1 =,95
Point estimate of the population mean μ—-2.1761
Two-sided confidence interval:
2. 176 10. 267 3$2. 176 1+0. 267 3That is,
1. 908 8≤p2. 443 4
21,761 =2. 176 1
48.610521751/10=0.1396
s/ /10-0. 118 2
From Appendix A, Table A. 2;
(s/ /10) 0. 975 (9) = 0. 118 2 × 2. 282 2 =0.267 3
Feature 1: The interval of the assumptions in this example is significantly larger than that of the variance [1]. This is because the sample standard deviation s-0.3736 is larger than the standard deviation (g-0.3315) used in the example in 5.2.1. Due to different confidence levels, different conclusions are obtained. Note 2: When there is a question of whether the variance obtained from past experience is still applicable, it is more reliable to use the estimated value of the variance. Note 3: If a confidence interval with a larger confidence level is required, 1-a-0.99 can be used, and the interval will be longer. From Appendix A Table A.2, we can get t,99 (9) =3.219 8, thus: (s/ /10)to.95 0.118 2X3.249 8= 0.384 0 Therefore, the confidence interval with a confidence level of 0.99 is 2.176 1-0.384 0≤±≤2,176 1+0,384 0 That is,
6 Comparison of means of two normal populations
6.1 Test for comparison of means of two populations
1.792 1s2.560 1bzxZ.net
6.1.1 Test for comparison of two population means with known variances Table C Comparison of two population means (with known variances) Population 1
Technical characteristics
Population 2
Sample 1
Sample 2
Sample 1
Sample 2
Technical characteristics
Observations excluded or corrected
Statistical data
Sample maximum:
Observations and:
Sample 1
Known population variance:
Significance level-
Sample 2
GB/T 4889—2008
Two-sided test:
Table C (continued)
When -->- holds, the hypothesis that the two means are equal is rejected. One-sided test:
a) When tt=u., reject the hypothesis that the first mean is not less than the second mean; b) When tt=u., reject the hypothesis that the first mean is not greater than the second mean. Proof: d = V/m, +7n basis n; and n2 two groups of observations are drawn separately to obtain the difference between the two sample means d--, a standard deviation of ± 2Example Still consider the problem of cotton yarn breaking strength, randomly draw two kinds of cotton yarn (respectively "cotton yarn 1\ and "cotton yarn 2\) and measure their breaking strength, the data are as follows:
Cotton yarn 1:2.2972.5821.9492.3622.0402.1331.9861.6422.915
Cotton yarn 2:2.2862.3272,3883,1723.1582.7512.2222.3672.2472.5122.1042.707Assume that the population variance has been obtained from the historical data: =0.1099,o,=0.3315
2 -0, 096 9,02 =0. 311 3
To test the comparison of the mean values of the breaking strength of the tested cotton yarns, the method provided in Table C is used to obtain: Overall technical characteristics: Manufacturer A delivered a batch of cotton yarn, totaling 10,000 spools; Manufacturer B also delivered a batch of cotton yarn, totaling 12,000 spools. These cotton yarns were packed in boxes that could accommodate 1CO spools. Technical characteristics of the sample: 10 and 12 boxes were randomly selected from the two batches of goods, and one spool was randomly selected from each box. A 50cm long cotton yarn was cut about 5m from the end of the spool as the test material. In actual experiments, only the middle 25cm was taken. The unit of breaking strength is N. Observed values that were excluded or corrected: None.
Statistical data
Sample 1
Sample size: n,=10
Sample 2
Sample total: 211=21.761
≥22, 30. 241
Population variance: -0. 109 9, -0, 096 9Significance level: α—C. 05
Double case:
—2, 176 1
0. 109 9 1 0. 096 ?
Ws. 975. 0d=1. 96 ×0. 138 1 =0. 27U 7[2. 176 12. 520 1. =0. 314 0>0. 271When the significance level is C.05, the source hypothesis is rejected, that is, the means of the two populations are not equal. From the sample mean, the breaking strength of the first cotton yarn is higher.
Note: If we are unwilling to take the risk of making a wrong decision as large as 0.05, we can take @-0.01, then: o 95 * 0 =2. 575 8× 0. 138 1-0. 355 7In the case of double cutting:
[2. 176 12. 520 1| =0. 344 00.355 7 Therefore, the null hypothesis cannot be rejected at the significance level of -C.01. 8
6.1.2 Test table for comparison of means of two populations with unknown variance but assumed equal variances Comparison of means of two populations (with unknown variance but assumed equal variances) Population
Population 2』
Technical characteristics of sample 11
Technical characteristics of sample 2
Sample 1
>Observations excluded or corrected
Sample 2
Statistical data
Sample size:
Sample 1
Observation sum:
Sample 2
Observation sum of squares:
Degrees of freedom:
Significance level:
Two-sided test:
u= -n—2-
2(-)3 12(2)*
GB/T 4889—2008
[n, +n2 2(1 -)*+2( -3))
t.--(s --
When |—>/（), reject the hypothesis that the two means are equal. One-sided test:
mt-+-nz 2
or ti-a/()—
a) When one-to-one:, (sa), reject the hypothesis that the first mean is not less than the second mean; b) When one-to-one>-（, reject the hypothesis that the -th mean is not greater than the -th mean. Note: 5: is the difference d-work, one, the standard deviation of the two sample means obtained by extracting two groups of observations, i.e., the standard error of d. Example Consider the same as 6.1.1 Example The same problem, in reality, the population variance is rarely different from the known situation, so it is necessary to obtain an estimate of the variance through sample data. It should be noted that only when the two population means are equal, the method of Table C is appropriate to test the comparison of the two population means. Here, we assume that the variances of the two cotton yarns in the example in 6.1.1 are equal. For the two cotton yarn samples given in the example in 6.1.1, using the method provided in Table C', we have: Technical characteristics of the population: Manufacturer A delivered a batch of cotton yarn, a total of 10,000 spools + Manufacturer B also delivered a batch of cotton yarn, a total of 12,000 spools. These cotton yarns are packed in a container that can accommodate 100 The sample was in a box with spools of thread. Technical characteristics of the sample: 10 and 12 boxes were randomly selected from two batches of goods respectively, and a spool of thread was randomly selected from each box. A 50cm long cotton yarn was cut about 5m away from the end of the spool as the test material. In the actual experiment, only the middle 25cm was taken. The unit of breaking strength is N. Observations to be replaced or corrected: None.
Statistical data
Sample size:
Sample 1
Sample 2
Sum of observations: 21-21.7612=30.241
Semi-sum of observations: Xi—1.256 4—1.389 8 Degrees of freedom:
+=10+12-220
Significance level: 0=0.05
Two-sided case:
, - 2. 175 1, x2 = 2. 520 1
2(x1. -1)3 -1-2(T2: -22)*-2. 646 1222.6461=0.155 7
10×1220
5utu.35 (20) --0. 155 7×2. 086=0. 324 8[2. 176 1 -2. 520 1 =0. 344 0>0. 324 H When the significance level is 0.05, we can reject the null hypothesis, that is, we believe that the means of the two populations are not equal. From the sample mean, the second type of cotton yarn has a greater breaking strength.
Note: If we do not want to take the risk of making a wrong decision as large as 0.05, we can take = 0.0t, then: to.995(20)5g = 2.8453X0.1557-0.4430 In the two-sided case:
2. 176 12. 520 1| = 0. 344 0 < 0. 443 0 Therefore, the original hypothesis cannot be rejected at the significance level α = 0.01. 9
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