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海关总署公告2007年第9号－－对原产于印度和台湾地区的进口壬基酚征收反倾销税【法规类别】关税【发文字号】海关总署公告2007年第9号【发布部门】海关总署【发布日期】2007.03.28【实施日期】2007.03.29【时效性】现行有效【效力级别】部门规范性文件海关总署公告（2007年第9号）根据《中华人民共和国反倾销条例》的规定，国务院关税税则委员会决定自2007年3月29日起，对原产于印度和台湾地区的进口壬基酚征收反倾销税，期限为5年。
商务部为此发布了中华人民共和国商务部2007年第11号公告（详见附件1）。
现将执行中的有关问题公告如下：一、自2007年3月29日起，对原产于印度和台湾地区的进口壬基酚（税则号列：29071310），除按现行规定征收进口关税外，还应区别不同的供货厂商，按照本公告附件2所列的适用税率和下述计算公式征收反倾销税及进口环节增值税。
反倾销税税额＝海关完税价格×反倾销税税率进口环节增值税税额＝（海关完税价格＋关税税额＋反倾销税税额）×进口环节增值税税率对应征收反倾销税产品的详细描述详见本公告附件1。
二、凡申报进口壬基酚的，应当向海关提交原产地证明，如果原产地为印度或台湾地区的，还需提供原厂商发票。
对于申报进口壬基酚时不能提供原产地证明，且经查验也无法确定货物的原产地不是印度或台湾地区的，海关按照本公告附件2所列的最高反倾销税税率征收反倾销税；对于能够确定货物的原产地是印度或台湾地区，但进口经营单位不能提供原厂商发票，且通过其他合法、有效的单证也无法确定原生产厂商的，海关按照本公告附件2所列的“其他印度公司”或“其他台湾地区公司”适用的反倾销税税率征收反倾销税。
240 Chapter 7: Sampling DistributionsCHAPTER 77.1 PHstat output:Common DataMean 100Standard Deviation 2Probability for a RangeProbability for X <= From X Value 95 X Value 95To X Value 97.5 Z Value -2.5Z Value for 95 -2.5 P(X<=95) 0.0062097Z Value for 97.5 -1.25P(X<=95) 0.0062 Probability for X > P(X<=97.5) 0.1056 X Value 102.2P(95<=X<=97.5) 0.0994 Z Value 1.1P(X>102.2) 0.1357Find X and Z Given Cum. Pctage.Cumulative Percentage 35.00%Z Value -0.38532X Value 99.22936(b) P(95 < X < 97.5) = P(– 2.50 < Z < – 1.25) = 0.1056 – 0.0062 = 0.0994(c) P(X > 102.2) = P(Z > 1.10) = 1.0 – 0.8643 = 0.1357(d) P(X > A) = P(Z > – 0.39) = 0.65X = 100 – 0.39(1025) = 99.227.2 PHStat output:Common DataMean 50Standard Deviation 0.5Probability for a RangeProbability for X <= From X Value 47 X Value 47To X Value 49.5 Z Value -6Z Value for 47 -6 P(X<=47) 9.866E-10Z Value for 49.5 -1P(X<=47) 0.0000 Probability for X > P(X<=49.5) 0.1587 X Value 51.5P(47<=X<=49.5) 0.1587 Z Value 3P(X>51.5) 0.0013Find X and Z Given Cum. Pctage.Cumulative Percentage 65.00% Probability for X<47 or X >51.5 Z Value 0.38532 P(X<47 or X >51.5) 0.0013X Value 50.19266(b) P(47 < X < 49.5) = P(– 6.00 < Z < – 1.00) = 0.1587 – 0.00 = 0.1587(c) P(X > 51.1) = P(Z > 2.20) = 1.0 – 0.9861 = 0.0139(d) P(X > A) = P(Z > 0.39) = 0.35 X = 50 + 0.39(0.5) = 50.195Solutions to End-of-Section and Chapter Review Problems 241 7.3 (a) For samples of 25 customer receipts for a supermarket for a year, the samplingdistribution of sample means is the distribution of means from all possible samples of 25customer receipts for a supermarket for that year.(b) For samples of 25 insurance payouts in a particular geographical area in a year, thesampling distribution of sample means is the distribution of means from all possiblesamples of 25 insurance payouts in that particular geographical area in that year.(c) For samples of 25 Call Center logs of inbound calls tracking handling time for a creditcard company during the year, the sampling distribution of sample means is thedistribution of means from all possible samples of 25 Call Center logs of inbound callstracking handling time for a credit card company during that year.7.4 (a) Sampling Distribution of the Mean for n = 2 (without replacement)iX1 1, 31X = 22 1, 62X = 3.53 1, 73X = 44 1, 94X = 55 1, 105X = 5.56 3, 66X = 4.57 3, 77X = 58 3, 98X = 69 3, 109X = 6.510 6, 710X = 6.511 6, 911X = 7.512 6, 1012X = 813 7, 913X = 814 7, 1014X = 8.515 9, 1015X = 9.5(a) Mean of All Possible Sample Means: Mean of All Population Elements:μX =9015=6136791066μ+++++==Both means are equal to 6. This property is called unbiasedness.242 Chapter 7: Sampling Distributions7.4 (b) Sampling Distribution of the Mean for n = 3 (without replacement) cont.X1 1, 3, 61X = 3 1/32 1, 3, 72X = 3 2/33 1, 3, 93X = 4 1/34 1, 3, 104X = 4 2/35 1, 6, 75X = 4 2/36 1, 6, 96X = 5 1/37 1, 6, 107X = 5 2/38 3, 6, 78X = 5 1/39 3, 6, 99X = 610 3, 6, 1010X = 6 1/311 6, 7, 911X = 7 1/312 6, 7, 1012X = 7 2/313 6, 9, 1013X = 8 1/314 7, 9, 1014X = 8 2/315 1, 7, 915X = 5 2/316 1, 7, 1016X = 617 1, 9, 1017X = 6 2/318 3, 7, 918X = 6 1/319 3, 7, 1019X = 6 2/320 3, 9, 1020X = 7 1/3μX =12020=6This is equal toμ, the population mean.(c) The distribution for n = 3 has less variability. The larger sample size has resulted in samplemeans being closer toμ.(d) (a) Sampling Distribution of the Mean for n = 2 (with replacement)Solutions to End-of-Section and Chapter Review Problems 243 7.4cont.XiX1X = 22 1, 32X = 3.53 1, 63X = 44 1, 74X = 55 1, 95X = 5.56 1, 106X = 27 3, 17X = 38 3, 38X = 4.59 3, 69X = 510 3, 710X = 611 3, 911X = 6.512 3, 1012X = 3.513 6, 113X = 4.514 6, 314X = 615 6, 615X = 6.516 6, 716X = 7.517 6, 917X = 818 6, 1018X= 419 7, 119X = 520 7, 320X = 6.521 7, 621X = 722 7, 722X = 823 7, 923X = 8.524 7, 1024X = 525 9, 125X = 626 9, 326X = 7.527 9, 627X = 828 9, 728X = 929 9, 929X = 9.530 9, 1030X = 5.531 10, 131X = 6.532 10, 332X = 833 10, 633X = 8.534 10, 734X = 9.535 10, 935X = 1036 10, 1036244 Chapter 7: Sampling Distributions 7.4 (d) (a) Mean of All Possible Mean of All cont. Sample Means: Population Elements:216636X μ== μ=1+3+6+7+7+126=6Both means are equal to 6. This property is called unbiasedness. (b) Repeat the same process for the sampling distribution of the mean for n = 3 (withreplacement). There will be 36216= different samples.6X μ= This is equal to μ, the population mean.(c) The distribution for n = 3 has less variability. The larger sample size has resulted inmore sample means being close to μ. 7.5(a)Because the population diameter of tennis balls is approximately normally distributed, the sampling distribution of samples of 9 will also be approximately normal with a mean ofX μμ== 2.63 and X σ== 0.01.(b)X(c)Upper bound: X = 2.6384Solutions to End-of-Section and Chapter Review Problems 2457.6 (a)When n = 4 , the shape of the sampling distribution of X should closely resemble the shape of the distribution of the population from which the sample is selected. Because the mean is larger than the median, the distribution of the sales price of new houses is skewed to the right, and so is the sampling distribution of X although it will be less skewed than the population.(b)If you select samples of n = 100, the shape of the sampling distribution of the sample mean will be very close to a normal distribution with a mean of $322,100 and a standarddeviation of X σ== $9,000.(c) X σ =900010090000==nσ(d)7.7 (a) X σ1==246 Chapter 7: Sampling Distributions 7.7 (b) PHStat output: cont.(c) X σ0.5==(d) With the sample size increasing from n = 25 to n = 100, more sample means will becloser to the distribution mean. The standard error of the sampling distribution of size 100 is much smaller than that of size 25, so the likelihood that the sample mean will fall within ±0.5 minutes of the mean is much higher for samples of size 100 (probability = 0.6827) than for samples of size 25 (probability = 0. 3829).7.8(b) P (X < A ) = P (Z < 1.0364) = 0.85 X = 27 + 1.0364 (1) = 28.0364(c) To be able to use the standard normal distribution as an approximation for the area underthe curve, we must assume that the population is symmetrically distributed such that the central limit theorem will likely hold for samples of n = 16.(d)X = 27 + 1.0364 (0.5) = 27.5182Solutions to End-of-Section and Chapter Review Problems 2477.9 (a)p = 48/64 = 0.75(b)p σ =64)30.0(70.0 = 0.05737.10 (a)p = 20/50 = 0.40(b)p σ= 7.11 (a) p = 14/40 = 0.35(b)p σ =40)70.0(30.0 = 0.07257.12 (a) 0.501p μπ==, 0.05p σ===Partial PHstat output:Probability for X > X Value 0.55Z Value 0.98P(X>0.55) 0.1635P (p > 0.55) = P (Z > 0.98) = 1 – 0.8365 = 0.1635(b) 0.60p μπ==, 0.04899p σ===Partial PHstat output:Probability for X > X Value 0.55Z Value -1.020621P(X>0.55) 0.8463P (p > 0.55) = P (Z > – 1.021) = 1 – 0.1539 = 0.8461(c) 0.49p μπ==, 0.05p σ===Partial PHstat output:Probability for X > X Value 0.55Z Value 1.2002401P(X>0.55) 0.1150P (p > 0.55) = P (Z > 1.20) = 1 – 0.8849 = 0.1151(d) Increasing the sample size by a factor of 4 decreases the standard error by a factor of 2.248 Chapter 7: Sampling Distributions7.12 (d) (a) Partial PHstat output:cont.Probability for X >X Value 0.55Z Value 1.9600039P(X>0.55) 0.0250P(p > 0.55) = P (Z > 1.96) = 1 – 0.9750 = 0.0250(b) Partial PHstat output:Probability for X >X Value 0.55Z Value -2.041241P(X>0.55) 0.9794P(p > 0.55) = P (Z > – 2.04) = 1 – 0.0207 = 0.9793(c) Partial PHstat output:Probability for X >X Value 0.55Z Value 2.4004801P(X>0.55) 0.0082P(p > 0.55) = P (Z > 2.40) = 1 – 0.9918 = 0.0082If the sample size is increased to 400, the probably in (a), (b) and (c) is smaller,larger, and smaller, respectively because the standard error of the samplingdistribution of the sample proportion becomes smaller and, hence, the samplingdistribution is more concentrated around the true population proportion.7.13 (a) Partial PHstat output:Probability for a RangeFrom X Value 0.5To X Value 0.6Z Value for 0.5 0Z Value for 0.6 2.828427P(X<=0.5) 0.5000P(X<=0.6) 0.9977P(0.5<=X<=0.6) 0.4977P(0.50 < p < 0.60) = P(0 < Z < 2.83) = 0.4977(b) Partial PHstat output:Find X and Z Given Cum. Pctage.Cumulative Percentage 95.00%Z Value 1.644854X Value 0.558154P(– 1.645 < Z < 1.645) = 0.90p = .50 – 1.645(0.0354) = 0.4418 p = .50 + 1.645(0.0354) = 0.5582(c) Partial PHstat output:Probability for X >X Value 0.65Z Value 4.2426407P(X>0.65) 0.0000P(p > 0.65) = P (Z > 4.24) = virtually zeroSolutions to End-of-Section and Chapter Review Problems 2497.13 (d)Partial PHstat output:cont.Probability for X > X Value 0.6Z Value 2.8284271P(X>0.6) 0.0023If n = 200, P (p > 0.60) = P (Z > 2.83) = 1.0 – 0.9977 = 0.0023Probability for X > X Value 0.55Z Value 3.1622777P(X>0.55) 0.00078If n = 1000, P (p > 0.55) = P (Z > 3.16) = 1.0 – 0.99921 = 0.00079More than 60% correct in a sample of 200 is more likely than more than 55% correct in asample of 1000.7.14 (a) ==πμp 0.80, ()np ππσ−=1(b)(c)250 Chapter 7: Sampling Distributions 7.14 (d) ==πμp 0.80, ()n p ππσ−=1cont.(b)(c)7.15 (a) ==πμp 0.57, ()np ππσ−=1 = 0.0495(b)Solutions to End-of-Section and Chapter Review Problems 2517.15 (c)(d) ==πμp 0.57, ()np ππσ−=1 = 0.0248(a)(b)(c)7.16 (a)population proportion and, hence,π=0.15. Also the sampling distribution of the sample proportion will be close to a normal distribution according to the central limit theorem.252 Chapter 7: Sampling Distributions7.16p μπ==0.15,p σ=== 0.0252cont. P (0.12 < p < 0.18) = P (-1.1882< Z < 1.1882) = 0.7652A = 0.1085B = 0.1915A = 0.1005B = 0.1995 7.17 (a) p μπ==0.49, p σ=== 0.0500A = 0.4078B = 0.5722Solutions to End-of-Section and Chapter Review Problems 2537.17 (c)Partial PHStat output:cont.A = 0.3920B = 0.5880(d) (a) p μπ==0.49, p σ=== 0.02500.6007A = 0.4489B = 0.5311 (c)A = 0.4410B = 0.5390254 Chapter 7: Sampling Distributions 7.18 (a) ==πμp 0.36, ()n p ππσ−=1= 0.0480(b)= 0.0240(c)Increasing the sample size by a factor of 4 decreases the standard error by a factor of . The sampling distribution of the proportion becomes more concentrated around the true proportion of 0.36 and, hence, the probability in (b) becomes smaller than that in (a).7.19 Because the average of all the possible sample means of size n is equal to the population mean.7.20 The variation of the sample means becomes smaller as larger sample sizes are taken. This is dueto the fact that an extreme observation will have a smaller effect on the mean in a larger sample than in a small sample. Thus, the sample means will tend to be closer to the population mean as the sample size increases.7.21 As larger sample sizes are taken, the effect of extreme values on the sample mean becomessmaller and smaller. With large enough samples, even though the population is not normally distributed, the sampling distribution of the mean will be approximately normally distributed.7.22 The population distribution is the distribution of a particular variable of interest, while thesampling distribution represents the distribution of a statistic.7.23 When the items of interest and the items not of interest are at least 5, the normal distribution canbe used to approximate the binomial distribution.Solutions to End-of-Section and Chapter Review Problems 2557.24 0.753X μ=5004.0==nX σσ = 0.0008PHStat output:Common DataMean 0.753 Standard Deviation 0.0008Probability for a RangeProbability for X <=From X Value 0.75 X Value 0.74 To X Value 0.753 Z Value -16.25 Z Value for 0.75 -3.75 P(X<=0.74) 1.117E-59 Z Value for 0.753 0P(X<=0.75) 0.0001 Probability for X >P(X<=0.753) 0.5000 X Value 0.76 P(0.75<=X<=0.753) 0.4999 Z Value 8.75P(X>0.76) 0.0000 Find X and Z Given Cum. Pctage.Cumulative Percentage 7.00% Probability for X<0.74 or X >0.76 Z Value -1.475791 P(X<0.74 or X >0.76) 0.0000X Value 0.751819Probability for a Range From X Value 0.74 To X Value 0.75 Z Value for 0.74 -16.25 Z Value for 0.75 -3.75 P(X<=0.74) 0.0000 P(X<=0.75) 0.0001 P(0.74<=X<=0.75) 0.00009(a) P (0.75 < X < 0.753) = P (– 3.75 < Z < 0) = 0.5 – 0.00009 = 0.4999 (b) P (0.74 < X < 0.75) = P (– 16.25 < Z < – 3.75) = 0.00009 (c) P (X > 0.76) = P (Z > 8.75) = virtually zero (d) P (X < 0.74) = P (Z < – 16.25) = virtually zero (e) P (X < A ) = P (Z < – 1.48) = 0.07 X = 0.753 – 1.48(0.0008) = 0.7518256 Chapter 7: Sampling Distributions 7.25 2.0X μ=505.0==nX σσ = 0.01PHStat output:Common DataMean 2 Standard Deviation 0.01Probability for a RangeProbability for X <= From X Value 1.99X Value 1.98 To X Value 2Z Value -2 Z Value for 1.99 -1P(X<=1.98) 0.0227501 Z Value for 2 0P(X<=1.99) 0.1587Probability for X >P(X<=2) 0.5000X Value 2.01 P(1.99<=X<=2) 0.3413Z Value 1P(X>2.01) 0.1587 Find X and Z Given Cum. Pctage.Cumulative Percentage 1.00%Probability for X<1.98 or X>2.01Z Value-2.326348P(X<1.98 or X >2.01) 0.1814 X Value 1.976737Find X and Z Given Cum. Pctage.Cumulative Percentage 99.50% Z Value 2.575829X Value2.025758(a) P (1.99 < X < 2.00) = P (– 1.00 < Z < 0) = 0.5 – 0.1587 = 0.3413 (b) P (X < 1.98) = P (Z < – 2.00) = 0.0228(c) P (X > 2.01) = P (Z > 1.00) = 1.0 – 0.8413 = 0.1587(d) P (X > A ) = P ( Z > – 2.33) = 0.99 A= 2.00 – 2.33(0.01) = 1.9767 (e) P (A < X < B ) = P (– 2.58 < Z < 2.58) = 0.99A = 2.00 – 2.58(0.01) = 1.9742B = 2.00 + 2.58(0.01) = 2.0258Solutions to End-of-Section and Chapter Review Problems 2577.26 4.7X μ=0.400.085X σ===PHstat output:Common DataMean 4.7 Standard Deviation 0.08Probability for X >Find X and Z Given Cum. Pctage. X Value 4.6 Cumulative Percentage 23.00% Z Value -1.25 Z Value -0.738847 P(X>4.6) 0.8944 X Value 4.640892Find X and Z Given Cum. Pctage. Find X and Z Given Cum. Pctage. Cumulative Percentage 15.00% Cumulative Percentage 85.00% Z Value -1.036433 Z Value 1.036433 X Value 4.6170853X Value 4.782915(a) P (4.60 < X ) = P (– 1.25 < Z ) = 1 – 0.1056 = 0.8944 (b) P (A < X < B ) = P (– 1.04 < Z < 1.04) = 0.70A = 4.70 – 1.04(0.08) = 4.6168 ounces X = 4.70 + 1.04(0.08) = 4.7832 ounces(c) P (X > A ) = P (Z > – 0.74) = 0.77A = 4.70 – 0.74(0.08) = 4.6408 7.27X μ=5.00.400.085X σ===(a)Partial PHStat output:Probability for X > X Value 4.6 Z Value -5 P(X>4.6)1.0000P (4.60 < X ) = P (– 5 < Z ) = essentially 1.0 (b) Partial PHStat output:Find X and Z Given Cum. Pctage. Cumulative Percentage 15.00% Z Value -1.036433 X Value 4.917085A = 5.0 – 1.0364(0.08) = 4.9171 ounces X = 5.0 + 1.0364(0.08) = 5.0829 ounces258 Chapter 7: Sampling Distributions7.27 (c) Partial PHStat output: cont.Find X and Z Given Cum. Pctage. Cumulative Percentage 23.00% Z Value -0.738847 X Value4.940892P (X > A ) = P (Z > – 0.7388) = 0.77 A = 5.0 – 0.7388(0.08) = 4.94097.28 μμ=X = 15.23, X σ=(a)(b)(c)7.29 (a) μ=3.17 σ = 10Solutions to End-of-Section and Chapter Review Problems 2597.29 (b)Partial PHStat output:cont.(c)(d) μμ=X= 3.17, X σ== 5(e)(f)(g)Since the sample mean of returns of a sample of stocks is distributed closer to the population mean than the return of a single stock, the probabilities in (a) and (b) are higher than those in (d) and (e) while the probability in (c) is lower than that in (f).。
海关总署公告2007年第11号－－关于批准发布海关化验标准正文：---------------------------------------------------------------------------------------------------------------------------------------------------- 海关总署公告（2007年第11号）根据《中华人民共和国海关行业标准管理办法（试行）》，海关总署批准《硅铝处理的金红石型钛白粉的鉴定方法》和《有机表面处理的二氧化硅的鉴定方法》等14项海关化验标准，现予以公布（标准名称见附件，标准文本由中国海关出版社出版），自2007年4月1日起实施。
特此公告。
附件：海关标准编号名称表二○○七年三月二十八日附件海关标准编号名称表┌───┬──────┬────────────────┬───────┬───────┐│序号│海关标准编号│海关标准名称│批准日期│实施日期│├───┼──────┼────────────────┼───────┼───────┤│ 1 │HS/T 3-2006 │《硅铝处理的金红石型钛白粉的鉴定│2007年3月28日│2007年4月1日││││方法》│││├───┼──────┼────────────────┼───────┼───────┤│ 2 │HS/T 4-2006 │《有机表面处理的二氧化硅的鉴定方│2007年3月28日│2007年4月1日││││法》│││├───┼──────┼────────────────┼───────┼───────┤│ 3 │HS/T 5-2006 │《有机改性碳酸钙的鉴定方法》│2007年3月28日│2007年4月1日│├───┼──────┼────────────────┼───────┼───────┤│ 4 │HS/T 6-2006 │《复合纸及纸板的鉴别方法》│2007年3月28日│2007年4月1日│├───┼──────┼────────────────┼───────┼───────┤│ 5 │HS/T 7-2006 │《柴油中1,4-二羟基蒽醌的定性分析│2007年3月28日│2007年4月1日││││分光光度法》│││├───┼──────┼────────────────┼───────┼───────┤│ 6 │HS/T 8-2006 │《2-甲基-1-丁烯与2-甲基-2-丁烯的│2007年3月28日│2007年4月1日││││定量分析气相色谱法》│││├───┼──────┼────────────────┼───────┼───────┤│7 │HS/T 9-2006 │《乙烯-乙酸乙烯酯共聚物中乙酸乙│2007年3月28日│2007年4月1日││││烯酯的含量分析热重法》│││├───┼──────┼────────────────┼───────┼───────┤│8 │HS/T 10-2006│《初级形状线性低密度聚乙烯的鉴别│2007年3月28日│2007年4月1日││││方法》│││├───┼──────┼────────────────┼───────┼───────┤│9 │HS/T 11-2006│《低熔点复合涤纶短纤维的鉴别方法│2007年3月28日│2007年4月1日││││》│││├───┼──────┼────────────────┼───────┼───────┤│10 │HS/T 12-2006│《滑石、绿泥石、菱镁石混合相的定│2007年3月28日│2007年4月1日││││量分析X射线衍射仪法》│││├───┼──────┼────────────────┼───────┼───────┤│11 │HS/T 13-2006│《牛、羊、鹿源性成分鉴定方法实时│2007年3月28日│2007年4月1日││││荧光PCR方法》│││├───┼──────┼────────────────┼───────┼───────┤│12 │HS/T 14-2006│《人造刚玉的鉴定方法》│2007年3月28日│2007年4月1日│├───┼──────┼────────────────┼───────┼───────┤│13 │HS/T 15-2006│《微化锆英砂的鉴定方法》│2007年3月28日│2007年4月1日│├───┼──────┼────────────────┼───────┼───────┤│14 │HS/T 16-2006│《无尘纸与绒毛浆的鉴别方法》│2007年3月28日│2007年4月1日│└───┴──────┴────────────────┴───────┴───────┘——结束——。
Q7a（中英文对照）FDA原料药GMP指南Table of Contents 目录1. INTRODUCTION 1. 简介1.1 Objective 1.1目的1.2 Regulatory Applicability 1.2法规的适用性1.3 Scope 1.3范围2. QUALITY MANAGEMENT 2.质量管理2.1 Principles 2.1总则2.2 Responsibilities of the Quality Unit(s) 2.2质量部门的责任2.3 Responsibility for Production Activities 2.3生产作业的职责2.4 Internal Audits (Self Inspection) 2.4内部审计（自检）2.5 Product Quality Review 2.5产品质量审核3. PERSONNEL 3. 人员3.1 Personnel Qualifications 3.人员的资质3.2 Personnel Hygiene 3.2 人员卫生3.3 Consultants 3.3 顾问4. BUILDINGS AND FACILITIES 4. 建筑和设施4.1 Design and Construction 4.1 设计和结构4.2 Utilities 4.2 公用设施4.3 Water 4.3 水4.4 Containment 4.4 限制4.5 Lighting 4.5 照明4.6 Sewage and Refuse 4.6 排污和垃圾4.7 Sanitation and Maintenance 4.7 卫生和保养5. PROCESS EQUIPMENT 5. 工艺设备5.1 Design and Construction 5.1 设计和结构5.2 Equipment Maintenance and Cleaning 5.2 设备保养和清洁5.3 Calibration 5.3 校验5.4 Computerized Systems 5.4 计算机控制系统6. DOCUMENTATION AND RECORDS 6. 文件和记录6.1 Documentation System andSpecifications6.1 文件系统和质量标准6.2 Equipment cleaning and Use Record 6.2 设备的清洁和使用记录6.3 Records of Raw Materials, Intermediates, API Labeling and Packaging Materials 6.3 原料、中间体、原料药的标签和包装材料的记录6.4 Master Production Instructions (MasterProduction and Control Records)6.4 生产工艺规程（主生产和控制记录）6.5 Batch Production Records (BatchProduction and Control Records)6.5 批生产记录（批生产和控制记录）6.6 Laboratory Control Records 6.6 实验室控制记录6.7 Batch Production Record Review 6.7批生产记录审核7. MATERIALS MANAGEMENT 7. 物料管理7.1 General Controls 7.1 控制通则7.2 Receipt and Quarantine 7.2接收和待验7.3 Sampling and Testing of IncomingProduction Materials7.3 进厂物料的取样与测试7.4 Storage 7.4储存7.5 Re-evaluation 7.5复验8. PRODUCTION AND IN-PROCESSCONTROLS8. 生产和过程控制8.1 Production Operations 8.1 生产操作8.2 Time Limits 8.2 时限8.3 In-process Sampling and Controls 8.3 工序取样和控制8.4 Blending Batches of Intermediates orAPIs8.4 中间体或原料药的混批8.5 Contamination Control 8.5 污染控制9. PACKAGING AND IDENTIFICATIONLABELING OF APIs ANDINTERMEDIATES9. 原料药和中间体的包装和贴签9.1 General 9.1 总则9.2 Packaging Materials 9.2 包装材料9.3 Label Issuance and Control 9.3 标签发放与控制9.4 Packaging and Labeling Operations 9.4 包装和贴签操作10. STORAGE AND DISTRIBUTION 10.储存和分发10.1 Warehousing Procedures 10.1 入库程序10.2 Distribution Procedures 10.2 分发程序11. LABORATORY CONTROLS 11.实验室控制11.1 General Controls 11.1 控制通则11.2 Testing of Intermediates and APIs 11.2 中间体和原料药的测试11.3 Validation of Analytical Procedures 11.3 分析方法的验证11.4 Certificates of Analysis 11.4 分析报告单11.5 Stability Monitoring of APIs 11.5 原料药的稳定性监测11.6 Expiry and Retest Dating 11.6 有效期和复验期11.7 Reserve/Retention Samples 11.7 留样12. V ALIDATION 12.验证12.1 Validation Policy 12.1 验证方针12.2 Validation Documentation 12.2 验证文件12.3 Qualification 12.3 确认12.4 Approaches to Process Validation 12.4 工艺验证的方法12.5 Process Validation Program 12.5 工艺验证的程序12.6 Periodic Review of Validated Systems 12.6验证系统的定期审核12.7 Cleaning Validation 12.7 清洗验证12.8 Validation of Analytical Methods 12.8 分析方法的验证13. CHANGE CONTROL 13.变更的控制14. REJECTION AND RE-USE OFMATERIALS14.拒收和物料的再利用14.1 Rejection 14.1 拒收14.2 Reprocessing 14.2 返工14.3 Reworking 14.3 重新加工14.4 Recovery of Materials and Solvents 14.4 物料与溶剂的回收14.5 Returns 14.5 退货15. COMPLAINTS AND RECALLS 15.投诉与召回16. CONTRACT MANUFACTURERS(INCLUDING LABORATORIES)16.协议生产商（包括实验室）17. AGENTS, BROKERS, TRADERS, DISTRIBUTORS, REPACKERS, AND RELABELLERS 17.代理商、经纪人、贸易商、经销商、重新包装者和重新贴签者17.1 Applicability 17.1适用性17.2 Traceability of Distributed APIs andIntermediates17.2已分发的原料药和中间体的可追溯性17.3 Quality Management 17.3质量管理17.4 Repackaging, Relabeling, and Holding of APIs and Intermediates 17.4原料药和中间体的重新包装、重新贴签和待检17.5 Stability 17.5稳定性17.6 Transfer of Information 17.6 信息的传达17.7 Handling of Complaints and Recalls 17.7 投诉和召回的处理17.8 Handling of Returns 17.8 退货的处理18. Specific Guidance for APIs Manufactured by Cell Culture/Fermentation 18. 用细胞繁殖/发酵生产的原料药的特殊指南18.1 General 18.1 总则18.2 Cell Bank Maintenance and RecordKeeping18.2细胞库的维护和记录的保存18.3 Cell Culture/Fermentation 18.3细胞繁殖/发酵18.4 Harvesting, Isolation and Purification 18.4收取、分离和精制18.5 Viral Removal/Inactivation steps 18.5 病毒的去除/灭活步骤19.APIs for Use in Clinical Trials 19.用于临床研究的原料药19.1 General 19.1 总则19.2 Quality 19.2 质量19.3 Equipment and Facilities 19.3 设备和设施19.4 Control of Raw Materials 19.4 原料的控制19.5 Production 19.5 生产19.6 Validation 19.6 验证19.7 Changes 19.7 变更19.8 Laboratory Controls 19.8 实验室控制19.9 Documentation 19.9 文件20. Glossary 20. 术语Q7a GMP Guidance for APIs Q7a原料药的GMP指南1. INTRODUCTION 1. 简介1.1 Objective 1.1目的This document is intended to provide guidance regarding good manufacturing practice (GMP) for the manufacturing of active pharmaceutical ingredients (APIs) under an appropriate system for managing quality. It is also intended to help ensure that APIs meet the quality and purity characteristics that they purport, or are represented, to possess. 本文件旨在为在合适的质量管理体系下制造活性药用成分（以下称原料药）提供有关优良药品生产管理规范（GMP）提供指南。
