周彤2012全国赛论文翻译汇总.doc

Abstract The thesis mainly describes that wine tasters give a core of classification indicators of the wine to make an assessment of the problem after a wine tasting. In response to question 1, we use the paired t test methods firstly and find out the t value to get the significant differences of the results of the evaluation between the wine tasting group;and after computing and comparing the variance of the sores of each sample of the two wine tasting groups respectively, consequently,we confirm that the score results of the second group is more credible. In response to question 2, we apply the software to conduct the principal component analysis to simplify the physical and chemical indicators of wine grapes;and then we divide 27 different varieties of wine grapes into 3 categories by using clustering analysis,; we determine the level according to the average score of wine grapes of each type of wine grape varieties grapes. In response to question 3, at first,we consult the same physical and chemical indicators of wine grapes and wine,and then obtains functional relationship between the various varieties of grapes and wine in the same the physical chemical indicators, by utilizingdata fitting.In response to question 4, we establish multiple linear regression model by taking wine grape and wine of the physical and chemical indicators as independent variables and defing the score of wine as dependent variable.Next,using the software to sove it, and through the analysis and appraisal we can obtain that applying the physical and chemical indicators of wine grapes and wine can evaluate the wine quality. Key words:t test;software; clustering analysis; data fitting; multivariate linear regression model1.Problem restatementGenerally,determining a wine quality depends on the evaluation by a batch of qualified wine tasters who were hired. Each wine taster give the final score of the classification of indicators after tasting it. After that, Summing the total score and ascertain the wine quality.There is a direct relationship between wine and the wine grapes. The physical and chemical indicators of wine grapes and wine will reflect the quality of the wine and grape to a certain extent. Attachment 1 gives the results of the evaluation of the wine at a given year. Attachment 2 and Attachment3 appears compositional data of the wine grapes and wine respectively. Try to establish mathematical models to discuss the following questions:1.Analyzing whether there are significant differences between the results given by the two group in Annex 1, which result is more credible? 2.Based on the the physical and chemical indicators of wine grapes and the quality of wine,please rank these wine grapes.3.Analyzing the relationship between the physical and chemical indicators of wine grapes and wine. 4.Analyzing the influence on the quality of wine by the physical and chemical indicators of wine grapes and wine and whether or not assessing the wine quality can use the physical and chemical indicators of wine grapes and wine. 2.Model Assumptions. 2.1 Assuming thatindicates,namely, the two methods of measuring results are the same. indicates, that is, the two methods of measuring results are different.2.2 Confidence level is assumed as .2.3 In the process of simplification, wine-making techniques and other steps have no influence on the quality of wine.3.NotationsSymbolMeaningThe confidence level, the default is .Under the indicators , the difference between the average value of each sample score results of the first and second groups ;Under the indicators , the standard deviation of the difference between the average value of each sample score results of the first and second groups ;The average value of ; the degrees of freedom, =4. The design of model4.1 model one4.1.1 analysis of model one.Because paired samples is the two sets of data which obtained by the same samples, or obtained by the two identical sample tested under different conditions .In this problem, the sample does not change, but two different wine tasters give a core of classification indicators of the wine after tasting , which is on the same sample for the two test samples, therefore, it is paired samples . For the data of paired sample , you should calculate the mean of the difference firstly . When the two results of treatment are undifferentiated or some kind of treatment does not work , in theory, the population mean of difference should be 0 .So,we can view the hypothesis test for paired samples data as the comparison between sample average and the overall mean = 0, the method used is paired t-test. Due to of a huge amount of data that given by the subject , therefore, we analyze these data at first, this method as follows:In allusion to red wine, for each indicator of each wine sample,we take the average score of ten wine tasters,and then,we compress the 100 values of each sample into 10 values.After obtaining the average score of each index of the first group and the second group,we conduct Paired Sample T Test towards each indicator of the two groups respectively. Next, we analyze the results and judge whether the results of the evaluation of two team has significant differences or not. And then,seek which groups result is more reliable.For white wine, in order to simplify operation process, 10 indicators can be considered together and there is no much difference between the results and the results of the red wine.Therefore, conduct t test for overall average of the two groups each wine sample.4.1.2 The Establishment of Model OneThe value of t: Standard Deviation: 4.1.3 The Solution of Model OneAim at Red WineAccording to the data of Annex 1 of the theme , computing the average scores of the index of clarity, we can get the data shown in Table 1 below.Table 1Sample No.group 1Group 2Difference ()12.33.1- 0.80.6422.93.1- 0.20.0433.43.400450.490.167582.73.4- 0.70.4993.13.6- 0.50.2510416121.13.5- 2.45.76132.63.7- 1.11.210.160.09163.13.2- 0.10.010.25181.93.6- 1.72.890.160.010.090.25233.23.6- 0.40.160.36259263.63.7- 0.10.01273.73.700Total- 1.914.271.for clarity(1) establish assumptions: Assuming that the results of the evaluation between the two groups has no significant difference : Assuming that the results of the evaluation between the two groups has significant difference. 0.05(2) computing test statistic(3) Seeking the boundary value table to determine the value of , we can draw the following conclusions. ，And ,by the standard , Accepting , rejecting , namely, the evaluation results of red wine clarity by the first group and the second group wine tasters have no significant differences.2.In a similar way,we can get the other 9 results as follows:(1)For colors: ， ， by the standard , refusing ,accepting, that is, the evaluation results of red wine color by the first group and the second group have significant differences.(2) For aroma pure degree: ，,， by the standard, refusing ,accepting,that is, the evaluation results of red wine aroma pure degree by the first group and the second group have significant differences. (3)For fragrance concentration:，, by the standard, accepting, refusing, that is, the evaluation results of red wine fragrance concentration by the first group and the second group have no significant differences. (4)For fragrance quality: ，,by the standard, refusing , accepting , that is, the evaluation results of red wine fragrance quality by the first group and the second group have no significant differences. (5)For the taste pure degree: ，, by the standard, accepting , refusing, that is, the evaluation results of red wine taste pure degree by the the first group and the second group have no significant differences.(6)For taste concentration: ，,by the standard, accepting, refusing, that is, the evaluation results of red wine taste concentration by the the first group and the second group have no significant differences.(7)For persistence: ，, by the standard, accepting, refusing, that is, the evaluation results of red wine persistence by the the first group and the second group have no significant differences.(8)For taste quality: ，, by the standard, accepting, refusing, that is, the evaluation results of red wine taste quality by the the first group and the second group have no significant differences.(9)For balance or the whole evaluation: ，,by the standard, accepting, refusing, that is, the evaluation results of red wine balance or the whole evaluation by the the first group and the second group have significant differences.In conclusion, it can be seen that after the t test of each index of the red wine in the first group and the second group,we can summarize that in the ten index including clarity, aroma concentration, taste pure degree, taste concentration, persistance,taste quality, balance and the whole evaluation results of the evaluation have no significant differences.However,the results of color,aroma pure degree, aroma pure degree and aroma quality have significant differences.4.1 .3 .2 Aim at white wineAccording to the data of Attachment 1, we can see appendix I for obtained data that averaging numerical integral of the indicators of each wine sample in the two groups. 1、 Establish assumptions: Two groups of score results without significant difference: Two groups of score results in significant difference 0.052、calculate the test statistics:3.Searchboundary value table to determine whatvalues and reached the following conclusions: and , by the standard of , refuse ,accept, that is, the first group and the second group have significant differences on each samples of the results of the evaluation of white wine.4.1.4 Reliability analysis of the evaluation results of the the two wine tasters on red wine Because the variance show the fluctuations of data, therefore, the more gentle the fluctuation trend of the data is,and data is small, the higher the stability and the credibility is. Similarly, the more intense the fluctuation trend of the variance is,and data is large, the lower the stability and the credibility is. Therefore, when we conduct the t test , we calculate the variance of the sores of each sample of the two wine tasting groups at the same time,each group wine tasting the wine for each sample each of the indicators of the score , then sum the variance of the same indicator , we get various data in table 2 as follows:Table 2The first groupThe second group Clarity 15.6710.72Tones 62.8256.04Aroma pure degree22.2112.72Aroma concentration37.0031.77Aroma quality73.3348.04Taste pure degree21.8111.54Taste concentration 42.7834.51Persistence17.0715.88Taste Quality147.1393.60Balanced/ the whole evaluate17.3210.91Average amount41.6529.79In comparison with the sum of variance towards the same indicator of two groups , we obtained the Line Graph as figure 1 showed : Figure 1In figure 1, series 1 represents the sum of variance of evaluation scores towards the same indicator of the first group in allusion to 27 wine samples, series 2 represents the sum of variance of evaluation scores towards the same indicator of the second group in allusion to 27 wine samples. According to figure 1, compared the fluctuations variance of series 2 with the fluctuations variance of series 1,we can find that series 2 tend to more gentle. And the average values for variance of evaluation scores towards the same indicator of the first group in allusion to 27 wine samples is 41.65, the value is large,however, the average values for variance of evaluation scores towards the same indicator of the second group in allusion to 27 wine samples is 29.79 , the value is small. Therefore, the results of evaluation of the group 2 are more credible.4.1.5 Reliability analysis of the evaluation results of the the two wine tasters on white wine Because the variance show the fluctuations of data, therefore, the more gentle the fluctuation trend of the data is,and data is small, the higher the stability and the credibility is. Similarly, the more intense the fluctuation trend of the variance is,and data is large, the lower the stability and the credibility is. Therefore, when we conduct the t test , we calculate the variance of the sores of each sample of the two wine tasting groups at the same time,each group wine tasting the wine for each sample each of the indicators of the score , then sum the variance of the same indicator , we get various data in table 3 as follows:Table 3The second groupThe first groupClarity 13.7323.29Tones 50.22102.81Aroma pure degree14.7427.59Aroma concentration34.4035.71Aroma quality59.3994.01Taste pure degree13.8332.80Taste concentration 33.4655.06Persistence16.2733.20Taste Quality138.00257.10Balanced/ the whole evaluate15.4828.58Average amount38.9569.02In comparison with the sum of variance towards the same indicator of two groups , we obtained the Line Graph as figure 2 showed : Figure 2In figure 2, series 1 represents the sum of variance of evaluation scores towards the same indicator of the first group in allusion to 27 wine samples, series 2 represents the sum of variance of evaluation scores towards the same indicator of the second group in allusion to 27 wine samples. According to figure 2, compared the fluctuations variance of series 2 with the fluctuations variance of series 1,we can find that series 2 tend to more gentle. And the average values for variance of evaluation scores towards the same indicator of the first group in allusion to 27 wine samples is 69.02, the value is large,however, the average values for variance of evaluation scores towards the same indicator of the second group in allusion to 27 wine samples is 38.95 , the value is small. Therefore, the results of evaluation of the group 2 are more credible.4.2 Dividing the level of grapesUse software for factor analysis, through principal component analysis, and transform the 30 grapes targets given by the attachment into 8 integrated indicators, as shown in Table 4.Table 4 coefficient matrix of composition scorecomposition12345678Total amino acid0.0540.110.0040.16- 0.12- 0.1710.118- 0.007Protein0.088- 0.1010.0480.0960.097- 0.0740.057- 0.099Content VC- 0.02- 0.080.025- 0.003- 0.2730.075- 0.0160.126anthocyanin0.122- 0.022- 0.028- 0.1060.0480.113- 0.0650.049tartaric acid0.0550.020.0980.1360.156- 0.0840.142- 0.407Malic acid0.0560.0650.045- 0.2330.0430.211- 0.080.092Citric acid0.0440.0380.107- 0.1290.178- 0.0410.205- 0.338Polyphenol oxidase activity0.0450.017- 0.057- 0.2090.118- 0.194- 0.0050.13browning degree0.086- 0.0180.015- 0.249- 0.01- 0.032- 0.0570.086Free radicals DPPH0.109- 0.093- 0.0030.076- 0.0120.0650.150.089Total phenol0.124- 0.035- 0.0470.079- 0.010.106- 0.0070.069Tannins0.108- 0.031- 0.075- 0.024- 0.0830.1410.171- 0.046Grape flavonoids0.103- 0.058- 0.0530.10.0150.1710.0880.044Resveratrol0.009- 0.0120.2190.026- 0.1090.0940.2070.231flavonol0.080.0040.007- 0.025- 0.086- 0.2880.3350.17Total sugar0.0370.159- 0.040.0920.051- 0.023- 0.0510.234revertose0.0110.156- 0.0310.0450.058- 0.062- 0.0350.058soluble solid0.0350.154- 0.0850.0510.061- 0.027- 0.0170.21PH value0.039- 0.0560.0480.2450.065- 0.065- 0.2020.189titratable acid- 0.0430.093- 0.16- 0.002- 0.1650.1260.209- 0.11Solid acid than0.057- 0.010.115- 0.0010.267- 0.059- 0.2230.177Dry matter content0.0540.173- 0.0510.0330.048- 0.0140.0380.027Grain quality- 0.049- 0.093- 0.0590.0230.2990.0260.160.073The grain quality- 0.077- 0.072- 0.1260.0280.1340.0860.1570.152Stem than0.084- 0.0430.046- 0.076- 0.205- 0.2330.0620.032juice yield0.078- 0.037- 0.0720.0590.0070.229- 0.1010.011Peel quality- 0.037- 0.05- 0.164- 0.040.163- 0.0450.3370.173L- 0.081- 0.0670.082- 0.0140.025- 0.1760.0540.272Pericarp color a- 0.0480.0560.1970.018- 0.0110.1690.1820.167Pericarp color b- 0.020.0990.161- 0.0080.0950.2610.2080.07From the table,we can see that the content of Grape flavonoids, Free radicals DPPH, Total phenol, Tannins and anthocyanin are more than 0.1 in the comprehensive composition one.Hence,we can firmly believe that the comprehensive composition one consists of Grape flavonoids, Free radicals DPPH, Total phenol, Tannins and anthocyanin.Besides,there is a positive correlation between these composition and the comprehensive composition one. In a similar way,we can get the comprehensive composition two consists of five compositions that are amino acid,soluble solid,revertose,dry matter and total sugar. The comprehensive composition three consists of five compositions that is Pericarp color b,Resveratrol,Citric acid,Pericarp color a