Диссертация (1172916), страница 19
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= 48,5011Mean absolute error = 36,5948Durbin-Watson statistic = 0,879308 (P = 0,0036)Lag 1 residual autocorrelation = 0,351608The StatAdvisorThe output shows the results of fitting a linear model to describe the relationship betweenсp and T The equation of the fitted model isc p 440,563 32,6772 T .Since the P-value in the ANOVA table is less than 0,05, there is a statistically significantrelationship between cp and T at the 95,0 % confidence level.The R-Squared statistic indicates that the model as fitted explains 95,8477 % of the variabilityin cp.
The correlation coefficient equals 0,979018, indicating a relatively strong relationshipbetween the variables. The standard error of the estimate shows the standard deviation of theresiduals to be 48,5011. This value can be used to construct prediction limits for new observationsby selecting the Forecasts option from the text menu.The mean absolute error (MAE) of 36,5948 is the average value of the residuals. The DurbinWatson (DW) statistic tests the residuals to determine if there is any significant correlation basedon the order in which they occur in your data file.
Since the P-value is less than 0,05, there isan indication of possible serial correlation at the 95,0 % confidence level. Plot the residuals versusrow order to see if there is any pattern that can be seen.134Таблица А.8 – Результаты обработки экспериментальных данных по изменениюкоэффициента теплопроводности образцов из ТБ от температурыDependent variable: λ = f(T)ParameterInterceptSlopeEstimate0,72927629,2799Standard Error0,02618022,03302T Statistic27,85614,4022P-Value0,00000,0000Analysis of VarianceSourceModelResidualTotal (Corr.)Sum of Squares1,311670,02900361,38123Df11112Mean Square1,311670,00632361F-Ratio207,42P-Value0,0000Correlation Coefficient = 0,974494R-squared = 94,9639 percentR-squared (adjusted for d.f.) = 95,2025 percentStandard Error of Est.
= 0,0795211Mean absolute error = 0,0604325Durbin-Watson statistic = 1,46825 (P = 0,0965)Lag 1 residual autocorrelation = 0,201128The StatAdvisorThe output shows the results of fitting a reciprocal-Y logarithmic-X model to describe therelationship between λ and f(T). The equation of the fitted model isλ 0,73 29,28 / (T 273).Since the P-value in the ANOVA table is less than 0,05, there is a statistically significantrelationship between λ and f(T) at the 95,0 % confidence level.The R-Squared statistic indicates that the model as fitted explains 94,9639 % of the variabilityin λ. The correlation coefficient equals 0,974494, indicating a relatively strong relationship betweenthe variables.
The standard error of the estimate shows the standard deviation of the residuals to be0,0795211. This value can be used to construct prediction limits for new observations by selecting theForecasts option from the text menu.The mean absolute error (MAE) of 0,0604325 is the average value of the residuals. TheDurbin-Watson (DW) statistic tests the residuals to determine if there is any significant correlationbased on the order in which they occur in your data file. Since the P-value is less than 0,05, there isan indication of possible serial correlation at the 95,0 % confidence level.135Таблица А.9 – Результаты обработки экспериментальных данных по изменениюплотности образцов из Б от температурыDependent variable: ρ = f(T)ParameterCONSTANTTT2Estimate2090,64-0,427760,000186736Standard Error5,843450,02667670,0000239734T Statistic357,775-16,0357,78929P-Value0,00000,00000,0000Analysis of VarianceSourceModelResidualTotal (Corr.)Sum of Squares89339,9801,70990141,6Df21012Mean Square44669,980,1709F-Ratio557,18P-Value0,0000R-squared = 99,1106 percentR-squared (adjusted for d.f.) = 98,9327 percentStandard Error of Est.= 8,95382Mean absolute error = 5,55512Durbin-Watson statistic = 1,87275 (P = 0,1587)Lag 1 residual autocorrelation = 0,0633386The StatAdvisorThe output shows the results of fitting a second order polynomial model to describe therelationship between ρ and f(T).
The equation of the fitted model isρ = 2090,64 – 0,43(Т – 273) + 0,00019(Т – 273)2.Since the P-value in the ANOVA table is less than 0,05, there is a statistically significantrelationship between ρ and f(T) at the 95 % confidence level.The R-Squared statistic indicates that the model as fitted explains 99,1106 % of the variabilityin ρ. The adjusted R-squared statistic, which is more suitable for comparing models with differentnumbers of independent variables, is 98,9327 %. The standard error of the estimate shows the standard deviation of the residuals to be 8,95382.
This value can be used to construct prediction limits fornew observations by selecting the Forecasts option from the text menu. The mean absolute error(MAE) of 5,55512 is the average value of the residuals. The Durbin-Watson (DW) statistic tests theresiduals to determine if there is any significant correlation based on the order in which they occur inyour data file. Since the P-value is greater than 0,05, there is no indication of serial autocorrelationin the residuals at the 95 % confidence level.In determining whether the order of the polynomial is appropriate, note first that the P-valueon the highest order term of the polynomial equals 0,0000148647.
Since the P-value is less than 0,05,the highest order term is statistically significant at the 95 % confidence level. Consequently, youprobably don't want to consider any model of lower order.136Таблица А.10 – Результаты обработки экспериментальных данных по изменениютемпературопроводности образцов из Б от температурыDependent variable: a = f(T)ParameterInterceptSlopeEstimate-0,03676440,477595Standard Error0,1180540,0199865T Statistic-0,31142223,8958P-Value0,76130,0000Analysis of VarianceSourceModelResidualTotal (Corr.)Sum of Squares3,992860,07691894,06978Df11112Mean Square3,992860,00699263F-Ratio571,01P-Value0,0000Correlation Coefficient = 0,990505R-squared = 98,11 percentR-squared (adjusted for d.f.) = 97,9382 percentStandard Error of Est.
= 0,083622Mean absolute error = 0,0861908Durbin-Watson statistic = 0,989262 (P = 0,0082)Lag 1 residual autocorrelation = 0,379551The StatAdvisorThe output shows the results of fitting a reciprocal-Y logarithmic-X model to describe therelationship between a and f(T). The equation of the fitted model isa = 1/(0,11 + 0,44ln(T – 273)).Since the P-value in the ANOVA table is less than 0,05, there is a statistically significantrelationship between a and f(T) at the 95,0 % confidence level.The R-Squared statistic indicates that the model as fitted explains 98,11 % of the variabilityin a. The correlation coefficient equals 0,990505, indicating a relatively strong relationship betweenthe variables.
The standard error of the estimate shows the standard deviation of the residuals to be0,083622. This value can be used to construct prediction limits for new observations by selecting theForecasts option from the text menu.The mean absolute error (MAE) of 0,0634689 is the average value of the residuals. TheDurbin-Watson (DW) statistic tests the residuals to determine if there is any significant correlationbased on the order in which they occur in your data file.
Since the P-value is less than 0,05, there isan indication of possible serial correlation at the 95,0 % confidence level. Plot the residuals versusrow order to see if there is any pattern that can be seen.137Таблица А.11 – Результаты обработки экспериментальных данных по изменениюудельной теплоемкости образцов из Б от температурыDependent variable: cp = f(T)ParameterInterceptSlopeEstimate382,35234,1068Standard Error69,99412,49587T Statistic5,4626313,6653P-Value0,00020,0000Analysis of VarianceSourceModelResidualTotal (Corr.)Sum of Squares650701,038329,8689031,0Df11112Mean Square650701,03484,53F-Ratio186,74P-Value0,0000Correlation Coefficient = 0,971788R-squared = 94,4371 percentR-squared (adjusted for d.f.) = 93,9314 percentStandard Error of Est.
= 59,0299Mean absolute error = 35,5587Durbin-Watson statistic = 0,975284 (P = 0,0072)Lag 1 residual autocorrelation = 0,223053The StatAdvisorThe output shows the results of fitting a linear model to describe the relationship betweenсp and T The equation of the fitted model isс p 382,352 34,1068 T .Since the P-value in the ANOVA table is less than 0,05, there is a statistically significantrelationship between cp and T at the 95,0 % confidence level.The R-Squared statistic indicates that the model as fitted explains 94,4371 % of the variabilityin cp.
The correlation coefficient equals 0,971788, indicating a relatively strong relationshipbetween the variables. The standard error of the estimate shows the standard deviation of the residuals to be 59,0299. This value can be used to construct prediction limits for new observations byselecting the Forecasts option from the text menu.The mean absolute error (MAE) of 35,5587 is the average value of the residuals.
The DurbinWatson (DW) statistic tests the residuals to determine if there is any significant correlation based onthe order in which they occur in your data file. Since the P-value is less than 0,05, there is an indication of possible serial correlation at the 95,0 % confidence level. Plot the residuals versus row orderto see if there is any pattern that can be seen.138Таблица А.12 – Результаты обработки экспериментальных данных по изменениюкоэффициента теплопроводности образцов из Б от температурыDependent variable: λ = f(T)ParameterInterceptSlopeEstimate0,70775332,9208Standard Error0,02913712,26263T Statistic24,290414,5498P-Value0,00000,0000Analysis of VarianceSourceModelResidualTotal (Corr.)Sum of Squares1,658150,08615941,74431Df11112Mean Square1,658150,00783267F-Ratio211,70P-Value0,0000Correlation Coefficient = 0,97499R-squared = 95,0606 percentR-squared (adjusted for d.f.) = 94,6115 percentStandard Error of Est.
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