1. The weight of a product is measured in pounds. A sample of 64 units is taken from a batch. The sample yielded the following results: X = 70 lbs. and s=16 lbs. Calculate a 99 percent confidence interval for the mean 3. In an early study, researchers at an Ivy University found that 33% of the freshmen had received at least one A in their first semester. Administrators are concerned that grade inflation has caused this percentage to increase. In a more recent study, of a random sample of 500 freshmen, 190 had at least one A in their first semester Calculate the appropriate test statistic and test the hypotheses related to the concern at 5% and 1%. 5. The mid-distance running coach, Zdravko Popovich, for the Olympic team of an eastern European country claims that his six-month training program significantly reduces the average time to complete a 1500-meter run. Five mid-distance runners were randomly selected before they were trained with coach Popovich's six-month training program and their completion time of 1500-meter run was recorded (in minutes). After six months of training under coach Popovich, the same five runners' 1500 meter run time was recorded again. The results are given below. Runner Completion time before training Completion time after training 1 6.1 2 7.5 3 6.1 4 6.9 5 8.3 5.5 7.0 6.3 6.7 7.9 At alpha levels of .05, test whether there has been a significant decrease in the average time per mile? 8. Consumer Reports provided extensive testing and ratings for more than 100 HDTVs. An overall score, based primarily on picture quality, was developed for each model. In general, a higher overall score indicates better performance. The following data show the price and overall score for the ten 42-inch plasma televisions (Consumer Report data slightly changed here): Brand Dell Hisense Hitachi JVC LG Maxent Panasonic Phillips Proview Samsung Price 3500 2800 3200 3500 3300 2000 4000 3000 2500 3000 Score 60 55 45 50 55 30 70 55 35 40 Use the above data to develop an estimated regression equation and interpret results. Compute the Coefficient of Determination and the correlation coefficient and show their relation. Interpret the explanatory power of the model. Estimate the overall score for a 42-inch plasma television with a price of $2900. (6 points) 9. A member of the state legislature has expressed concern about the differences in the mathematics test scores of high school freshmen across the state. She asks her research assistant to conduct a study to investigate what factors could account for the differences. The research assistant looked at a random sample of school districts across the state and used the factors of percentage of mathematics teachers in each district with a degree in mathematics, the average age of mathematics teachers and the average salary of mathematics teachers (in thousand dollars per year): Regression Output Predictor Constant Math Degree Age Salary Coef. 20.25 0.30 0.35 0.15 SE Coef. 7.850 0.096 0.180 0.077 DF 3 31 SS 1025.50 630.89 Analysis of Variance Source Regression Residual Error Write the least squares prediction equation. What is the number of observations in the sample? Based on the multiple regression model given above, estimate the mathematics test score and calculate the value of the residual, if the percentage of teachers with a mathematics degree is 50.0, the average age is 40 and the average salary is 45,500. If the actual mathematics test score for these factors is 65.50, what is the error for this observation? What is the total sum of squares? What is the explained variation? What is the mean square error? Calculate the Coefficient of Determination and the Adjusted coefficient of Determination and Test for the overall usefulness of the model using F-Statistic at 5% and 1% significance levels. Finally, test the usefulness (or significance of the three independent variables using t-test for 5% and 1% significance levels. 10. It was felt by the local utility company that fuel consumption (natural gas) in their small town would be affected by temperature and chill index. The following data was collected for 8 weeks during the winter. Week Fuel Consumption, Average Hourly Chill Index, Y(MMcf) Temperature, X1 X2 (0F) 1 15.4 30.0 18 2 11.7 28.0 15 3 12.4 32.5 25 4 10.8 39.0 20 5 9.4 40.9 10 6 6.5 50.8 16 7 10.0 40.1 5 8 9.5 50.5 0 Conduct a Multiple Regression among Fuel consumption (Y), temperature (X1), and chill index (X2). Solve by excel or MegaStat. Using alpha = 0.05 and 0.01 answer the following: What is the multiple regression equation? Interpret it and predict the fuel consumption when temperature is 40 degrees and the chill index is 10. Are the coefficients of the multiple regression model statistically significant? How do you know? Discuss indicating the test statistics. What are R2, adjusted R2 and se? Interpret R2 and Adjusted R2 and compare these values. Perform F-test, show the ANOVA table and conclude about the overall 1. The weight of a product is measured in pounds. A sample of 64 units is taken from a batch. The sample yielded the following results: X = 70 lbs. and s=16 lbs. Calculate a 99 percent confidence interval for the mean z* = 2.58 CI = 70 2.58 * 16 / 64 = (64.84,75.16) 3. In an early study, researchers at an Ivy University found that 33% of the freshmen had received at least one A in their first semester. Administrators are concerned that grade inflation has caused this percentage to increase. In a more recent study, of a random sample of 500 freshmen, 190 had at least one A in their first semester Calculate the appropriate test statistic and test the hypotheses related to the concern at 5% and 1%. Testing for increase in percentage, so one-tailed. H0: Percentage of freshmen receiving at least one A is less than or equal to previous years: p1 p2 Ha: Percentage of freshmen receiving at least one A is more than 21previous years: p1 > p2 Difference in proportion: Assume 500 in first sample, as well. p1 = 0.33, so Y1 = 165, p2 = 190/500 = 0.38 n1 & n2 = 500. ^p= 165+ 190 =0 .355 ( 500+ 5 00 ) Zcalculated = (0.33 - 0.38) / [(0.355)*(1 - 0.355) * (1/500 + 1/500)] = -1.65214 For 5% zcritical (one-tailed) = -1.645 For 1% zcritical (one-tailed) = -2.326 At the 5% level, zcalculated is more extreme than zcritical so we would reject the null hypothesis and conclude that p1 > p2 and that grade inflation has occurred. At the 1% level, zcalculated is not more extreme than zcritical so we would not reject the null hypothesis and conclude that p1 p2 and that grade inflation has NOT occurred. 5. The mid-distance running coach, Zdravko Popovich, for the Olympic team of an eastern European country claims that his six-month training program significantly reduces the average time to complete a 1500-meter run. Five mid-distance runners were randomly selected before they were trained with coach Popovich's six-month training program and their completion time of 1500-meter run was recorded (in minutes). After six months of training under coach Popovich, the same five runners' 1500 meter run time was recorded again. The results are given below. Runner 1 2 3 4 5 Mean STD DEV Completion time before 6.1 7.5 6.1 6.9 8.3 training Completion time after 5.5 7.0 6.3 6.7 7.9 training difference 0.6 0.5 -0.2 0.2 0.4 0.3 0.3162 At alpha levels of .05, test whether there has been a significant decrease in the average time per mile? H0: 1500-meter run times have not been reduced. d 0 Ha: 1500-meter run times have been reduced. d < 0 The test statistic is t: t= d sd/ n = 0.3 / (0.3162 / 5) = 2.1213 t critical for this test would be a one-tailed test with = 0.05 and df = (n - 1) = 4 => 2.1319 Because the t stat is not more extreme than the t critical, we do not reject the null hypothesis and conclude that the 1500 meter run times have not been reduced with this program. 8. Consumer Reports provided extensive testing and ratings for more than 100 HDTVs. An overall score, based primarily on picture quality, was developed for each model. In general, a higher overall score indicates better performance. The following data show the price and overall score for the ten 42-inch plasma televisions (Consumer Report data slightly changed here): Brand Dell Hisense Hitachi JVC LG Maxent Panasonic Phillips Proview Samsung Price 3500 2800 3200 3500 3300 2000 4000 3000 2500 3000 Score 60 55 45 50 55 30 70 55 35 40 Use the above data to develop an estimated regression equation and interpret results. Compute the Coefficient of Determination and the correlation coefficient and show their relation. Interpret the explanatory power of the model. Estimate the overall score for a 42-inch plasma television with a price of $2900. (6 points) The regression equation is: Score = -7.0098 + Price*0.01834. This can be interpreted that as the price goes up, the score should go up slightly, as well. The Coefficient of Determination (R2) = 0. 0.73 and the correlation coefficient (R) = 0.85. R2 = R*R. The model predicts 73% of the variation in the score of the televisions. Score = -7.0098 * $2900 * 0.01834 = 46.2 (46). 9. A member of the state legislature has expressed concern about the differences in the mathematics test scores of high school freshmen across the state. She asks her research assistant to conduct a study to investigate what factors could account for the differences. The research assistant looked at a random sample of school districts across the state and used the factors of percentage of mathematics teachers in each district with a degree in mathematics, the average age of mathematics teachers and the average salary of mathematics teachers (in thousand dollars per year): Regression Output Predictor Constant Math Degree Age Salary Coef. 20.25 0.30 0.35 0.15 SE Coef. 7.850 0.096 0.180 0.077 Analysis of Variance (Given) Source Regression Residual Error Total DF 3 31 34 SS 1025.50 630.89 1656.39 MS 341.8333 20.3512 F 16.80 Write the least squares prediction equation. What is the number of observations in the sample? Based on the multiple regression model given above, estimate the mathematics test score and calculate the value of the residual, if the percentage of teachers with a mathematics degree is 50.0, the average age is 40 and the average salary is 45,500. If the actual mathematics test score for thes e factors is 65.50, what is the error for this observation? What is the total sum of squares? What is the explained variation? What is the mean square error? Calculate the Coefficient of Determination and the Adjusted coefficient of Determination and Test for the overall usefulness of the model using F-Statistic at 5% and 1% significance levels. Finally, test the usefulness (or significance of the three independent variables using t-test for 5% and 1% significance levels. Test scores = 20.25 + MathDegree*0.3 + AverageAge*0.35 + AverageSalary*0.15 N (number of observations) = Total df + 1 = 35. Test score = 20.25 + 50*0.3 + 40*0.35 + 45.5*0.15 = 56.08 Residual: = 7.85+(50*.096)+(40*.18)+(45.5*0.077) = 23.35 Error = 65.5 - 56.08 = 9.42 Total Sum of Squares = 1656.39 Explained variation = SSR = 1025.5 Mean Square Error = 20.3512 Coefficient of Determination (R2) = SSR/SST = 0.6191 (Adjusted R2) = 1 - (1 - 0.6191)[(34-1)/(34-(4+1)) = 0.5683 F = 16.80 Fcritical (5%) = 2.91 Because F calculated is greater than F critical, the model is useful at the 5% level. Fcritical (1%) = 4.48 Because F calculated is greater than F critical, the model is useful at the 1% level. 10. It was felt by the local utility company that fuel consumption (natural gas) in their small town would be affected by temperature and chill index. The following data was collected for 8 weeks during the winter. Week Fuel Consumption, Average Hourly Chill Index, Y(MMcf) Temperature, X1 X2 (0F) 1 15.4 30.0 18 2 11.7 28.0 15 3 12.4 32.5 25 4 10.8 39.0 20 5 9.4 40.9 10 6 6.5 50.8 16 7 10.0 40.1 5 8 9.5 50.5 0 Conduct a Multiple Regression among Fuel consumption (Y), temperature (X1), and chill index (X2). Solve by excel or MegaStat. Using alpha = 0.05 and 0.01 answer the following: What is the multiple regression equation? Interpret it and predict the fuel consumption when temperature is 40 degrees and the chill index is 10. Are the coefficients of the multiple regression model statistically significant? How do you know? Discuss indicating the test statistics. What are R2, adjusted R2 and se? Interpret R2 and Adjusted R2 and compare these values. Perform F-test, show the ANOVA table and conclude about the overall model. Y (MMcf) = 21.54 - 0.267*X1 - 0.03*X2 Y = 21.54 - 0.267*40 - 0.03*10 = 10.6 MMcf The intercept and temperature coefficients are statistically significant because their p values are 0.004 and 0.028 respectively. The chill index is not statistically significant because its p value is 0.76. R2 = 0.7033 means that 70.33% of the variation in the model can be explained by the regression equation. The adjusted R2 = 0.5847. This is much lower than R2 in this model because the chill index does not add value to the prediction. The F-test shows that, at the 0.05 level, the model is significant, however, it is not significant at the 0.01 level, as the p value = 0.0479. Average s t tcrit 0.6 0.5 -0.2 0.2 0.4 0.3 0.3162278 2.1213203 2.13185 Brand Price Score Dell 3500 Hisense 2800 Hitachi 3200 JVC 3500 LG 3300 Maxent 2000 Panasonic 4000 Phillips 3000 Proview 2500 Samsung 3000 SUMMARY OUTPUT 60 55 45 50 55 30 70 55 35 40 Regression Statistics Multiple R 0.8526177018 R Square 0.7269569453 Adjusted R Square 0.6928265635 Standard Error 6.7184395489 Observations 10 ANOVA df Regression Residual Total Coefficients Standard Error t Stat -7.0098039216 12.4274218692 -0.564059 0.0183473389 0.0039754776 4.615128 Intercept Price $2,900 Score SS MS 1 961.400560224 961.4006 8 361.099439776 45.13743 9 1322.5 46.20 F Significance F 21.299408514 0.001721105 P-value Lower 95% Upper 95%Lower 95.0% Upper 95.0% 0.5881734549 -35.66749014 21.64788 -35.66749 21.64788 0.001721105 0.0091798711 0.027515 0.00918 0.027515 Predictor Coef. SE Coef. Constant 20.25 7.85 Math Degree 0.3 0.096 Age 0.35 0.18 Salary 0.15 0.077 Source DF Regression Residual Error Total R2 Adj R2 Fcritical (5%) Fcritical (1%) SS 3 31 34 0.6191 0.5683 2.91 4.48 MS F 1025.5 341.8333 630.89 20.3512 1656.39 16.80 Week Fuel Consumption, Average Hourly Temperature X1 Y(MMcf) 1 2 3 4 5 6 7 8 15.4 11.7 12.4 10.8 9.4 6.5 10 9.5 Chill Index, X2 30 28 32.5 39 40.9 50.8 40.1 50.5 18 15 25 20 10 16 5 0 temperature is 40 degrees and the chill index is 10 Y: 10.5469691157 10.6 SUMMARY OUTPUT Regression Statistics Multiple R 0.8386558272 R Square 0.7033435965 Adjusted R Square 0.5846810352 Standard Error 1.6743144414 Observations 8 ANOVA df Regression Residual Total index is 10 Intercept X1 X2 SS MS F Significance F 2 33.2321057574 16.61605 5.927258 0.0479329676 5 14.0166442426 2.803329 7 47.24875 Coefficients 21.536685762 -0.2672659046 -0.0299080463 Standard Error t Stat 4.3039771014 5.003903 0.0880744897 -3.034544 0.0927026774 -0.322623 P-value Lower 95% Upper 95% 0.004091 10.472960404 32.60041112 0.028931 -0.4936685881 -0.0408632211 0.760035 -0.268207865 0.2083917724 SUMMARY OUTPUT Regression Statistics Multiple R 0.8386558272 R Square 0.7033435965 Adjusted R Square 0.5846810352 Standard Error 1.6743144414 Observations 8 ANOVA df Regression Residual Total Intercept X1 X2 SS MS F Significance F 2 33.2321057574 16.61605 5.927258 0.0479329676 5 14.0166442426 2.803329 7 47.24875 Coefficients 21.536685762 -0.2672659046 -0.0299080463 Standard Error t Stat 4.3039771014 5.003903 0.0880744897 -3.034544 0.0927026774 -0.322623 P-value Lower 95% Upper 95% 0.004091 10.472960404 32.60041112 0.028931 -0.4936685881 -0.0408632211 0.760035 -0.268207865 0.2083917724 Lower 95.0% Upper 95.0% 10.472960404 32.60041112 -0.4936685881 -0.0408632211 -0.268207865 0.2083917724 Lower 99.0% Upper 99.0% 4.1824346914 38.890936833 -0.6223948404 0.0878630312 -0.4036984966 0.343882404 1. The weight of a product is measured in pounds. A sample of 64 units is taken from a batch. The sample yielded the following results: X = 70 lbs. and s=16 lbs. Calculate a 99 percent confidence interval for the mean z* = 2.58 CI = 70 2.58 * 16 / 64 = (64.84,75.16) 3. In an early study, researchers at an Ivy University found that 33% of the freshmen had received at least one A in their first semester. Administrators are concerned that grade inflation has caused this percentage to increase. In a more recent study, of a random sample of 500 freshmen, 190 had at least one A in their first semester Calculate the appropriate test statistic and test the hypotheses related to the concern at 5% and 1%. Testing for increase in percentage, so one-tailed. H0: Percentage of freshmen receiving at least one A is less than or equal to previous years: p1 p2 Ha: Percentage of freshmen receiving at least one A is more than 21previous years: p1 > p2 Difference in proportion: Assume 500 in first sample, as well. p1 = 0.33, so Y1 = 165, p2 = 190/500 = 0.38 n1 & n2 = 500. ^p= 165+ 190 =0 .355 ( 500+ 5 00 ) Zcalculated = (0.33 - 0.38) / [(0.355)*(1 - 0.355) * (1/500 + 1/500)] = -1.65214 For 5% zcritical (one-tailed) = -1.645 For 1% zcritical (one-tailed) = -2.326 At the 5% level, zcalculated is more extreme than zcritical so we would reject the null hypothesis and conclude that p1 > p2 and that grade inflation has occurred. At the 1% level, zcalculated is not more extreme than zcritical so we would not reject the null hypothesis and conclude that p1 p2 and that grade inflation has NOT occurred. 5. The mid-distance running coach, Zdravko Popovich, for the Olympic team of an eastern European country claims that his six-month training program significantly reduces the average time to complete a 1500-meter run. Five mid-distance runners were randomly selected before they were trained with coach Popovich's six-month training program and their completion time of 1500-meter run was recorded (in minutes). After six months of training under coach Popovich, the same five runners' 1500 meter run time was recorded again. The results are given below. Runner 1 2 3 4 5 Mean STD DEV Completion time before 6.1 7.5 6.1 6.9 8.3 training Completion time after 5.5 7.0 6.3 6.7 7.9 training difference 0.6 0.5 -0.2 0.2 0.4 0.3 0.3162 At alpha levels of .05, test whether there has been a significant decrease in the average time per mile? H0: 1500-meter run times have not been reduced. d 0 Ha: 1500-meter run times have been reduced. d < 0 The test statistic is t: t= d sd/ n = 0.3 / (0.3162 / 5) = 2.1213 t critical for this test would be a one-tailed test with = 0.05 and df = (n - 1) = 4 => 2.1319 Because the t stat is not more extreme than the t critical, we do not reject the null hypothesis and conclude that the 1500 meter run times have not been reduced with this program. 8. Consumer Reports provided extensive testing and ratings for more than 100 HDTVs. An overall score, based primarily on picture quality, was developed for each model. In general, a higher overall score indicates better performance. The following data show the price and overall score for the ten 42-inch plasma televisions (Consumer Report data slightly changed here): Brand Dell Hisense Hitachi JVC LG Maxent Panasonic Phillips Proview Samsung Price 3500 2800 3200 3500 3300 2000 4000 3000 2500 3000 Score 60 55 45 50 55 30 70 55 35 40 Use the above data to develop an estimated regression equation and interpret results. Compute the Coefficient of Determination and the correlation coefficient and show their relation. Interpret the explanatory power of the model. Estimate the overall score for a 42-inch plasma television with a price of $2900. (6 points) The regression equation is: Score = -7.0098 + Price*0.01834. This can be interpreted that as the price goes up, the score should go up slightly, as well. The Coefficient of Determination (R2) = 0. 0.73 and the correlation coefficient (R) = 0.85. R2 = R*R. The model predicts 73% of the variation in the score of the televisions. Score = -7.0098 * $2900 * 0.01834 = 46.2 (46). 9. A member of the state legislature has expressed concern about the differences in the mathematics test scores of high school freshmen across the state. She asks her research assistant to conduct a study to investigate what factors could account for the differences. The research assistant looked at a random sample of school districts across the state and used the factors of percentage of mathematics teachers in each district with a degree in mathematics, the average age of mathematics teachers and the average salary of mathematics teachers (in thousand dollars per year): Regression Output Predictor Constant Math Degree Age Salary Coef. 20.25 0.30 0.35 0.15 SE Coef. 7.850 0.096 0.180 0.077 Analysis of Variance (Given) Source Regression Residual Error Total DF 3 31 34 SS 1025.50 630.89 1656.39 MS 341.8333 20.3512 F 16.80 Write the least squares prediction equation. What is the number of observations in the sample? Based on the multiple regression model given above, estimate the mathematics test score and calculate the value of the residual, if the percentage of teachers with a mathematics degree is 50.0, the average age is 40 and the average salary is 45,500. If the actual mathematics test score for thes e factors is 65.50, what is the error for this observation? What is the total sum of squares? What is the explained variation? What is the mean square error? Calculate the Coefficient of Determination and the Adjusted coefficient of Determination and Test for the overall usefulness of the model using F-Statistic at 5% and 1% significance levels. Finally, test the usefulness (or significance of the three independent variables using t-test for 5% and 1% significance levels. Test scores = 20.25 + MathDegree*0.3 + AverageAge*0.35 + AverageSalary*0.15 N (number of observations) = Total df + 1 = 35. Test score = 20.25 + 50*0.3 + 40*0.35 + 45.5*0.15 = 56.08 Residual: = 7.85+(50*.096)+(40*.18)+(45.5*0.077) = 23.35 Error = 65.5 - 56.08 = 9.42 Total Sum of Squares = 1656.39 Explained variation = SSR = 1025.5 Mean Square Error = 20.3512 Coefficient of Determination (R2) = SSR/SST = 0.6191 (Adjusted R2) = 1 - (1 - 0.6191)[(34-1)/(34-(4+1)) = 0.5683 F = 16.80 Fcritical (5%) = 2.91 Because F calculated is greater than F critical, the model is useful at the 5% level. Fcritical (1%) = 4.48 Because F calculated is greater than F critical, the model is useful at the 1% level. 10. It was felt by the local utility company that fuel consumption (natural gas) in their small town would be affected by temperature and chill index. The following data was collected for 8 weeks during the winter. Week Fuel Consumption, Average Hourly Chill Index, Y(MMcf) Temperature, X1 X2 (0F) 1 15.4 30.0 18 2 11.7 28.0 15 3 12.4 32.5 25 4 10.8 39.0 20 5 9.4 40.9 10 6 6.5 50.8 16 7 10.0 40.1 5 8 9.5 50.5 0 Conduct a Multiple Regression among Fuel consumption (Y), temperature (X1), and chill index (X2). Solve by excel or MegaStat. Using alpha = 0.05 and 0.01 answer the following: What is the multiple regression equation? Interpret it and predict the fuel consumption when temperature is 40 degrees and the chill index is 10. Are the coefficients of the multiple regression model statistically significant? How do you know? Discuss indicating the test statistics. What are R2, adjusted R2 and se? Interpret R2 and Adjusted R2 and compare these values. Perform F-test, show the ANOVA table and conclude about the overall model. Y (MMcf) = 21.54 - 0.267*X1 - 0.03*X2 Y = 21.54 - 0.267*40 - 0.03*10 = 10.6 MMcf The intercept and temperature coefficients are statistically significant because their p values are 0.004 and 0.028 respectively. The chill index is not statistically significant because its p value is 0.76. R2 = 0.7033 means that 70.33% of the variation in the model can be explained by the regression equation. The adjusted R2 = 0.5847. This is much lower than R2 in this model because the chill index does not add value to the prediction. The F-test shows that, at the 0.05 level, the model is significant, however, it is not significant at the 0.01 level, as the p value = 0.0479. Average s t tcrit 0.6 0.5 -0.2 0.2 0.4 0.3 0.3162278 2.1213203 2.13185 Brand Price Score Dell 3500 Hisense 2800 Hitachi 3200 JVC 3500 LG 3300 Maxent 2000 Panasonic 4000 Phillips 3000 Proview 2500 Samsung 3000 SUMMARY OUTPUT 60 55 45 50 55 30 70 55 35 40 Regression Statistics Multiple R 0.8526177018 R Square 0.7269569453 Adjusted R Square 0.6928265635 Standard Error 6.7184395489 Observations 10 ANOVA df Regression Residual Total Coefficients Standard Error t Stat -7.0098039216 12.4274218692 -0.564059 0.0183473389 0.0039754776 4.615128 Intercept Price $2,900 Score SS MS 1 961.400560224 961.4006 8 361.099439776 45.13743 9 1322.5 46.20 F Significance F 21.299408514 0.001721105 P-value Lower 95% Upper 95%Lower 95.0% Upper 95.0% 0.5881734549 -35.66749014 21.64788 -35.66749 21.64788 0.001721105 0.0091798711 0.027515 0.00918 0.027515 Predictor Coef. SE Coef. Constant 20.25 7.85 Math Degree 0.3 0.096 Age 0.35 0.18 Salary 0.15 0.077 Source DF Regression Residual Error Total R2 Adj R2 Fcritical (5%) Fcritical (1%) SS 3 31 34 0.6191 0.5683 2.91 4.48 MS F 1025.5 341.8333 630.89 20.3512 1656.39 16.80 Week Fuel Consumption, Average Hourly Temperature X1 Y(MMcf) 1 2 3 4 5 6 7 8 15.4 11.7 12.4 10.8 9.4 6.5 10 9.5 Chill Index, X2 30 28 32.5 39 40.9 50.8 40.1 50.5 18 15 25 20 10 16 5 0 temperature is 40 degrees and the chill index is 10 Y: 10.5469691157 10.6 SUMMARY OUTPUT Regression Statistics Multiple R 0.8386558272 R Square 0.7033435965 Adjusted R Square 0.5846810352 Standard Error 1.6743144414 Observations 8 ANOVA df Regression Residual Total index is 10 Intercept X1 X2 SS MS F Significance F 2 33.2321057574 16.61605 5.927258 0.0479329676 5 14.0166442426 2.803329 7 47.24875 Coefficients 21.536685762 -0.2672659046 -0.0299080463 Standard Error t Stat 4.3039771014 5.003903 0.0880744897 -3.034544 0.0927026774 -0.322623 P-value Lower 95% Upper 95% 0.004091 10.472960404 32.60041112 0.028931 -0.4936685881 -0.0408632211 0.760035 -0.268207865 0.2083917724 SUMMARY OUTPUT Regression Statistics Multiple R 0.8386558272 R Square 0.7033435965 Adjusted R Square 0.5846810352 Standard Error 1.6743144414 Observations 8 ANOVA df Regression Residual Total Intercept X1 X2 SS MS F Significance F 2 33.2321057574 16.61605 5.927258 0.0479329676 5 14.0166442426 2.803329 7 47.24875 Coefficients 21.536685762 -0.2672659046 -0.0299080463 Standard Error t Stat 4.3039771014 5.003903 0.0880744897 -3.034544 0.0927026774 -0.322623 P-value Lower 95% Upper 95% 0.004091 10.472960404 32.60041112 0.028931 -0.4936685881 -0.0408632211 0.760035 -0.268207865 0.2083917724 Lower 95.0% Upper 95.0% 10.472960404 32.60041112 -0.4936685881 -0.0408632211 -0.268207865 0.2083917724 Lower 99.0% Upper 99.0% 4.1824346914 38.890936833 -0.6223948404 0.0878630312 -0.4036984966 0.343882404