13. PLEASE HELP!!!

Assignment Submission For this assignment, you submit answers by question parts. The number of submissions remaining for each question part only changes if you submit or change the answer. Assignment Scoring Your last submission is used for your score. 1. [-/2 Points] MendStat15 13.R.001. Groups of 10-day-old chicks were randomly assigned to seven treatment groups in which a basal diet was supplemented with 0, 50, 100, 150, 200, 250, or 300 micrograms/kilogram (Hg/kg) of biotin. The table gives the average biotin intake (x) in micrograms per day and the average weight gain (y) in grams per day. Added Biotin Biotin Intake, x Weight Gain, y 0 0.13 8.0 50 2.00 17.1 100 6.05 22.4 150 6.33 24.4 200 7.16 26.5 250 9.66 23.5 300 12.51 23.4 In the MINITAB printout, the second-order polynomial model E(y) = Bo + Bix + B2x2 is fitted to the data. Use the printout to answer the questions. (Use the exact values found in the MINITAB output.) Regression Analysis: y versus x, x-sq Analysis of Variance Source DF SS MS F-Value P-Value Regression 2 226.66 113.331 35.10 0.003 Residual Error 4 12.92 3.229 Total 6 239.58 Model Summary S R-Sq R-Sq (adj) 1.79689 94.61% 91.91% Coefficients Term Coef SE Coef T-Value P-Value Constant 8.62 1.60 5.38 0.006x 3.816 0.555 6.87 0.002 xsq -0.2157 0.0429 -5.03 0.007 Regression Equation y = 8.62 + 3.816 x - 0.2157 xsq Versus Fits (response is 1;} Residual I: ll} 12 14 '16 15 21] 22 24 2'6 Fitted Value A residual plot titled "Versus Fits (resp0nse is y)" is shown. The horizontal axis is labeled "Fitted value" with values from about 8 to 26. o The vertical axis is labeled "Residual" with values from 2.1 to 2.1. . There are 7 points. . A horizontal line at 0 on the vertical axis spans the width of the plot. . There are 3 points below the line and 4 points above the line. . The approximate locations of the points are as follows. 0 (9.2, 71.1), o (15.2,1.7), o (22.8, 0.8), o (24, 71.4), o (24, 0.3), o (24.8, 1.6), o (25.2, 71.9). Normal Probability Plot (response is y) 99 Percent 4 6 5868838 -3 -2 0 3 Residual A normal probability plot titled "Normal Probability Plot (response is y)" is shown. The horizontal axis is labeled "Residual" with values from about -4 to 4. . The vertical axis is labeled "Percent" with values from 1 to 99. There are 7 points. The points are plotted from left to right in an upward, diagonal direction starting from the bottom left of the plot. The points are between approximately -1.9 and 1.7 on the horizontal axis and between approximately 7 and 93 on the vertical axis. A line with positive slope is imposed onto the plot. The line begins at about (-3.5, 1), goes up and right, then ends at about (3.5, 99). There are 4 points below the line and 3 points above the line. The points tend to be fairly close to the line. (a) What is the fitted least-squares line? (Round all numerical values to two decimal places.) y = (b) Find R2. R2 = % Interpret its value. R2 is the percent of the total variation accounted for by using x and x in the model. R2 is the percent of the y-values that fit the model perfectly. R2 is the percent of the totalresiduals less than 1. R2 is the percent of the total residuals greater than 1. R2 is the percent of the total variation unaccounted for by using x and x in the model. (c) Do the data provide sufficient evidence to conclude that the model contributes significant information for predicting y? Based on the p-value for the x term, the model contributes significant information for the prediction of y at the a = 0.01 significance level. Based on the p-value for the F-test, the model does not contribute significant information for the prediction of y at the a = 0.01 significance level. Based on the p-value for the F-test, the model contributes significant information for the prediction of y at the a = 0.01 significance level. Based on the p-value for the x term, the model does not contribute significant information for the prediction of y at the a =0.01 significance level. Based on the p-value for the constant term, the model contributes significant information for the prediction of y at the a = 0.01 significance level. (d) Find the results of the test of Ho: B2 = 0. Is there sufficient evidence to indicate that the quadratic model provides a better fit to the data than a simple linear model does? Based on the p-value for the x term, the quadratic model does provide a better fit than the simple linear model at the a = 0.01 significance level. Based on the p-value for the x term, the quadratic model does not provide a better fit than the simple linear model at the a = 0.01 significance level. Based on the p-value for the x term, the quadratic model does not provide a better fit than the simple linear model at the a = 0.01 significance level. Based on the p-value for the F-test, the quadratic model does provide a better fit than the simple linear model at the a = 0.01 significance level. Based on the p-value for the x2 term, the quadratic model does provide a better fit than the simple linear model at the a = 0.01 significance level. (e) Do the residual plots indicate that any of the regression assumptions have been violated? Explain. The residual plot is unusual, indicating a violation in the assumption of normality. The normal probability plot is unusual, indicating a violation in the assumption of normality. There are no obvious violations in the assumptions based on the patterns shown in the diagnostic plots. The normal probability plot is unusual, indicating a violation in the assumption of constant variance. The residual plot is unusual, indicating a violation in the assumption of constant variance. 2.[12 Points] MendStatlS 13.R.005. The mercury concentrations in striped dolphins were measured as part of a marine pollution study. This concentration is expected to differ in males and females because the mercury in a female is apparently transferred to her offspring during gestation and nursing. The study involved 28 males between the ages of 0.21 and 39.5 years, and 17 females between the ages of 0.80 and 34.5 years. For the data in the table, x1 = age of the dolphin (in years), X2 = 0 if female 1 if male, and y = mercury concentration (in micro grams/gram) in the liver. y x1 x2 y x1 X: 1.70 0.21 1 480.00 22.50 1 1.72 0.33 1 484.00 24.50 1 8.80 2.00 1 212.00 24.50 1 5.90 2.20 1 398.00 25.50 1 101.00 8.50 1 259.00 26.50 1 85.40 11.50 1 323.00 26.50 1 118.00 11.50 1 320.00 26.50 1 177.00 13.50 1 445.00 26.50 1 178.00 16.50 1 278.00 27.50 1 218.00 16.50 1 286.00 28.50 1 180.00 17.50 1 315.00 29.50 1 264.00 20.50 1 y x1 x2 y x1 x2 249.00 31.501 145.00 17.50 0 403.00 31.50 1 180.00 17.50 0 208.00 36.50 1 174.00 18.50 0 314.00 37.50 1 246.00 19.50 0 318.00 39.501 217.00 21.500 2.50 0.80 0 162.00 21.500 9.35 1.58 0 148.00 25.500 4.01 1.75 0 177.00 25.500 29.80 5.50 0 475.00 32.500 45.30 7.50 0 342.00 34.500 101.00 8.05 0 140.00 11.50 0 (a) Write a second-order model relating y to 161 and X2. Allow for curvature in the relationship between age and mercury concentration, and allow for an interaction between gender and age. If\" y=o+1x1+2x2+4x1m+ rt y = 50 + 183an + 4x1x2 + sxlzxz + 8 r y = 30 + 3m + 32.762 + 4xix2 + 357612162 + S t" y = [30 + lxr + m + gxlz + 34mm + sxrzxz + 8 Use a computer software package to perform the multiple regression analysis. Refer to the printout to answer these questions. 0)) Comment on the t of the model, using relevant statistics from the printout. State the tted prediction model. (Enter 0 as the coefficient for any term not used in the model. Round our answers to two decimal places.) 'E r = + Baa.2 x1 X2 State the F test statistic for the model. (Round your answer to two decimal places.) F =1 State the approximate p-value for the model. p-value 0.100 State the R' value for the model. (Round your answer to three two decimal places.) R2 = 1 Does the model fit well? Yes No (c) What is the prediction equation for predicting the mercury concentration in a female dolphin as a function of her age? (Enter 0 as the coefficient for any term not used in the model. Round your answers to two decimal places.) y = L + x1 + (d) What is the prediction equation for predicting the mercury concentration in a male dolphin as a function of his age? (Enter 0 as the coefficient for any term not used in the model. Round your answers to two decimal places.)=| + l; T , x12 (6) Does the quadratic term in the prediction equation for females contribute significantly to the prediction of the mercury concentration in a female dolphin? State the null and alternate hypotheses. Find the test statistic. (Round your answer to two decimal places.) ,=| Approximate the p-value for the test. r p-Value 0.100 State your conclusion. r. H0 is not rejected. The quadratic term is important in predicting mercury concentration for female dolphins. r' H0 is not rejected. The quadratic term is not important in predicting . . f" . . . . mercury concentration for female dolphins. Ho 18 rejected. The quadratic term is not important in predicting mercury concentration for female dolphins. r H0 is rejected. The quadratic term is important in predicting mercury concentration for female dolphins. (1) Are there any other important conclusions that you feel were not considered regarding the tted prediction equation? 3. [12 Points] MendStatlS 13.R.006. The quantitative reasoning scores on the Graduate Record Examination (GRE) were recorded for students admitted to three different graduate programs at a local university. These data were analyzed in Chapter 11 using the analysis of variance for a completely randomized design. Graduate Program Life Sciences Physical Sciences Social Sciences 640 650 670 750 450 530 630 650 650 660 340 460 480 470 710 710 680 560 590 640 700 720 580 540 590 700 540 460 600 640 (a) Write the theoretical model relating the GRE score to the qualitative variable "graduate program" using two dummy (indicator) variables to represent the three graduate programs. (Let y = score, x1 = 1, if Physical Sciences; 0 otherwise, and x2 = 1, if Social Sciences; 0 otherwise.) (II. y = 30 +31x1+ Bzxrxz r' y = [5'0 + 3m + 'mzxzz It" y = [30 +31X1 + 322162 If\" y = 30 + B1131 + 22x1x2 r. y = o+m + zxz (b) Use a computer package to analyze the data with a multiple regression analysis. Is there sufcient evidence to indicate a difference in the average scores between the students who have been admitted to the three graduate programs? Use a = 0.05. State the null and alternative hypotheses. Ho: At least one of B1, B2 is not 0. Ha: B1 = P2 = 0 Ho: At least one of B1, B2 is not 0. Ha: B1> B2> 0 Ho: BI 0.100 State your conclusion. Ho is not rejected. There is sufficient evidence to indicate there is a difference in the average scores between the students who have been admitted to the three graduate programs. Ho is rejected. There is insufficient evidence to indicate there is a difference in the average scores between the students who have been admitted to the three graduate programs. Ho is rejected. There is sufficient evidence to indicate there is a difference in the average scores between the students who have been admitted to the three graduate programs. Ho is not rejected. There is insufficient evidence to indicate there is a difference in the average scores between the students who have been admitted to the three graduate programs. C Comment on the fit of the model and any regression assumptions that may have been violated. Summarize your results in a report, including printouts and graphs if possible. You may need to use the appropriate appendix table or technology to answer this question. 4. [-/2 Points] MendStat15 13.R.007. Performance tires used to be fitted mostly on sporty or luxury vehicles. Now they come standard on many standard vehicles. The data that follow are abstracted from a report on all-season tires by Consumer Reports/ in which several aspects of performance were evaluated for n = 16different tires where y = Score x1 = Dry braking x2 = Wet braking x3 = Handling x4 = Hydroplaning x5 = Tread life (1,000 miles) Tire Price ($) y x1X2X3X4 X5 Michelin Defender 120.00 70 4 3 4 4 90 Continental TrueContact 106.04 68 4 3 4 4 60 General Altimax RT43[T] 90.50 664 3 3 4 65 Pirelli P4 Four Seasons Plus 100.00 66 4 3 3 4 100 Nexen Aria AH7 119.00 64 4 3 3 3 75 Goodyear Assurance Triple Tred All-Season[T] 121.10 624 3 4 4 80 Kuhmo Solus TAll 108.00 62 4 2 3 4 55 Cooper CS5 Grand Touring 91.50 62 4 3 3 4 70 Yokohama Avid Ascend[T] 92.95 60 4 2 3 4 90 BFGoodrich AdvantageT/A 101.94 584 2 3 4 75 Uniroyal Tiger Paw Touring 90.44 564 2 3 4 65 Sumitomo HTR Enhance L/X[T] 77.82 56 4 2 3 4 70 Toyo Extensa A/S 79.50 54 4 2 3 3 60 Firestone Precision Touring 94.45 54 4 3 3 3 55 Firestone Precision FR710 98.00 52 4 3 3 3 55 GT Radial Champiro VPI [T] 63.98 503 3 3 4 45 The variables X1 through X4 are coded using the scale 5 = excellent, 4 = very good, 3 = good, 2 = fair, and 1 =poor. (a) Use a program of your choice to find the correlation matrix for the variables under study including price. (Use a = 0.05.) Is price significantly correlated with any of the study variables? (Select all that apply.) y x x2 x3 X4 x Which variables appear to be highly correlated with the score y? (Select all that apply.) X1 X2 X3 x4 X5 (b) Write a model to describe y, the score, as a function of the variables x1 = Dry braking, x2 = Wet braking, x3 = Handling,x4 = Hydroplaning, and x5 = Tread life (1,000 miles). I!\" E(y) =30 +1x1+2x2 + 53x3 +b'm +[J'sxs r' E(y) = 30 + 31x1 + 3m + SXlxz + 34x12 + 35x22 1* EO') = 30 + 3m + 32262 + 33x12 + 34ch2 + sxlxzz r. E(y) = g + 1x1+ zxz + 33m + [34.36110 + sxlxzxs (I. E0') = 30 + IXI + 32x2 + 53x3 + ulxz + shxs (C) Use a regression program to t the full model using all predictors. Discuss the adequacy of the model based on your results. (Use a = 0.05. Round your g-value to three decimal places and all other values to two decimal places.) Since F = I with p-value = I . This indicates that the model signicant information for the prediction of y. (d) Use the best subsets program to detcmline which variables produce the largest value for R2(adj). Fit the appropriate model based on the results of a best subsets program. What proportion of the variation is explained by the retted model? Comment on the adequacy of this reduced model in comparison to the full model. (Use a = 0.05. Round your g-value to three decimal places and all other values to two decimal places.) % of the total variatiOn is accounted for by using this model and the F -test yields Since test ields F = i withp-value = I . This indicates that the model signicant information for the prediction of y. You may need to use the appropriate appendix table or technology to answer this question. % of the total variation is accounted for by using this retted model and the F- 5' [12 Points] MendStatlS 13.R.008. The tuna sh data below were analyzed as a completely randomized design with four treatments. However, we could also view the experimental design as a 2 x 2 factorial experiment with unequal replications. The data are shown below. Oil Water Light Tuna 2.57 1.91 0.98 1.92 1,301.79 1.23 0.85 1.23 0.62 0.65 0.53 0.66 0.62 1.41 1.12 0.65 0.60 0.63 0.67 0.67 0.69 0.60 0.60 0.66 White Tuna 1.28 1.50 1.30 1.22 1.29 1.00 1.19 1.27 1.27 1.22 1.35 1.28 The data can be analyzed using the model y=o+lx1+2x2+3x1x2+ where .761 = 0 if oil, 1 if water X2 = 0 if light tuna, 1 if white tuna. (a) Show how you would enter the data into a computer spreadsheet, entering the data into columns for y, x1, x2, and x1362. _..l' (b) The printout generated by MINITAB is shown below. (Use the exact values found in the MINI TAB output.) Regression Analysis: y versus x1, x2, x1 x2 Analysis of Variance Source DF Adj SS Adj MS F-Value P-Value Regression 3 0.9366 0.3122 1.51 0.230 Error 33 6.8272 0.2069 Total 36 7.7638 Model Summary S R-Sq R-Sq (adj) 0.454845 12.1% 4.1% Coefficients Term Coef SE Coef T-Value P-Value Constant 1.147 0.137 8.37 0.000 -0.252 0.183 1.37 0.179 X2 0.080 0.266 0.30 0.764 x1x2 0.307 0.333 0.92 0.365 Regression Equation y = 1.147 - 0.252 x1 + 0.080 x2 + 0.307 x1x2 What is the least-squares prediction equation? y = (c) Is there an interaction between type of tuna and type of packing liquid? The p-value is 0.179. There is insufficient evidence to indicate that the interaction is significant at the a = 0.05 level. The p-value is 0.000. There is sufficient evidence to indicate that the interaction is significant at the a = 0.05 level. The p-value is 0.764. There is insufficient evidence to indicate that the interaction is significant at the a = 0.05 level. The p-value is 0.365. There is insufficient evidence to indicate that the interaction is significant at the a = 0.05 level. The p-value is 0.230. There is insufficient evidence to indicate that the interaction is significant at the a = 0.05 level. (d) Which, if any, of the main effects (type of tuna and type of packing liquid) contribute significant information for the prediction of y? (Use a = 0.05.) C X1 * 2 x1 and x2 neither main effect (e) How well does the model fit the data? Explain. The combination of a high p-value and high R value indicate that the model is not a good fit. The combination of a high p-value and low R value indicate that the model is not a good fit. The combination of a low p-value and low R2 value indicate that the model is not agood fit. The combination of a high p-value and high R' value indicate that the model is a good fit. The combination of a low p-value and high R value indicate that the model is a good fit. 6. [-/2 Points] MendStat15 13.R.010. A manufacturer recorded the number of defective items (y) produced on a given day by each of 10 machine operators and also recorded the average output per hour (x1) for each operator and the time in weeks from the last machine service (x2). y x1 X2 13 19 3.0 1 14 2.0 1 1 22 1.5 2 9 4.0 20 29 1.0 15 20 3.5 27 37 0.0 5 172.0 26 23 5.0 1 15 1.5 The printout that follows resulted when these data were analyzed using the MINITAB package using the model. (Use the exact values found in the MINITAB output.) E(y) = Bo + PIXI + B2x2. Regression Analysis: y versus x1, x2 Analysis of Variance Source DF Adj SS Adj MS F-Value P-Value Regression 2 884.795 442.397 1470.841 0.000 Error 7 2.105 0.301 Total 9 886.900 Model Summary S R-Sq R-Sq (adj) 0.548433 99.76% 99.69% Coefficients Term Coef SE Coef T-Value P-Value Constant -26.928 0.803 -33.55 0.000 x1 1.4631 0.0270 54.20 0.000x2 3.845 0.143 26.97 0.000 Regression Equation y = -26.928 + 1.4631 x1 + 3.845 x2 (a) Interpret R2 and comment on the fit of the model. Since of the variation is explained by the model, the fit of the model is good. (b Is there evidence to indicate that the model contributes significantly to the prediction of y at the a = 0.01 level of significance? State the null and alternative hypotheses. C Ho: B1 = B2 Ha: B1 # B2 Ho: B1 B2> O Ho: B1 = B2 = 0 Ha: At least one of B1, B2 is not O. Ho: At least one of B1, B2 is not 0. Ha: B1 = B2 = 0 Find the test statistic. F = Find the approximate p-value for the test. p-value 0.100 State your conclusion. C Ho is not rejected. There is insufficient evidence to indicate that the model contributes information for the prediction of y. Ho is not rejected. There is sufficient evidence to indicate that the model contributes information for the prediction of y. Ho is rejected. There is insufficient evidence to indicate that the model contributes information for the prediction of y. Ho is rejected. There is sufficient evidence to indicate that the model contributes information for the prediction of y. (c) What is the prediction equation relating y and X1 when X2 = 2? y =(d Use the fitted prediction equation to predict the number of defective items produced for an operator whose average output per hour is 24 and whose machine was serviced 2 weeks ago. Round your answer to the nearest integer.) defects (e) What do the residual plots tell you about the validity of the regression assumptions? (Select all that apply.) Normal Probability Plot (response is y) 99 95 Percent -1.5 -0.5 0 0.5 1.5 Residual A normal probability plot titled "Normal Probability Plot (response is y)" has a horizontal axis labeled "Residual" with values from -1.5 to 1.5 and a vertical axis labeled "Percent" with values from 1 to 99. There are 10 points. . The points are plotted from left to right in an upward, diagonal direction starting from the bottom left of the plot A line with positive slope is imposed onto the plot. The line begins at about (-1.2, 1), goes up and right, then ends at about (1.2, 99). There are 2 points below the line, 6 points on or nearly on the line, and 2 points above the line. The points tend to be fairly close to the line.Versus Fits (reap unse is 3!] Residual 1.0 {1.0 {'.|.5 Fitted 1.1) {J 5 10 15 2D 25 Svalue A plot titled "Versus Fits (response in y)" has a horizontal axis labeled "Fitted value" with values from 0 to 30 and a vertical axis labeled "Residual" with values from 1 to l. . There are 10 points. . A horizontal line at 0 on the vertical axis spans the width of the plot. . There are 5 points above the line, 1 point on or nearly on the line, and 4 points below the line. . The approximate locations of the points are as follows. (0.8, 0.21), (1.2, 70.24), (1.6, 0.38), (5.6, 70.63), (11.0, 70.03), (12.4, 0.60), (15.8, 0.79), (19.3, 0.65), (25.9, 0.05), (27.2, 70.21) OOOOOOOOOO I l The residual plots show a violation the regression assumption of independence. The I residual plots show a violation the regression assumption of common variance. The residual plots show a violation of the regression assumption of normality. ' show any violations of the regression assumptions. The residual plots do not [-/2 Points] MendStat15 13.R.004. The effect of mean monthly daily temperature xi and cost per kilowatthour x2 on the mean daily household consumption of electricity (in kilowatt-hours, kWh) was the subject of a short-term study. The investigators expected the demand for electricity to rise in cold weather (due to heating), fall when the weather was moderate, and rise again when the temperature rose and there was need for air-conditioning. They expected demand to decrease as the cost per kilowatt- hour increased, reflecting greater attention to conservation. Data were available for 2 years, a period in which the cost per kilowatt-hour x2 increased because of the increasing cost of fuel. The company fitted the model E()) = Bo + PIXI + B2x12 + B3x2 + B4x1x2+ BSx12x2 to the data shown in the table. The Excel printout for this multiple regression problem is also provided Price per Daily Temperature Mean Daily Consumption kWh, x2 and Consumption (k Wh) per Household Mean daily 30 35 39 42 47 56 temperature ( F), x1 62 66 68 71 76 78 Mean daily 56 49 46 47 40 43 consumption, y 41 46 44 51 62 73 Mean daily 32 36 39 42 48 56 temperature, x1 62 66 68 72 75 79 10c Mean daily 50 44 42 42 38 40 consumption, y 39 44 40 44 50 55 SUMMARY OUTPUT Regression Statistics Multiple R 0.949 R Square 0.901 Adjusted R Square 0.874 Standard Error 2.887 Observations 24 ANOVA df SS MS Significance F Regression 5 1365.810 273.162 32.774 0.000 Residual 18 150.023 8.335 Total 23 1515.833 Coefficients Standard Error t Stat P Value Intercept 315.429 80.699 3.909 0.001 XI -10.892 3.142 -3.467 0.003 x1-sq 0.108 0.029 3.794 0.001 x2 -20.681 8.999 2.298 0.034x1x2 0.824 0.350 2.356 0.030 xIsqx2 -0.008 0.003 -2.631 0.017 (a) Do the data provide sufficient evidence to indicate that the model contributes information for the prediction of mean daily kilowatt-hour consumption per household? Test at the 5% level of significance. State the null and alternative hypotheses. C Ho: BI B2 > B3 > B4> Bs > O Ho: B1 = B2 = B3 = B4 = Bs Ha: At least one of B1, B2, B3, B4, Bs is different from the others. Ho: B1 = B2 = B3 = B4 = B5 = 0 Ha: At least one of B1, B2, B3, B4, Bs is not O. " Ho: Bi > B2 > B3 > B4 > Bs > O Ha: B1 = B2 = B3 = B4 = Bs =0 Find the test statistic. (Round your answer to three decimal places.) F =1 Approximate the p-value for the test. p-value 0.00 State your conclusion. C Ho is rejected. There is insufficient evidence to indicate that the model contributes information for the prediction of y. Ho is not rejected. There is insufficient evidence to indicate that the model contributes information for the prediction of y. Ho is not rejected. There is sufficient evidence to indicate that the model contributes information for the prediction of y. Ho is rejected. There is sufficient evidence to indicate that the model contributes information for the prediction of y. (b) Graph the curve depicting y as a function of temperature x1 when the cost per kilowatt-hour is X2 = 84. Construct a similar graph for the case when x2 = 10 per kilowatt-hour. Are the consumption curves different? Yes, they are different. No, they are not different. (c) If cost per kilowatt-hour is unimportant in predicting use, then you do not need the terms involving x2 in the model. Therefore, the null hypothesis Ho: X2 does not contribute information for the prediction of y is equivalent to the null hypothesisH0133=134=s=0 (3=54=s=0 the terms involving 152 disappear from the model). The Excel printout, obtained by tting the reduced model E01) = 50 + rst] + y)\"2 to the data, is shown here. SUMMARY OUTPUT Regression Statistics Multiple R 0.8405 R Square 0.7064 Adjusted R Square 0.67 85 Standard Error 4.6033 Observations 24 ANOVA df SS MS F Signicance F Regression 2 1070.839 535.419 25.267 0.000 Residual 21444.995 21.190 Total 23 1515.833 Coefcients Standard Error t Stat P Value Intercept 130.583 14.271 9.150 0.000 x1 73.518 0.555 76.340 0.000 xl-sq 0.033 0.005 6.647 0.000 Determine whether price per kilowatt-hour x2 contributes significant information for the prediction of y. (Test at the 5% level of significance.) Find the test statistic. (Round your answer to two decimal places.) F=| Find the rejection region. (Round your answer to two decimal places.) F > State your conclusion. in There is evidence that x2 is important in predicting usage and should not be eliminated from the model. r There is evidence that x2 is important in predicting usage and should be eliminated from the model. in There is no evidence that x2 is important in predicting usage, and it should be eliminated from the model. 1" There is no evidence that x; is important in predicting usage, and it should not be eliminated from the model. (60 Compare the values of R2(adj) for the two models t in this exercise. Which of the two models would you recommend? t" Since the R2(adj) values are very close, one model is not better than the other. If. Since R2(adj) is lower in the complete model than in the reduced model, the reduced model is better. If Since R2(adj) is higher in the complete model than in the reduced model, the complete model is better. F Since R2(adj) is lower in the complete model than in the reduced model, the complete model is better. r" Since R2(adj) is higher in the complete model than in the reduced model. the reduced model is better. You may need to use the appropriate appendix table or technology to answer this question. 8' [12 Points] MendStatlS 13.R.009. The tuna sh data below were analyzed as a completely randomized design with four treatments. However, we could also view the experimental design as a 2 x 2 factorial experiment with unequal replications. The data are shown below. 011 Water Light Tuna 2.57 1.91 0.98 1.92 1.301.79 1.23 0.85 1.23 0.62 0.65 0.53 0.66 0.62 1.41 1.12 0.65 0.60 0.63 0.67 0.67 0.69 0.60 0.60 0.66 White Tuna 1.28 1.50 1.28 1.22 1.29 1.00 1.19 1.27 1.27 1.22 1.35 1.28 The data can be analyzed using the model y = o+m + 62x2 +ax1x2 + where xi = 0 if oil, 1 ifwater xz = 0 if light tuna, 1 if white tuna. The hypothesis tested earlier in the bookthat the average prices for the four types of tuna are the sameiis equivalent to saying that EU) will not change as x1 and X2 change. This can only happen when B1 = B2 = B3 = 0 Use the MINITAB printout for the one-way ANOVA shown below to perform the test for equality of treatment means. (Use the exact values found in the MINITAB output.) One-Way ANOVA: Light Water, White Oil, White Water, Light Oil Analysis of Variance Source DF Adj. SS Adj MS F-Value P-Value Factor 3 0.929 0.310 1.50 0.234 Error 33 6.827 0.207 Total 36 7.756 Model Summary S R-Sq R-Sq (adj) R-Sq (pred) 0.4548 11.98% 3.98% 0.00% State the null and alternate hypotheses. C Ho: M1 = M2 = M3 = M4 Ha: At least two of the treatment means are different. Ho: All four treatment means are different. Ha: MI = M2 = M3 = M4 Ho: HI 12 13 M2 > M3 > M4 Ho: At least two of the treatment means are different. Ha: MI = M2 = M3 = M4 Ho: M1 = M2 = M3 = M4 Ha: All four treatment means are different. Find the test statistic. F = What can be said about the p-value for this test? p-value 0.100 State your conclusion. Ho is rejected. There is insufficient evidence to conclude that the treatment means differ. Ho is not rejected. There is sufficient evidence to conclude that the treatment means differ. Ho is rejected. There is sufficient evidence to conclude that the treatment means differ. C Ho is not rejected. There is insufficient evidence to conclude that the treatment means differ. Given below is the MINITAB output for the analysis of the previously described regression equation. Regression Analysis: y versus x1, x2, x1x2 Analysis of VarianceSource DF Adj SS Adj MS F-Value P-Value Regression 3 0.9290 0.3097 1.50 0.234 Error 33 6.8268 0.2069 Total 36 7.7558 Model Summary S R-Sq R-Sq (adj) 0.454834 12.0% 4.0% Coefficients Term Coef SE Coef T-Value P-Value Constant 1.147 0.137 8.37 0.000 -0.252 0.183 -1.37 0.179 X2 0.080 0.266 0.30 0.764 x1x2 0.304 0.333 0.91 0.368 Regression Equation y = 1.147 - 0.252 x1 + 0.080 x2 + 0.304 x1x2 Is the test for equality of treatment means that you performed identical to the test for significant regression? Yes, the tests are identical. No, the tests are different. You may need to use the appropriate appendix table or technology to answer this question. 9. [-/2 Points] MendStat15 13.R.501.XP. Suppose that E(y) is related to two predictor variables, X1 and X2, by the equation E(y) = 4 +x1 -2x2. (a) Graph the relationship between E(y) and x1 when x2 = 2. Repeat for 12 = 1 and for 12 = 0.\fX2 = 0 X2 = 1 E(y) HOHNWAUG X2 = 2 1 N X1 (b What relationship do the lines in part (a) have to one another? The lines are all vertical. The lines are parallel. The lines intersect each other at the same point. The lines are all horizontal. 10. [-/2 Points] MendStat15 13.R.502.XP. Suppose that is related to two predictor variables, X1 and X2, by the equation E(y) = 6 + x1 -4x2. (a) Graph the relationship between and x2 when x1 = 0. Repeat for x1 = 1 and for x1 = 2.\f45'" 1'2 (b) What relationship do the lines in part (a) have to one another? I" The lines intersect each other at the same point. t" The lines are all horizontal. r The lines are all vertical. t" The lines are parallel. (C) Suppose, in a practical situation, you want to model the relationship between EU) and two predictor variables x1 and x2. What is the implication of using the rst-order model E(y) =30 +31)\" + 32362? f. The rst-order model E(y) = n + 31x1 + 15' 212 will be a twisted surface in three-dimensional Space. rt The rst-order model E(y) = [90 + 3m + zxz will be a plane in three-dimensional space. If" The rst-order model E(y) = e + 31x1 + [32x2 will be a straight line in two-dimensional space. If. The rst-order model E0!) = e + B 1x1 + zxz will be a set of intersecting lines in two- dimensional space