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A study was conducted on the effect of water temperature and time in solution on the amount of dye absorbed by a certain kind of fabric. A standard amount of dye (200 milligrams/mg) was added to a fixed amount of water. The three temperature levels used in the experiment were 105, 120, and 135' C. The fabric was left in the water 15, 30, or 60 minutes (min). For each of these temperature-time combinations the amount of dye left inside the fabric was measured. The experiment yielded the following data: Dye in yarn (mg) Time in solution (min) Temperature of H,O(C) (v) 136 15 105 153 30 105 186 105 182 120 175 120 187 120 170 135 179 135 183 60 135 If a dye solution is prepared with the temperature at 125: C and the fabric left in the solution for 20 min, how much dye, on the average, would you predict to be left in the fabric?A study on the burn time for the common Fourth of July sparkler was conducted. The variables considered were the length of the chemical coating that covers the tip of the sparkler (in inches) and the bum time (in seconds) of the sparkler. Seventeen sparklers were burned, and the values of the chemical coating (x) and the bum time ()) are given below. Ohs. no. Chemical length (in.) Burn time (sec.) 4.5 29 3.6 26 4.0 25 3.7 25 4.0 27 3.7 27 4.0 28 4.0 25 3.8 25 4.0 28 3.8 24 4.1 15 3.9 22 4.1 25 3.9 24 16 4.2 26 17 3.8 24 (a) Estimate the simple linear regression model with response variable burn time (y) and predictor variable chemical length (x). (b) Test for significant regression at o = 0.05. Is this a good model to predict burn time? (c) Can you think of other variables that should be included in a future study to improve prediction ability?A study was conducted on the effect of nitric acid dilutions on the accelerated weathering of wood. The acid pH levels of 2.0, 2.5, 3.0, 3.5, and 4.0 were tested with distilled water (pH 5.6) used as a control. An accelerated weathering chamber was used with times of 200, 400, 600, 800, and 1000 hours. Red cedar wafers of approximately 700 mp were obtained and weighed accurately for the start weight at the beginning of the experiment. At the end of the accelerated time, the wafers are again weighed for the final weight. The resulting data, including the difference in "start weight" and "final weight" are given in the following table. Obs no. PH Time Start wi. Final wi. Difference 2.0 200 693 669 24 2.01 400 696 647 40 2.0 700 621 2.0 800 692 601 91 2.0 1000 693 STS 18 2.5 200 697 617 20 2.5 400 698 656 42 2.5 600 698 632 66 2.5 693 612 86 2.5 1000 697 585 12 3.0 200 698 677 21 3.0 400 699 656 43 3.0 600 690 636 63 3.0 800 695 617 78 3.0 1000 598 597 101 3.5 2010 698 679 19 17 3.5 400 696 662 34 18 3.5 60O 698 641 ST 19 3.5 699 622 37 20 35 1000 698 598 100 21 4.0 200 695 677 18 22 4.0 400 661 37 23 4.0 GOD 604 543 51 24 4.0 BOD GOR 672 76 25 4.0 1000 700 602 26 5.6 201 695 677 5.6 400 695 662 28 5.6 600 642 56 29 5.6 800 698 625 73 30 5.6 1000 698 606 92 For the model you selected in Exercise 70, calculate RY, and test for significance at o = 0.05.Consider the model = Bot Bix + Brixi + B2x2 Express this model in the general linear form of (12.1).In Exercise 3 we considered the model Anna = Bo + Box, + Box; based on these data: 00 10 (a) Find the model specification matrix. (b) Find XX. (c) Find Xy. (d) Find the normal equations, and compare them to those found in Exercise 3. (e) Show that 1680 -4 -200 16 16 16 (XX)= -4 2 0 16 16 -200 0 24 16 16 9 (f) Verify that h = -5 0Use the data of Example 12.2.5 to find 95% confidence intervals 8, and B. Is there evidence that 8, # 0? That B, # 0? Explain. Example 12.2.5 These data are available on X, the number of units of drug pro- duced, and Y the cost per unit of producing the drug. (See Example 12.1.1.) 5 5 10 10 15 15 20 20 25 25 y 14.0 125 7.0 5.0 2.1 1.8 6.2 4.9 13.2 14.6 The model specification matrix for a quadratic model is 5 25 5 25 10 100 10 100 X = 15 225 15 225 20 400 20 400 1 25 625 25 625 10 150 2750 X'X = 150 2750 56.250 2750 56.250 1,223,750 81.3 Xy = 1228 24,555 Apart from round-off error 2.3 -.33 (xx)-= -.33 .05342857 -.00171429 -01 -00171429 .00005714286 The least-squares estimates for B. 81. B, are 27.3 b - (xx) ry= -3.313 -111 The estimated model is Are - 27.3-3313x + Illx The predicted unit cost of producing 12 units of the drug is y = 27.3 - 3.313(12) + . 111(12) = 3.528 Example 12.1.1 Suppose that we want to develop an equation with which we can predict the gasoline mileage of an automobile based on its weight and the temperature at the time of operation. We might pose the model Arina na = BotBixt + Byxz Here the response variable is Y, the mileage obtained. There are two independent or predictor variables. These are X, the weight of the car, and X2. the temperature. The values assumed by these variables are denoted by x, and a,, respectively. For example, we might want to predict the gas mileage for a car that weighs 1.6 tons when it is be- ing driven in 85" F weather. Here x, = 1.6 and x2- 85. The unknown parameters in the model are By 81, and By. Their values are to be estimated from the data gathered.Use the information given in Example 12.3.3 to find a 90% confidence interval on the mean gasoline mileage obtained by cars weighing 1.5 tons when operated on a 40 F day. Find a 90% prediction interval on the gasoline mileage obtained by a specific automobile weighing 1.5 tons when operated on a 40* F day. Which interval is wider? Example 12.3.3 In Example 12.2.3 we found that the estimated regression equation for predicting the gasoline mileage of a car based on its weight and the temperature at the time of operation is MyIsany = 24.75 -4.16x1 - 014897x2 Since 130 X'y = 282.405 8887 we already have available most of the information needed to compute SSE. The only other term needed is 3/-, y/. A quick computation yields a value of 2900.46 for this term. Substituting, we have Spy = 105y-x) /10 - [2900.46 - (170)31/10 = 10.46 SSR - bo En + b Sry+ box- (x)/ 10 = 24.75(170) - 4.16(282.405) - .014897(8887) - (170)'/10 = 10.31 By subtraction SSE = S,, - SSR = 10.46 -10.31 = .15 Hence 61 - s' = SSF/m-k- D) = .15/10-2-1) = 10214 In Example 12.2.3 we found that the matrix (XX) for these data is 6.070769 -3.02588 -.0171838 (XX) -3.02588 1.738599 002166306 -.0171888 .002166306 .0002582903 Since Var B = o'(X'X)", to find the estimates for the variances of Bo, B1, and By, we multiply each number on the main diagonal of (XX)-' by o'. Thus Var By - 6.070769(.0214) =.1299 Var B = 1.738599(.0214) = 0372 Var By = .0002582903(.0214) - .000005 Example 12.2.3 The matrix XX with which we are working is 10 16.75 525 X'X = 16.75 28.6375 874.5 525 874.5 31,475 We shall let the computer find (X'X) ' for us! You can verify that the inverse of this matrix is, apart from round-off error, 6.070769 -3.02588 -,017188 (xx]-1= -3.02588 1.738599 002166306 -.0171888 .002166306 .0002582903 The vector of parameter estimates, apart from round-off error, is b = (XX) -X'y 6.070769 -3.02588 -.0171888 170 -3.02588 1.738599 002166306 282.405 -.0171888 .002166306 0002582903 8887 24.75 -4.16 -.014897 The estimated model is Mylxx,- 24.75-4.16x - .014897x, Based on this equation, we estimate the mileage for a car weighing 1.5 tons on a 70' F day to be y = 24.75 - 4.16(1.5) - ,014897(70) = 17.47 miles per gallon.A study was conducted to study energy consumption versus household income and home ownership status. Let Y (in units of 10" Btu's) denote energy consumption, x, (in units of $1000/year) denote income, and X2 (X2 = 1, 0) for ownership versus rental, respectively, denote ownership status. The following data were obtained: Consumption Income Ownership 1.8 20.0 4.7 25.0 3.0 30.5 5.8 35.1 4.8 40.0 7.1 48.2 5.0 55.1 8.0 60.5 7.0 74.9 9.9 80.3 9.0 88.4 113 90.1 9.2 95.2 (a) Assume that an appropriate model is Myisnt, = Bot Bix + B2x2 Find the model specification matrix. (b) For these data 5416711 -.00690843 -.151114 (XX]I= -.00690843 0001 196709 .0001430353 -.151114 0001430353 .3096948 and 86.6 Xy = 5751.14 46.8 Use this information to estimate the model. (c) For the given data 62=.13417107 Use this information to test Ho: By = 0. (d) State the estimated models that describe the relationship between energy consumption and income for homeowners; for renters. (e) Find a point estimate for the difference in intercepts for the two models in part (d).An engineer is investigating the recovery of heat lost to the environment in the form of exhaust gases for two types of furnaces. The experiment is designed to fix flow speed past heat pipes (in meters per second) and then to measure the recovery ratio. Recovery Flow speed Type furnace .740 .745 1.5 .718 2 678 25 652 3 627 .607 4 .507 4.5 .545 5 .402 .321 .255 .175 .115 085 (a) Assume that it is not known whether or not flow speed affects each type of furnace in a similar way. The appropriate model is where *2 = 1 if furnace A is used and X2 = 0 if furnace B is used. Find the model specification matrix. (b) For these data 1 .285714 -1 .2857143 (XX)= -.285714 .09795918 .2857143 -.0979592 -1 .2857143 1.711111 -.485714 .2857143 -.0979592 -.485714 .1646259 and 7.172 19.665 X'S = 5.819 16.5735 Use this information to estimate the model. (c) Estimate the difference in slopes for the regression lines for the two furnaces. (d) For the given data 62 = .0007562574 Use this information to test Ho: By = 0. What practical conclusion can be drawn?Assume that we have available four possible predictor variables X1, X2, X3, X4. Suppose that our final model via forward selection contains only the variables x, and *, and that they entered the model in the order stated. Outline the steps taken in developing this model. Follow the format given in Example 12.7.1. Example 12.7.1. Assume that we have available three possible predictor variables X1 X2, and X,. Suppose that our final model via forward selection contains only the variables X, and X, and that they entered the model in the order stated. These are the steps that are taken by the computer: 1. The three single-variable models Mys = Bo+ 81-*1 Myn = Bo + 82x2 are fitted. The value of R? is found for each. The one with the highest R" is cho- sen and compared to the reduced model My = By. In this case we test He: My Ba (reduced model is appropriate) Hi:Mys = Bo + Byxy (full model is needed) and Ho is rejected. The variable X, is now included in our model. 2. The two two-variable models Myinn, = Bo+ Box + Byly are fitted. The value of R? is found for each. The one with the highest R' is cho- sen and compared to the reduced model pru, = Bo + Box,. In this case we test Ha Myin " Bo + Byty (reduced model is appropriate) H:Prison = Bot But, + Byty (full model is needed) and He is rejected. The variable X, is now included in our model. 3. The three-variable model is fitted, and we test He Myina, = Bot Bix + Box (reduced model is appropriate) HI: Myinamx, = But Bix + But, + Byty (full model is needed) In this case H, is not rejected. The variable X2 does not appear to be needed in our model. The final model that we obtain is the two-variable model HYlx x, = Bot Bix + Byx3Assume that we have four potential predictor variables and that via backward elimination we obtain a reduced model containing only the variables x, and *2- Assume that the variables %; and * are deleted in the order mentioned. Outline the steps taken in developing this model. Follow the format given in Example 12.7.2. Example 12.7.2. Assume that we have three potential predictor variables and that via backward elimination we obtain a reduced model containing only the variable X Assume that the variables X, and X, are deleted in the order mentioned. These are the steps that are taken: 1. The full model is fitted. The value of R' is found. 2. The three two-variable models are fitted. The value of R is found for each. The model with the largest R- is cho- sen and compared with the full model. In this case we test He MYlips, - Bot Bag + Byta (reduced model is adequate) Hi Mylandsx, - Bot But, + Byty + Bats (full model is needed) and are unable to reject #j- We delete the variable X, from the model, since it ap- pears that the reduced model is adequate. 3. The one-variable models MYLe, = Bot ByX3 are fitted. In this case we test Ho Priss = Bo + By*2 (reduced model is adequate) HI: Prison = But Bakz + Boxy (full model is needed) and are unable to reject , We delete the variable X, from the model, since it ap- pears to be unnecessary. 4. We now fit the model my = &, and test Hy My = Bo Hi: Mris, = But B2x2 In this case Ho is rejected, and we are left with the model that contains the one predictor variable X2
