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The three basic structural elements of a data processing system are files, flows, and processes. Files are collections of permanent records in the system, flows are data interfaces between the system and the environment, and processes are functionally defined logical manipulations of data. An investigation of the cost of developing software as related to files, flows, and processes was investigated. The following data are based on that study. Cost (in units of 1000) Files Flows Processes (v) (IJ) 22.6 44 15.0 33 15 78.1 80 80 28.0 24 21 80.5 50 24.5 18 20.5 41 13 147.6 187 137 4.2 19 15 48.2 50 21 20.5 48 17 Find a 95% confidence interval on Arx,, r, x, when x, = 12, x, = 40, and X, = 20.\fThe three basic structural elements of a data processing system are files, flows, and processes. Files are collections of permanent records in the system, flows are data interfaces between the system and the environment, and processes are functionally defined logical manipulations of data. An investigation of the cost of developing software as related to files, flows, and processes was investigated. The following data are based on that study. Cost (in units of 1000) Files Flows Processes ( v) 22.6 44 18 15.0 33 15 78.1 80 80 28.0 24 21 80.5 227 50 24.5 20 18 20.5 41 13 147.6 187 137 4.2 19 15 48.2 50 21 20.5 48 17 Find a 90% confidence interval on Ap.Consider the simple linear regression model Nyx = Bo + Bix Show that M (X'X)-I nS ATThe simple linear regression model is a polynomial model of what degree? Verify that, in this case, the normal equations given in (12.3) reduce to those given in Chap. 11 for the simple linear regression model. Equation (12.3) bantb, Exit by f + . ..+b. = IM Vi 1=1 1= 1 Exyi (12.3) i=1 1=1 1=1 12 bo +b. Exp = Exy; - 1 1=1 1=1 1= 1\fConsider the model .11.\" 1.. 1 = {30 + 315'\" + 3212. These data are available: 1': I: JJ 0 E 9 2 '9 E 4 S T {a} Find 3 3 3 3 [2'1 x\" 21 1'2: E Eula E In}; a 7: i=1 a: 3 3 3 3 to) Find the normal equations. to} Show that on = 9, is = _5, and 132 = t] are solutions to the normal equations. In Example 12.2.5, we developed a quadratic regression equation from which the unit cost of producing a drug can be predicted based on the number of units produced. Use the information given there to estimate Var Bo, Var By, and Var B2. Example 12.2.5 These data are available on X, the number of units of drug pro- duced, and I the cost per unit of producing the drug. (See Example 12.1.1.) 5 10 10 15 15 20 20 25 25 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 2.5 25 1 10 100 10 100 15 225 15 225 1 20 400 20 400 25 625 25 625 10 150 2750 X'X 150 2750 $6.250 2750 56.250 1,223,750 81.3 Xy= 1228 24.535 Apart from round-off enor 2.3 -,33 (xx)-= -33 .05342857 -.00171429 01 -,00171429 00005714286 The least-squares estimates for - Bj, By are 27.3 b = (xx) ry= -3.313 -111 The estimated model is Ary - 27.3-3313x + Illx The predicted unit cost of producing 12 units of the drug is 5 = 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 Arin. 12 = BotBixt + Batz 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 X,, 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 B1, and By. Their values are to be estimated from the data gathered.Use 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
