4315_Text_Problem_for_Midterm_(F09)

# 4315_Text_Problem_for_Midterm_(F09) - 4 Example 10.2 -—...

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Unformatted text preview: 4 Example 10.2 -— Symptomatic Recovery in Pediatric Dehydration A study was conducted to determine the degree of recovery that takes place 90 minutes following treatment of 36 children diagnosed at the clinic with moderate to severe dehydration. Upon diagnosis and study entry, patients were treated with an electrolytic solution at various dose levels (0, 0.5, 1.0, 1.5, 2.0, 2.5, 3.0 mEq/l) in a frozen, flavored, ice popsicle. The degree of rehydration was determined by using a subjective scale based on physical examination and parental input. These scores were converted to a 0 to I 00 point scale, representing the percent of recovery. The child’s age and weight were also recorded. Is recovery related to electrolyte dose based on the data shown in Table 10.2? _____________________.——————-—-—-—- Solution The regression procedure (PROC REG) in SAS is used to perform a multiple linear regression analysis, regressing the percent recovery at 90 minutes (y) on the concomitant variables Dose Level (X1), Age (X2), and Weight (X3). Although, PROC GLM can also be used for this analysis, PROC REG has diagnostic and model building advantages for regression analysis when there are no class variables. in addition, multiple MODEL statements can be used with PROC REG. In this example, the model is written y=l30 +l31X1 + [52X2 + Bsxs +8 where X] = dose (mEq/l), x2 = patient’s age (years), and X3 = patient’s weight (lbs) are the independent or ‘explanatory’ variables. The parameters are Bo, ‘which represents the overall average response, and Bi, which represents the average ‘ increase in the degree of recovery (y) for a one-unit increase in xi (i = 1, 2, 3). The ‘ objective is to investigate the effect of Dose Level on response and how age and weight might inﬂuence that effect. TABLE 10.2. Raw Data for Example 10.2 Degree of Patient Recovery (%) Dose (mEqu) Age (yrs) Weight (lbs) Number (Y) (X1) (X2) (X3) 1 77 0.0 r 4 28 2 65 1.5 5 35 3 75 I 2.5. 8 55 4 63 1.0 9 76 5 . 75 0.5 5 V 31 6 82 2.0 5 27 7 70 1.0 6 35 8 90 2.5 6 47 9 V ‘ 49 0.0 9 59 10 72 3.0 8 50 11 67 2.0 7 5O 12 100 ‘ 2.5 '“ 7 46 13 75 1.5 4 33 14 58 3.0 8 59 15. 58 1.5 6 2' 40, 16 ' ‘ 55‘ a V 0.0: 8 58; 17. 80 . 1.0‘ 7 55' 18 ‘ 55 2.0 ' 10 76 1 19 44 0.5 9 66 20 62 1.0 6 43 21 60 1.0 6 48 22 75 2.5 7 50 23 77 1.5 , 5 29 24 80 2.5 11 ' 64 25 68 3.0 9 61 26 71 2.5 10 71 27 90 I 1.5 4 26 28 - 80 a I , 2.0. 3 27} 29 70 ' 0.0 . , 9 5'6, 30 ' 58 2.5 8 57 31 88 1.0 3 22 32 68 0.5 5 37 33 60 0.5 6 44 34 90 3.0 5 45 35. v. 79 1.5 8 53 36 90‘ 2.0 4 29 OUTPUT 10.2. SAS Output for Example 10.2 (continued) Multiple Linear Regression Example 10.2: Recovery in Pediatric Dehydration The REG Procedure Correlation Variable AGE DOSE 0.1625 1.0000 WT 0.9367 Y —0.4652 The REG Procedure Model: MODELl Dependent Variable: Y Analysis of Variance ‘ Sum of Mean Source Squares Square Model 2667.66870 Error 3151.22019 Corrected Total 5818.88889 Root MSE 9.92349 R—Square Dependent Mean 71.55556 Adj R—Sq Coeff Var 13.86823 Parameter Estimates Parameter Standard Variable DF Estimate Intercept 85.47636 5.96528 DOSE 6.16969 1.79081 AGE 0.27695 2.28474 WT ~0.54278 0.32362 Parameter Estimates Variance Variable Inflation Intercept DOSE AGE WT 0 1.02833 8.16330 8.16745 Error t Value Pr>[tl Type I SS 184327 752.15170 638.51013 .77.00687 ...
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## This note was uploaded on 11/28/2009 for the course STAT 4315 taught by Professor Heuter during the Spring '09 term at Columbia.

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4315_Text_Problem_for_Midterm_(F09) - 4 Example 10.2 -—...

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