Homework08Solns
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Homework08Solns

Course Number: STAT stat512, Spring 2011

College/University: Purdue North Central

Word Count: 2084

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Stat 512 2 Solutions to Homework #8 Dr. Simonsen Due October 26, 2005 by 4:30pm The first three problems use the Filling Machines dataset from Problem 16.11 of KNNL described on page 725, and continue the analysis begun on Homework #7. 1. Use the Tukey multiple comparison method to determine which pairs of machines differ significantly. Summarize the results. The Tukey comparison method shows that machines 3...

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512 Stat 2 Solutions to Homework #8 Dr. Simonsen Due October 26, 2005 by 4:30pm The first three problems use the Filling Machines dataset from Problem 16.11 of KNNL described on page 725, and continue the analysis begun on Homework #7. 1. Use the Tukey multiple comparison method to determine which pairs of machines differ significantly. Summarize the results. The Tukey comparison method shows that machines 3 and 4 are significantly different from machines 1, 2, 5, and 6 (on a pairwise basis). The GLM Procedure Tukey's Studentized Range (HSD) Test for wtdev NOTE: This test controls the Type I experimentwise error rate, but it generally has a higher Type II error rate than REGWQ. Alpha 0.05 Error Degrees of Freedom 114 Error Mean Square 0.03097 Critical Value of Studentized Range 4.09949 Minimum Significant Difference 0.1613 Means with the same letter are not significantly different. Tukey Grouping Mean N machine A A A B B B B B B B 0.46000 0.36550 0.19050 0.15150 0.12500 0.07350 20 20 20 20 20 20 3 4 2 6 5 1 2. Suppose you want to compare the average of the first two machines with the average of the last four. Use the ESTIMATE and CONTRAST statements in PROC GLM to test the appropriate hypothesis. Report the estimated value of this contrast with its standard error; state the null and alternative hypotheses, the test statistic with degrees of freedom, the Pvalue and your conclusion. The contrast is L = 2 4 ^ ^ The estimated value of this contrast is L = -0.1435 with standard error s L = 0.03408 . We test ( 1 + 2 ) - ( 3 + 4 + 5 + 6 ) . {} the null hypothesis ( 1 + 2 ) - ( 3 + 4 + 5 + 6 ) = 0 vs H : H0: a 2 4 [These could also be correctly written as H0: L = 0 vs. Ha: L 0 or ( 1 + 2 ) - ( 3 + 4 + 5 + 6 ) 0 . 2 4 H0: ( 1 + 2 ) = ( 3 + 4 + 5 + 6 ) 2 4 vs. Ha: ( 1 + 2 ) ( 3 + 4 + 5 + 6 ) ]. 2 4 The test statistic is either F = 17.73 with 1,114 df or t = 4.21 with 114 df, and the P-value is P < 0.0001. We reject H0 and conclude that the mean for the first two machines is not the same as the mean for the last four machines. The GLM Procedure Dependent Variable: wtdev Contrast prob2 Parameter prob2 DF 1 Contrast SS 0.54912667 Mean Square F Value Pr > F 0.54912667 17.73 <.0001 Standard Error t Value Pr > |t| 0.03407905 -4.21 <.0001 Estimate -0.14350000 3. Check assumptions using the residuals. Turn in the plots/output you used to check the assumptions and state your conclusions. r esi d 0. 4 0. 3 0. 2 0. 1 0. 0 - 0. 1 - 0. 2 - 0. 3 - 0. 4 1 2 3 m achi ne 4 5 6 The assumptions of normality and constant variance appear to be satisfied, as shown by the residual plot and qqplot. [Grader: the residual plot may be plotted vs. machine or vs. predicted value. Only one is necessary, plus the qqplot.] The remaining problems use the Helicopter Service dataset from Problem 18.15 on page 804 of KNNL. Helicopter service. An operations analyst in a sheriff's department studied how frequently their emergency helicopter was used during a recent 20-day period, by time of day (shift 1: 2 am 8 am; shift 2: 8 am 2 pm; shift 3: 2 pm 8 pm; shift 4: 8 pm 2 am). The data follow (in time order). Since the data are counts, the analyst was concerned about the normality and equal variances assumptions of ANOVA model (16.2). 4. Here we will examine the equal variances assumption. (a) Fit a one-way ANOVA model and give the (four) predicted values.. The fitted values are the means for the four levels as shown below. The GLM Procedure Level of shift 1 2 3 4 -------------use------------Mean Std Dev 3.90000000 1.15000000 2.00000000 3.40000000 1.97084006 1.08942283 1.45095250 1.78885438 N 20 20 20 20 (b) Prepare suitable residual plots to study whether or not the error variances are equal for the four shifts. What do you conclude? r esi d 4 r esi d 4 3 3 2 2 1 1 0 0 -1 -1 -2 -2 -3 -3 -4 1 2 pr edi ct 3 4 -4 1 2 shi f t 3 4 The residual plot shows that the variances are not constant. The plot of residuals vs. predictors shows that the variances tend increase with the predicted value. [Grader: only one of these plots is required] (c) Use the modified Levene's test whether the error variances are equal. Report the Pvalue and a conclusion. Are your results consistent with your conclusion from (b)? The P-value of the modified Levene test is 0.1344, so we do not reject the null hypothesis of equal variances; however note that the P-value is fairly small This failure to reject does not support the conclusion of unequal variances reached in part b. The GLM Procedure Levene's Test for Homogeneity of use Variance ANOVA of Absolute Deviations from Group Means Sum of Squares 4.8394 64.0395 Mean Square 1.6131 0.8426 Source shift Error DF 3 76 F Value 1.91 Pr > F 0.1344 (d) Section 18.5 of KNNL gives a simple guide to selecting a suitable transformation for constant variance problems. Prepare a table similar to the one on page 791 where si2 si si , , are tabulated for each level of the factor. Choose an appropriate Yi Yi Yi2 transformation (square root, log, reciprocal) of Y depending on which of these is the most stable (i.e. approximately constant across levels). shift Yi i level 1 2 3 4 The ratio 3.90 1.15 2.00 3.40 si 1.971 1.089 1.451 1.789 si2 Yi i 0.996 1.032 1.053 0.941 si Yi i si Yi i2 0.130 0.824 0.363 0.154 0.505 0.947 0.725 0.526 si2 appears the most consistent across levels. This indicates that a square root Yi i transformation may be appropriate. m e a n u s e s q 15.2100 1.3225 4.0000 11.5600 l o g m e a n u s e 1.36098 0.13976 0.69315 1.22378 l o g s t d u s e 0.67846 0.08565 0.37222 0.58158 r o o t r a t i o 0.99595 1.03204 1.05263 0.94118 l o g r a t i o 0.50534 0.94732 0.72548 0.52613 r e c i p r a t i o 0.12958 0.82376 0.36274 0.15475 O b s 1 2 3 4 s h i f t 1 2 3 4 _ T Y P E _ 0 0 0 0 _ F R E Q _ 20 20 20 20 m e a n u s e 3.90 1.15 2.00 3.40 v a r u s e 3.88421 1.18684 2.10526 3.20000 s t d u s e 1.97084 1.08942 1.45095 1.78885 5. A rather simple approximation of the Box-Cox procedure is the following: (i) Compute the mean and standard deviation for each treatment factor level. (ii) Take the log of both the mean and standard deviation. (iii) Fit the regression model log ( i ) = 0 + 1 log ( i ) + using the observed means and standard deviations as the data for i and i respectively (there are 4 "observations" in this dataset). ^ (iv) Set = 1 - b1 where b1 is the estimate for 1 obtained in (iii). Use the Helicopter service data to perform this approximation. What value of appears reasonable according to this method? ^ The regression model gives the estimate b1 = 0.47, and so = 1 - 0.47 = 0.53 0.5 . This also suggests a square root transformation. The REG Procedure Model: MODEL1 Dependent Variable: logstduse Parameter Estimates Parameter Standard Estimate Error 0.02766 0.02261 0.47029 0.02305 Variable Intercept logmeanuse DF 1 1 t Value 1.22 20.40 Pr > |t| 0.3458 0.0024 6. Define a new response variable by adding 1 to the original response. (This will avoid 0's which mess up the log and reciprocal transformations.) Then use SAS's Box-Cox procedure to determine an appropriate transformation. Proc transreg can be used to perform ANOVA if we tell it shift is a class variable, as in the following: proc transreg data=helicopter; model boxcox(usesplus1) class(shift); The = TRANSREG Procedure Transformation Information for BoxCox(usep1) Lambda -3.00 -2.75 -2.50 -2.25 -2.00 -1.75 -1.50 -1.25 -1.00 -0.75 -0.50 -0.25 0.00 0.25 0.50 + 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00 R-Square 0.14 0.15 0.15 0.16 0.17 0.18 0.20 0.21 0.23 0.24 0.26 0.28 0.30 0.31 0.32 0.32 0.33 0.33 0.32 0.32 0.31 0.30 0.29 0.28 0.26 Log Like -182.106 -165.922 -150.178 -134.942 -120.291 -106.320 -93.144 -80.897 -69.735 -59.830 -51.364 -44.517 -39.442 -36.250 * -34.991 < -35.645 * -38.129 -42.313 -48.035 -55.124 -63.413 -72.748 -82.989 -94.017 -105.728 < - Best Lambda * - Confidence Interval + - Convenient The Box-Cox procedure suggests that = 0.5, i.e. a square root transformation, would be appropriate. 7. NKNW 18.16 (Omit the coefficient of correlation in part b). In part (c), Brown-Forsythe is the same as Modified Levene).): Refer to Helicopter service Problem 18.15. The analyst decided to apply the square root transformation Y = Y and examine its effectiveness. (a) Obtain the transformed response data, fit ANOVA model (16.2), and obtain the residuals. The GLM Procedure Dependent Variable: rootuse Source Model Error Corrected Total R-Square 0.288945 Sum of DF Squares Mean Square F Value 3 13.60854916 4.53618305 10.29 76 33.48888826 0.44064327 79 47.09743742 Coeff Var Root MSE rootuse Mean 46.66182 0.663810 1.422597 The GLM Procedure Level of shift 1 2 3 4 -----------rootuse----------Mean Std Dev 1.87136414 0.84265033 1.22925287 1.74712038 0.64725840 0.68051097 0.71740569 0.60486655 Pr > F <.0001 N 20 20 20 20 (b) Prepare suitable plots of the residuals to study the equality of the error variances of the transformed response variable for the four shifts. Also obtain a normal probability plot and the coefficient of correlation between the ordered residuals and their expected values under normality. What are your findings? Does the transformation appear to have been effective? r esi d 1 r esi d 1 0 0 -1 -1 -2 0. 8 0. 9 1. 0 1. 1 1. 2 1. 3 1. 4 1. 5 1. 6 1. 7 1. 8 1. 9 -2 1 2 shi f t 3 4 pr edi ct There has been an improvement in the variances of the transformed variable, but there are still signs that the variance increases with the predicted value. [Grader: only one of the two above plots is necessary, either vs. shift or vs. predictor] re d si 1 0 -1 -2 -3 -2 -1 Rn fo a k r 0 1 2 3 V ri a l e re d a b si The QQ-plot shows some skewness away from normality. The transformation stabilized the variance somewhat, however, it also appears to have destroyed the normality of the residuals. (c) Test by means of the modified Levene test whether or not the treatment error variances for the transformed response variable are equal; use = 0.10. State the alternatives, decision rule, and conclusion. Are your findings in part (b) consistent with your conclusion here? The GLM Procedure Levene's Test for Homogeneity of rootuse Variance ANOVA of Absolute Deviations from Group Means Sum of Squares 0.3652 11.3771 Mean Square 0.1217 0.1497 Source shift Error DF 3 76 F Value 0.81 Pr > F 0.4905 The Levene test is now very far from significant. The P-value is 0.49, which is a big increase over the 0.13 for the untransformed data. This gives some evidence that the errors are more homogeneous after the transformation. (However, the Levene test is sensitive to the normality assumption, and since we know normality is violated, we cannot really trust the results.) 8. Use the Tukey multiple comparison method for differences in means on both the untransformed and square-root transformed Helicopter service data (from Problem 7) to determine which shifts differ significantly. Summarize and compare the results. Using the transformed data we find that the following pairs are significantly different: 1 with 2 and 3 2 with 1 and 4 Tukey's Studentized Range (HSD) Test for rootuse Alpha 0.05 Error Degrees of Freedom 76 Error Mean Square 0.440643 Critical Value of Studentized Range 3.71485 Minimum Significant Difference Tukey Grouping Mean N A 1.8714 20 A B A 1.7471 20 B B C 1.2293 20 C C 0.8427 20 0.5514 shift 1 4 3 2 Using the untransformed data we find that shifts 1 and 4 differ significantly from shifts 2 and 3. Tukey's Studentized Range (HSD) Test for use Alpha 0.05 Error Degrees of Freedom 76 Error Mean Square 2.594079 Critical Value of Studentized Range 3.71485 Minimum Significant Difference 1.3379 Tukey Grouping Mean N shift A A A B B B 3.9000 3.4000 2.0000 1.1500 20 20 20 20 1 4 3 2 It is interesting that in the transformed data 3 and 4 are not significantly different but in the untransformed data they are. SAS Code *Homework 8 * Problems 1-3; title1 'detergent'; data detergent; infile 'H:\Stat512\Datasets\Ch16pr14.dat'; input wtdev machine j; proc glm data = detergent; class machine; model wtdev = machine; output out=det2 r=resid; means machine / tukey; estimate 'prob2' machine 0.5 0.5 -0.25 -0.25 -0.25 -0.25; contrast 'prob2' machine 0.5 0.5 -0.25 -0.25 -0.25 -0.25; symbol1 v=circle; proc gplot data=det2; plot wtdev*machine; plot resid*machine/vref=0; proc univariate data=det2 noprint; var resid; qqplot /normal(mu=est sigma=est L=1); run; *Problems 4 - 5; * Read in the data; title1 'helicopter'; data helicopter; infile 'h:\stat512\datasets\ch18pr15.dat'; input use shift; proc print data=helicopter; * Run ANOVA, store the residuals and predictors, do the modified Levene test; proc glm data=helicopter; class shift; model use=shift; output out=diags p=predict r = resid; means shift / hovtest=levene(type=abs); * Plot the residuals vs. factor or vs. predictor; symbol1 v = circle; proc gplot data=diags; plot resid*shift/vref=0; plot resid*predict/vref=0; proc univariate data=diags normal; var resid; qqplot / normal (mu=est sigma=est); * Store the means, variances, and standard deviations for each shift; proc means data=helicopter; by shift; output out=heli2 mean = meanuse var = varuse std = stduse; * Store variations on the basic variables; data heli2; set heli2; meanusesq = meanuse*meanuse; logmeanuse = log(meanuse); logstduse = log(stduse); rootratio = varuse/meanuse; logratio = stduse/meanuse; recipratio = stduse/meanusesq; proc print data=heli2; * Run regression on log(sigma) vs. log(mu) to get b1 = 1 - lambda; proc reg data=heli2; model logstduse = logmeanuse; * Problem 6; data helicopter; set helicopter; usep1 = use + 1; proc print data=helicopter; run; proc transreg data=helicopter; model boxcox(usep1) =class(shift); run; *Problem 7 (18.16); * Use the square root transformation on the data; title1 'helitrans'; data helitrans; set helicopter; rootuse = sqrt(use); * Run ANOVA, store the residuals and predictors, do the modified Levene test; proc glm data=helitrans; class shift; model rootuse = shift; output out=heli3 r = resid p = predict; means shift / hovtest = levene(type = abs); * Plot residuals vs. factor level or vs. predictor; symbol1 i = none; proc gplot data=heli3; plot resid*(shift predict)/vref=0; * Check normality with a qqplot; proc univariate data=heli3 normal; var resid; qqplot / normal (mu = est sigma = est L = 1); run; * Use Tukey to find which means differ significantly; proc glm data=helitrans; class shift; model rootuse = shift; means shift / tukey; run; proc glm data=helicopter; class shift; model use = shift; means shift / tukey; run; quit;

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Purdue - MA - 373
Purdue - MA - 373
Purdue - MA - 373
Purdue - MA - 373
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Purdue - MA - 373
Purdue - MA - 373
Purdue - MA - 373
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Purdue - MA - 373
Purdue - MA - 373
Purdue - MA - 373
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Purdue - MA - 373
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Purdue - MA - 373
Purdue - MA - 373
Purdue - MA - 373
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Purdue - MA - 373
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Purdue - MA - 373
Purdue - MA - 373
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