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Unformatted text preview: Stat 332 R.J. MacKay, University of Waterloo, 2005 Chapter 3 Solutions1 Chapter 3 Exercise Solutions 1. Practice with the F distribution. a) Suppose U F ~ , 6 24 . Find c so that Pr( ) . U c = 0 05. From the tables, we have c = 2 51 . b) Estimate Pr( ) U 3 . From the tables, interpolating we have Pr( ) . U 3 0 03 c) What is the distribution of 1/ U ? Since U K K = 6 2 24 2 , we have 1 24 6 / ~ , U F d) Find d so that Pr( ) . U d = 0 05. Using the result from c), Pr( ) Pr( / / ) U d U d = 1 1 and from the tables we get 1 385 / . d so d 0 26 . e) Show that if W t k ~ then W 2 has an F distribution. What are the degrees of freedom? W G K K K F k k k 2 2 2 1 2 2 1 0 1 ~ ( , ) , = = . Recall that K Z Z k k = + + 1 2 2 ... where Z G i ~ ( , ) 0 1 . 2. In a small investigation, three treatments A,B,C were compared by assigning 6 units at random to each treatment. The data are shown below. A B C 11.82 15.46 15.43 13.03 12.90 14.28 10.78 14.88 14.76 14.31 13.75 12.07 14.21 18.59 13.80 8.56 13.80 13.46 average 12.12 14.90 13.97 st. dev. 2.22 2.02 1.16 a) Calculate the ANOVA table. The treatment sum of squares 6 2 ( ) y y i i + ++ is 12 times the square of the sample standard deviation of the treatment averages. Hence the treatment sum of squares is 24.00. The total sum of squares is 17 times the square of the sample standard deviation of the 18 response variate values or 75.77. We calculate the residual sum of squares by subtraction. The completed table is shown below. b) Is there any evidence of a difference among the treatments? To test the hypothesis 1 2 3 = = = , the discrepancy is 3.48 and the pvalue is Pr( . ) . , F 2 15 348 0 057 = so there is weak evidence of a difference among the treatments. Stat 332 R.J. MacKay, University of Waterloo, 2005 Chapter 3 Solutions2 Source Sum of squares Degrees of freedom Mean square Ratio to residual ms Treatments 24.00 2 12.00 3.48 Residual 51.76 15 3.45 Total 75.765 17 c) Treatment A was a control. Is there any evidence that the average effect of treatments B and C exceeds the effect of the control? [Use a onesided test here  why?] From the ANOVA table, we have $ . = 186 . Consider the contrast = + 2 3 1 2 to compare the average of treatments 2 and 3 to the control. We want a onesided test to see if there is evidence that &gt; 0 . The estimate of is $ . = + = + + + y y y 2 2 1 2 2 315 and the corresponding estimator has standard deviation 1 24 1 24 1 6 2 + + = / . To test the hypothesis, suppose that = 0. The (onesided) discrepancy is d = = $ $ / .49 2 2 and the pvalue is Pr( .49) . t 15 2 0 013 = so there is strong evidence against the hypothesis. There is strong evidence that the average treatment effect exceeds the effect of the control....
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This note was uploaded on 06/27/2009 for the course STAT 332 taught by Professor Xu(sunny)wang during the Spring '09 term at Waterloo.
 Spring '09
 Xu(Sunny)Wang

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