Week_7_Practice_Problems_Solutions

Week_7_Practice_Problems_Solutions - Week 7 Practice...

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Unformatted text preview: Week 7 Practice Problems 1. In the production of a certain chemical, the yield (y) of the process (measured as %) is approximately a linear function of the amount of catalyst (x) used. In a laboratory experiment, 10 levels of catalyst were used with two replicates at each level. The data are Catalyst (x) Yield (y) Catalyst (x) Yield (y) 0.3 46.9 0.8 53.9 0.3 50.6 0.8 52.8 0.4 50 0.9 55.3 0.4 48.6 0.9 53.8 0.5 50.1 1 58.9 0.5 50.1 1 60.9 0.6 52.2 1.2 58.7 0.6 50 1.2 61.8 0.7 52.4 1.4 63.4 0.7 57.9 1.4 62.1 To save you calculation time, we provide the following summaries: 2 2 0.78, 54.42, ( ) 2.238, ( ) 30.458, 65.220 i i i i x y x x x x y r = = = = = catalyst yield 1.4 1.2 1.0 0.8 0.6 0.4 0.2 65 60 55 50 45 Yield vs Catalyst Level Consider the Gaussian response model to describe the experiment. ( ) , ~ (0, ), 1,...,20independent i i i i Y x x R R G i = + + = a) What do the parameters and represent? represents the average yield (in the study population) when the catalyst level is 0.78 x = . represents the increase in the average yield (in the study population) when the catalyst is increased by 1 unit. b) Find the MLEs of and (you can do this from scratch for practice if you wish otherwise use the formulas given in class). Add the fitted line to the scatterplot. We have 54.42 y = = and 2 ( ) 13.61 ( ) i i i x x y x x = = c) Find the estimate of . What are the degrees of freedom in this case? 2 1.806 20 2 i r = = with 18 degrees of freedom. d) Find a 95% confidence interval for . How would you interpret this interval formally? We have ~ ( , / ) G S x x where 2 ( ) 2.238 xx i S x x = = and 18 ~ / xx t S . From the tables we have . Hence we have 18 ( 2.10 2.10) 0.95 P t = 0.95 ( 2.10 2.10) ( 2.10 / 2.10 / ) / x x x xx P P S S = = + x S . Replacing the estimators by the corresponding estimates, we have the 95% confidence interval is (10.94, 16.28). Based on the above probability statement, if we repeated the study over and over, then 95% of the time, the interval we construct with this procedure will contain the unknown parameter . e) Find a 90% confidence interval for the expected yield if the catalyst level is 1.5. How would you interpret this interval informally? We want a confidence interval for (1.5 0.78) 0.72 = + = + . The estimate is and the corresponding estimator is 0.78 67.62 = + = 0.78 = + . To find the standard error, we need to find the standard deviation of . We have 2 2 ( ) [ ] [ 0.72 ] [ ] 20 i i i i i i xx Y x x Y Var Var Var cY c S = + = = where 2 2 2 1 1 0.72 ( 0.72 ) 0.281 20 20 i...
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This note was uploaded on 03/20/2011 for the course STATISTICS 231 taught by Professor Mckay during the Spring '10 term at Waterloo.

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Week_7_Practice_Problems_Solutions - Week 7 Practice...

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