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Unformatted text preview: Practice Final Exam December 14, 2009 1 New Material 1.1 ANOVA 1. A puri cation process for a chemical involves passing it, in solution, through a resin on which impurities are adsorbed. A chemical engineer is testing the e ciency of 3 di erent resins in collecting impurities; he breaks each resin into 5 pieces and measures the concentration of impurities after passing through the resins. The data are as follows: Concentration of impurities Resin 1 Resin 2 Resin 3 .046 .038 .031 .025 .035 .042 .014 .031 .020 .017 .022 .018 .043 .012 .039 Test the hypothesis that there is no di erence in the e ciency of the resins, using analysis of variance tech niques. Solution We want to test the hypothesis H : μ 1 = μ 2 = μ 3 . The analysis of variance table is Source d.f. SOS Mean Squares Fstatistic Treatments 2 SSTr = 1 . 4533 · 10 5 MSTr = 7 . 27 · 10 6 F = 0 . 0487 Error 12 SSE = 0 . 0018 MSE = 1 . 49 · 10 4 Total 14 SST = 0 . 0018 pvalue: . 953 Since our pvalue is . 953 , we can accept the null hypothesis. 2. Four standard chemical procedures are used to determine the magnesium content in a certain chemical com pound. Each procedure is used four times on a given compound with the following data resulting: Magnesium content Method 1 Method 2 Method 3 Method 4 76.42 80.41 74.20 86.20 78.62 82.26 72.68 86.04 80.40 81.15 78.84 84.36 78.20 79.20 80.32 80.68 Do the data indicate that the procedures yield equivalent results? Solution We want to test the hypothesis H : μ 1 = μ 2 = μ 3 = μ 4 . The analysis of variance table is Source d.f. SOS Mean Squares Fstatistic Treatments 3 SSTr = 135 . 7625 MSTr = 45 . 254 F = 7 . 474 Error 12 SSE = 72 . 66 MSE = 6 . 055 Total 15 208.4230 pvalue: . 0044 Since our pvalue is . 0044 , we can safely reject the null hypothesis. 1 3. For data x ij , i = 1 ,...,m , j = 1 ,...,m , show that ¯ x ·· = m X i =1 ¯ x i · /m = n X j =1 ¯ x · j /n where ¯ x ·· = 1 mn ∑ m i =1 ∑ n j =1 x ij is the sample mean of all x ij . Solution We can write ¯ x ·· = 1 mn m X i =1 n X j =1 x ij = 1 m m X i =1 1 n n X j =1 x ij  {z } =¯ x i · = m X i =1 ¯ x i · /m Also, ¯ x ·· = 1 mn m X i =1 n X j =1 x ij = 1 mn n X j =1 m X i =1 x ij = 1 n n X j =1 1 m m X i =1 x ij !  {z } =¯ x · j = n X j =1 ¯ x · j /n 4. Problem 11.1.21 in the text. Solution The rst con dence interval, for example, is μ 1 μ 2 ∈ ¯ x 1 · ¯ x 2 · ± sq α,k,ν √ 2 r 1 n 1 + 1 n 2 = 46 . 09 42 . 21 ± 4 . 33 · q . 05 , 5 , 40 √ 2 r 1 10 + 1 9 ! = ( 1 . 803 , 9 . 563) This establishes that the hypothesis μ 1 = μ 2 is plausible at the . 05 signi cance level, since the interval contains . Carrying out this procedure for all pairs, the con dence intervals that contain are μ 1 μ 2 , μ 2 μ 5 , and μ 3 μ 4 . The largest mean is μ 3 or μ 4 and the smallest mean is either μ 2 or μ 5 , which can be veri ed by looking at the values of ¯ x 1 · through ¯ x 5 · 1.2 Regression 1. The following table shows the number of units of a good that customers ordered, when the good was priced1....
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 Spring '08
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 Normal Distribution, Variance, Null hypothesis, Probability theory, probability density function

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