# 12 - STAT 350 Fall 2008 Homework #12 - Solution covers...

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STAT 350 – Fall 2008 Homework #12 - Solution covers through Lecture #19 1. A biologist is studying a population of snarks. She traps 25 snarks (consider this a random sample from the population) and measures the length (in cm) and weight (in kg) of each snark. Let x i denote the length of the i th snark and let y i denote the weight of the i th snark. She finds 25 1 i i x = = 1635, 25 1 i i y = = 3356, 25 2 1 i i x = = 117325, 25 2 1 i i y = = 468176, and ( 25 1 ii i ) x y = = 230003. The length of the snarks ranged from 12.7 to 93.8 cm and their weight ranged from 74.0 to 181.4 kg. Warning: do not round-off too much in intermediate calculations! a. Calculate SS xy . () 1635 3356 230003 25 = 10520.60 b. Calculate SS xx . 2 1635 117325 25 = 10396.00 c. Calculate SS yy . 2 3356 468176 25 = 17666.56 e. Calculate the sample standard deviation of snark weight. 17666.56 27.13128575 25 1 y s == f. Give a 95% CI for the true mean weight of snarks from this population. 3356/ 25 134.24 y 95% CI: N from Table IV 27.13128575 134.24 2.064 25 ± ⎝⎠ (123.04, 145.44) g. The biologist set out a trap to catch another snark. Give a 95% prediction interval for the weight of the snark she will catch. 95% PI: N from Table IV 1 134.24 2.064 27.13128575 1 25 ±+ (77.13, 191.35) h. Find the Pearson Correlation between the length and weight of these snarks. 10520.60 10396.00 17666.56 r = = 0.7763 i. Find the least squares regression equation to predict the weight of a snark from the length of a snark. (give a and b to at least 5 decimal places) b = 10520.60/10396.00 = 1.01199 a = (3356/25) – (1.01199)(1635/25) = 68.05616 = 68.05616 + 1.01199 x ˆ y Homework #12 - Solution Page 1 of 6

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j. Create the ANOVA table for the regression analysis (use Excel to get the p -value). Total Sum of Squares (SSTo) = SS yy = 17666.56 Residual Sum of Squares (SSResid) = SSTo – b SS xy = 17666.56 – (1.01199)(10520.60) = 7019.818006 Regression Sum of Squares (SSReg) = SSTo – SSResid = 17666.56 – 7019.818006 = 10646.74199 Source df Sum of Squares Mean Square F p Regression 1 10646.74199 10646.74199 34.8834 5.085 × 10 -6 Error 23 7019.8180006 305.2094785 Total 24 17666.56 k.
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## This note was uploaded on 02/23/2009 for the course STAT 350 taught by Professor Staff during the Fall '08 term at Purdue University.

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12 - STAT 350 Fall 2008 Homework #12 - Solution covers...

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