COMM 291 ASSIGNMENT 6

# E 95 confidence interval for mean when x 20 970133

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Unformatted text preview: calculation and Excel is just due to round-off error. e) 95% Confidence Interval for Mean when x = 20 ˆ µ = 9701.33 + 1213.45(20) = 33970.33 se2 = 7705.6262 = 59,376,672.05 SE(b1) = 338.10 x-bar = 17 ˆ SE( µ ) = sqrt[(338.102)(20-17)2+7705.6262/25] = 1844.96 t*(23 df) = 2.069 95% CI: 33,970.33 ± 2.069(1844.96) = 33,970.33 ± 3817.22 or (\$30,153.11 ,\$37,787.55) f) 90% Prediction Interval when x = 20. ˆ y = 9701.33 + 1213.45(20) = 33970.33 se2 = 7705.6262 = 59,376,672.05 SE(b1) = 338.10 x-bar = 17 ˆ SE( y ) = sqrt[(338.102)(20-17)2+7705.6262/25+ 7705.6262] = 7923.42 t*(23 df) = 1.714 (remember that 90% was asked for) 90% PI: 33,970.33 ± 1.714(7923.42) = 33,970.33 ± 13,580.74 or (\$20,389.59,\$47,551.07) Upper Bound: \$47,551 Note: Round-off errors will occur in e) and f) if the Excel values of t*(23) are used. Extra: Residual plot and Normal Probability plot confirm that the assumptions are reasonable. Question 2: Modeling PGA Golfers' Success a) Here is the output from fitting a regression model t...
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## This note was uploaded on 01/16/2014 for the course COMM 291 taught by Professor E.fowler during the Fall '10 term at UBC.

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