Question

Your task s to mid a prediction interval lor the y-value when x=2019. show 4 deerma Iditarod Modeling to Prediction Interval for single Y- Find x Win Years Day Hour Min Sec 2018 9 12 0 0 228.0 2017 8 3 40 13 195.6703 2016 8 11 20 16 203.3378 2015 8 18 13 6 210.2183 2014 8 13 4 19 205.0719 2013 9 7 39 56 223.6656 2012 9 4 29 26 220.4906 2011 8 18 46 39 210.7775 2010 8 23 59 9 215.9858 2009 9 21 38 46 237.6461 2008 9 11 46 48 227.78 2007 9 5 8 41 221.1447 2006 9 11 11 36 227.1933 2005 9 18 39 31 234.6586 Use the data from 2005 to 2018 to find a linear model to predict winning time in hours for 2018. r2- Show prediction for x*-2019. (all steps) Now find a 95% prediction interval and interpret. Show all values substituted into formula. 1 (x*-x)2 a +bx*+(t critical value)s , 1++

Answer 1

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Ans. Years(X) Win hours (X-mean)A2 (Y-meanY)2 (X-meanX)(Y-meanY)YA 2005 234.6586 42.25 255.0407 103.8050 230.797 14.9 Scatter Plot 2006 227.1933 72.3298 30.25 -46.7758 228.936 3.0 250.0000 20.25 2007 221.1447 6.0324 11.0524 227.075 35.2 2008 227.7800 12.25 82.6534 31.8199 225.214 6.6 200.0000 2009 237.6461 6.25 359.3865 47.3937 223.354 204.3 2010 215.9858 2.25 7.3052 4.0542 221.493 30.3 150.0000 2011 210.7775 0.25 62.5856 3.9556 219.632 78.4 2012 220.4906 0.25 3.2472 0.9010 217.772 -1.8607x+ 100.0000 2013 223.6656 2.25 24.7705 7,4655 215.911 60.1 R 0.3973 6.25 185.4147 -34.0418 214.05 80.6 2014 205.0719 50.0000 2015 210.2183 71.7461 12.25 29.6461 212.19 20.25 235.6473 2016 203.3378 -69.0786 210.329 48.9 0.0000 2017 195.6703 30.25 529.8425 126.6007 208.468 163.8 2004 2006 2008 2010 2012 2014 2016 2018 2020 457.6 42.25 86.7020 60.5241 206.607 2018 228.0000 2011.5 218.6886 227.50 1982.7038 -423.3137 1195.0 Sxx 227.5 1982.7 423.314 slope(b)1.86072 intercept coeff(alz 3961.52556 3961.525-1.861X value of y when X 2019 204.166 39.73% by using formula (R:Sxy42/(Sxx®Syy)) ritical value (t)2.17881283 SS(residuell- 1195.03594 9.97929 Se- 95% prediction confidence interval is give! as lower limit- 179.197658 Upper limit- 229.134342