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Solution; Given that, F = 138.5 1 0 - 43.7 f n=33 cool. confidence lovel the zis *=1-90%. = 1-0.90 = 0.10 9). At 2012 = 20.05 -1-645 10% CI for H = 3.1 km = 138.5 16.45 (1991) 9861, 151.0138) ( 2 ) ? is b) At i Lovey limit 15.1 upper limit - 151.0 Marryin of err = 19.5 quol. confidence level the x=1-0.95 - 0.05 *%=0.025 7042 = 20.035 = 196 = 7 1 2212500 188.5 1.96 louxrlinit.

123.58 upp linit = 153.4 margin of error = 14.9

©. At 99.1. confidence level the z is L- 1-0.99 = = 0.01 x2 = 0.005 ; 2K12 = 70.005 = 2.576 4 99% confidence interval Porite is - 13 Zapalov) = 138.5 1 2.576 ( ) Lower linet - 118.9 upper Duit = 158.0 margin of enor = 19.5

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