Question

Evaluation: Estimate the Difference of the Means of Two Normal Random Variables Each numerical entry must be accurate to the nearest 0.001 . Given a sample of size 200 of the normal random variable Z_5.1,3.8 , and a sample of size 250 of the normal random variable 26,5.8 , let X1 and X, denote the averages of the two samples. a. The difference of the averages Xi - X2 is a normal random variable with mean = and standard deviation or b. The symmetric about the mean 95% confidence interval for Xi - X2 has lower bound L= and upper bound U= If our samples yield averages -4.7868 for X and 5.5940 for X2 the difference (pick one) in the interval.

Answer 1

Samplo I 2~-Sl,3.8) maan = -S Inisa = 200 vur Sampu 2 NN6,5.81 meun = 6. vur S-8 Page i o samplo z Know N muun do Seum plel- mean sumple 2 - menn of damplez) = -5.1-6-0 = -11.1 mean of S1 varl - X2) 3.8 5.8 = 0.o422 Lo ant to Ainu confidone interva for - we we caun fino His ho m t- test Aur meun for eluferent Sumple Por avfferent 6umpo mean t-test INULL HYpothis ) (AIkinahve Hy.) HL= ( S59uo) i4 we almeady ljed te nall Hypokosis -) -(4- Me) ndey H, + (ne-1)0;2 nitne-2

(ages (200-1) *3.8+(2 50-1) *5 200 +2 50-2 M-L)=D-10-3808 2200.4 448 Emithe- (a) = 1.965 73 tuur,o.05 2.216215 %3D ह) Jo.009 = 0.og49 t. 1. 2.216 *0.09ug %31 O.21032 -L) = %3D (1.965273) *10-21082) 7 10-3R08 O'4133362+ 10.3808 (-3.964463 10 794136 ) UL uppr Umit - Lbwer U mit = 20.7615