Question

Calculate the 95% confidence interval for the difference (mul-mu2) of two population means given the following sampling results. Population 1: sample size = 15, sample mean = 30.95, sample standard deviation = 4.07. Population 2: sample size = 11, sample mean = 14.38, sample standard deviation = 4.07. Your answer: O 13.79 < mul-mu2 < 19.36 13.97 < mul-mu2 < 19.18 O 12.83 < mul-mu2 <20.32 11.14 <mul-mu2 < 22.01 O 16.31 <mul-mu2 < 16.84 O 11.33 < mul-mu2<21.82 o 12.42 < mul-mu2 <20.72 O 13.83 < mul-mu2 < 19.32 O 13.21 < mul-mu2 < 19.93 O 12.06 < mul-mu2 <21.08

Answer 1

g. Given population 1:- Sample size = 15, sample mean = 30.95 Sample standard deviation = 4.07 ie ni= 15, X = 30.959 01 = 4.07 population 28 - Sample mean = 14:38, sample size = )) » Sample standard deviation = 4.07. ie na=ll, ĭ = 14038, 02=4.07 To find 95% confidence interval for the difference (mui -mu 2) of two popis mean 801": We know that Let Xie X2 *. Xng and you Y2 - - you be the two candom sample N1419012), N (42902?) we have XON(u 01/01) TNN [42, 02/ou. el=mal & 42=MU2 X-7) UN/mui-muz, 0,² +02 D) 12 the standarized fandom Variable 2= (-∞)-(41-42). 012 + 022 N N (0,1) from n2

which is independent of paramelex : the value of 2 has a pivotal quanting in order to find 2 for (mur-mus, PAGE NO find a number (18 c2 Such that (1-0) Geom Po < (X-%) - (411-42) 012 + 02² Di 1 2 As N(0,1) we take equal tail prob me symmetric to get shortest vength CI= -20/2 C2 = 20/2 ta) 2012 The inequality becomes P1 - 242) < (3-)-(mại - Du2) < (4/2 012 +02² Di D2 וס D2 (X-ý) - 2012/0² +02² s (mul-muo) Į (x-7)+Za12 101²+02² =(1-4). ni na This follows 100 (1-0) % confidence interval mur-mu). where 014033 are known is Random interval (T1912) TI = (-y) - - Zal2 01² +02² ni 02 T2 = (x-7)+ Zala ol? † 822 ni 12

DATE T = ( 30 35 - 14:38) -(1-96) | (407)?t (107) 11 = (16.5)) - 1.96 x .16.5649 + 16-5649 15 (16.57) - 1.96 1.105 + 1.5059 16.57-1.96 2.6109 16.57 - 1.96 x 1.61582 16.57 - 3-1670 T = 13.403 Now, Ta=(x-7)+ Z0/2 0124 023 NI n2 = (30.95-1438) + 1.96 (4:07) + (4:07)? 15 - 16.57 tlo96 x 1061582 = 16.57 + 3.1670 T2 = 19.737 Concluion The 95% confidence interval for the difference (mul - MU of the population mean is 13.403 kmur-muz (19.737 all 13 31 { gl - g2 4 19 93 .