Question

Suppose that we randomly select 50 billing statements from each of the computer databases of the Hotel A, the Hotel B, and the Hotel C chains, and record the nightly room rates. The means and standard deviations for the data are given in the table.

	Hotel A	Hotel B	Hotel C
Sample Average ($)	155	180	115
Sample Standard Deviation	17.7	22.6	12.3

(a)

Find a 95% confidence interval for the difference in the average room rates for the Hotel A and the Hotel B chains. (Use Hotel A − Hotel B. Round your answers to two decimal places.)

$__________ to  $___________

(b)

Find a 99% confidence interval for the difference in the average room rates for the Hotel B and the Hotel C chains. (Use Hotel B − Hotel C. Round your answers to two decimal places.)

$__________ to  $___________

Answer 1

Soln Here we first create a confidence interual of (1-2) Range and use it to solve both problems. Suppose Mi and My be the population meam, M e M be be the samble mean, si e s2² be the sample variance of Hotel and a respectively. This 71 - X2-(M1-12) t= s V ni 2 where [ (11-1) si? + Co, -1) si nith2-2 If t t (nithe -2 then 2-2 ) 7 I et n2 Pl- to (nitr2-2) < t < td, 601491-2)) = 1-0 Pl-ta, miten som <a - (14,- (4) <a hyan-235/) = to (11772-2) 1 1 2 3 4 in < M, - the Linini 5 / + 1) = -2 the confidence s Plau-dr-to + tą Thus interval of Mi-M2 is given by

7-12 I S ㅗ I , 0492-0) hา Hotel A and 44 Here Let Hotel 1 = Hotel 2 = Hotel B Hotel B, thus T = 155 TL 180 nie M2 = 50 Si = 17.7. S = 22.6 & a= 0.05 ㅋ +0.025,98 = 7.9845 (By t-li ti bution Thus 15 10 table confidence interval és (155-180) I (2.9845) 5 2 50 sa ba (49.413-23 +(49) (2206)") = 412:02 / 98 S = 20-298 =1 a) 6 Here Let confidence Interval = (-33.056, -16.940) Hotel i Hotel B and Hotel 2 z Hole C This n = 180 ; T = 115, n = 12=50, S = 22.6 S2 = 12.3 e x = 0.01 and this gives to.005,98 = 2.63 2. 63 and s? + ((49) (22.679+(49) (1203)2) = 331.05 98 SF 18.1920 T Baby

So 99% Confidence Interval for Mi-M2 x's (180- |15) (2:63) (18:19) Z (55.432 74.568)