Question

QUESTION 9 Cloud seeding has been studied for many decades as a weather modification procedure for an interesting study of this subject, see the article in Technometrics, "A Bayesian Analysis of a Multiplicative Treatment Effect in Weather Modification", Vol. 17, pp. 161-166). The rainfall in acre-feet from 20 clouds that were selected at random and seeded with silver nitrate follows: 18.5, 31.2, 20.3, 27.6, 22.8, 19.3, 32.3, 23.9, 21.7, 28.4. 32.4, 27.6. 25.5, 25.2, 27.4. 22.3, 29.7, 35.3, 27.2, 32.1 (a) Can you support a claim that mean rainfall from seeded clouds exceeds 25.5 acre-feet? Use alpha=0.01 There is no evidence to indicate that the true mean rainfall is greater than 25.5 acre- feet. QUESTION 10 A textile fiber manufacturer is investigating a new drapery yarn, which the company claims has a mean thread elongation of 12 kilograms with a standard deviation of 0.5 kilograms. The company wishes to test the hypothesis HO: mu >=12 against H1: mu < 12 using a random sample of n=14 specimens. Calculate the p-value if the observed sample mean is 11.1.. Round your final answer to five decimal places (e.g. 98.76543).

Answer 1

I Queshon @ we have given n=20 observations. ler x = Observation = 18.5,31-2,20.3, 27.6,22.8, 19.3 7, 32.3, 23.9, 21:7, 28.4 32-4, 27-6,.25.5, 25.2, 27.4 22.3, 29.7, 35-3, 272, 32-1) a) clavem :- that mean rainfall from seeded Clouds exceeds 25.5 acre feet 3 1=0.011 To Test : Ho: M = 25.5 w Hii > 25.5 Test stulbs hic : t= (2-M) Sn whese zasomple mean

4 sa cumple standard deviation 1 < žē = 2= } = [ 18.5 + 31.2 + 20-34 20 = 13.5+ 31:2 Trio ---+35.3+27-2+32.), = 59.4 1 28.525 1. Now Now s²t Elli- sel? n- at # 19" (18.5-26.53512 +0312 - 26.525)?+ - - -+ (32-1 – 26.535)?) to * (4134.9855) (22.89397 1.5= 4 7847 C 22

+= (26.535 – 25.5) 520 1 0.96737 4.7847 td, df = toolia E2539_] Decision - we reject Ho at (1-2) +100 % level of significance if it) taide here IH <ta, df fail to reject Ho at d=0.01 hence we conclusion the - true There mean is no evidence to indicate that rainfall is greater than 25.5 acre-feet.

pueshon lol : we have ne 14, M4 =12, E = 11.1 s= TO 0.5 Test .. Ho: mu= 12 muc12 NS H: 05 Test statistie i ta (7-mu) so = (1.462) 314 7=2634 tt = 1-6.7341 E 6-73498) : P value 70.00001 from table Decision :- we reject Ho if p valueca here p value that conclusion : mur12 There is enough . at x = given evidence to conclude revel. (5%, 1. 10%.