Question

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: 17.0, 29.7. 18.8. 26.1, 21.3, 17.8, 30.8. 22.4, 20.2, 26.9. 30.9.26.1, 24.0, 23.7, 25.9, 20.8, 28.2, 33.8, 25.7, 30.6 (a) Can you support a claim that mean rainfall from seeded clouds exceeds 24.0 acre-feet? Use a =0.01. There is insufficient evidence to indicate that the true mean rainfall is greater than 24.0 acre-feet at a = 0.01. (b) Compute the power of the test if the true mean rainfall is 26.0 acre-feet. Round your answer to one decimal place (e.g. 98.7), (c) What sample size would be required to detect a true mean rainfall of 26.5 acre-feet if we wanted the power of the test to be at least 0.9? ηΣ Round your answer to the nearest integer. Statistical Tables and Chart

Answer 1

to the given геел даса, де Solution Accoъдия да » и – 80, х = S 035 (а s? 22. 8434 аид - 0.0 / We want to test Но е: 24 от Н : 240

~ Test statistici ta x-ce suu 25 035-24 An-1 under Ho 4.7848/850 t = 0.9674

Guidical value: to= k11-4.n-1) = 16.9919) - 2.5395 Heren Eala so accept Ho conchiede thad is insufficiend evidence to indicate the true mean rainfall is than 24.0 acre refeed an a these Therefore We greated las X = 0.01

6 Let Hii (l=26.0 llen, f x-cl under Hi sun 25.03.-26.0 4.78481130 =-0.9019 The power of the testy as day = P( t (1977 T = P( {n-13>) >-0.9019 0.0108 Itence pourer of the test af fl=26-0 = 0.010Q = as 2

6 Let Hi = 26.5 Geiven, power of the test, t= 0.9 Then we kuolo Thaibo 7² 5-ce P ( &(n-1) (5) 0 p(469) 3 4610.19) 0.92 25.035-265 4.7848(vh (25.035-265754 7.7848 -1.465 Ju -1.3277 7.7848 Th= 1.465 vh 3 4.3364 1:3277x4.7848 a

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