Question

(10%) In Relief from Arthritis published by Thorsons Publishers, Ltd., John E. Croft claims that over 40% of those who suffer from osteoarthritis receive measurable relief from an ingredient produced by a particular species of mussel found off the coast of New Zealand. To test this claim, the mussel extract is to be given to a group of 10 osteoarthritic patients. If 3 or more of the patients receive relief, we shall not reject the null hypothesis that p > 0.4; otherwise, we conclude that p <0.4. (a) Evaluate , assuming that p-0.4. (b) Evaluate B for the alternative p=0.3. (c) At a 0.05, how many osteoarthritic patients out of 10 receive relief will we reject Ho? (d) At a 0.05, what is the needed sample size if the true p -0.3 and we want B -0.05 ? (e) If 30 osteoarthritic patients out of 81 receive relief, estimate the 95% confidence interval of population proportion p.

Answer 1

Binomial distribution, level of significance, power, and conference interval for proportion

Croft claims that over 40% of oneoarthritis recire an easwaste relief from an ingredient-pfloduced. To test this claim, the mussel extract is to be giren to a group of 10 patients, if 3 or more racire relief, we shan not regeer Ho: P70.4, otherwise, we conclude that (H1 :) p < 09. a) a= level of tuznificance a Prov. of rejecting null when wall is true. = P(XE W l Ho) [ne is the critical region

= P(x (31 p = 0.1) [x nolf parents got relief . ) (1+)** follows a Poin (10, 1) dism X 20, 1, 2 Tn210 / 2014 (0) (0.9) 10.6) + ((0) (014) (15) + (12) 10.07 06² 2 0.167289 ~ 0.1673 is the evaluated d. ß 2 power of a test. & Porobability of rejecting Ho while. Hi is true. PCXEWI ) [W = criticatrazion] - I C4%)(0.99" (01770-2 120,1,2 (C) 2 (0) (013) 20.7) "() 0.9 0.9?-(3 2.0. 28278 is the power of me test, At d20.05, do know how many patients out of 10 rexire nerief will werejut Ho, we compute upto which x, i lů pq (1-P) long <005 ise, the sum does not exceed d20.05. (0) (0.9% (0.6) 0+ (0) (019' (01679 20.046357 <0.05 Hence me cubicat region for vejerting to be 10:13

(d) Al- Q20.05 we need the sample in ze if the true p20.3 & we want 8 20:05, her this be no 720 7- Ź (2) (6. 177 (0.63 94 {0105 7 Hone K is que 2 (017) *-* = 0,05 we have to maine (0) 0,07***005 it was said that 3 or more pasient. kad to accept to -X We have to soare for since the critical region is set before as {0, 1, 2} . & 2013 We see that for Nz 18 18-- 7 0.05995 (18) (613) * (0.28x = 0.05995 & for 12 19 (1) (013) (0.239-x 0.04622366 -X z 004622366 0.04622 368 is closer to 0.05, as the distribution is discrete we cannot reach at the exant Rower, so we mat the sample size Comlude is 19 what we need to infer pollowing me giren critical region, arthritic patiems out of 8 receive confidence interval of popla the 95% relief, estimate proportion p

number We know that 30 is the rate of swuess out of stars, so it is Binomial distributed is ample pe 30 if for Bin (hip) proportion I EX) = np V(x) = np (14) 20.370 37 >E(B) = P (D) hp (+1) n2 5 $ - sample propation - By Central Limit Theorem i p Now Rom as we donot know the variame were estimate it him the sample variance, Torozy is the upper cutoff point- for (011), 700025 196, n 281 1 p- p iller 30.95. -79-4 Php 45 < = 14 for 2016 4P[-8652 <150149557 -0,95. 967. a toris (0.2652, 014935).