University of South Florida**We aren't endorsed by this school
Course
EGN 3443
Subject
Statistics
Date
Dec 26, 2024
Pages
4
Uploaded by AgentFreedom10128
Wi A/7 Vo f a"/s" Ul b / 3 21.In @ of 1000 randomly selected consumers who had opportunities to send in a rebate claim form "'/g after purchasing a product, 250 of these people said they never did so (“Rebates: Get What You Deserve,’| joms a (f Consumer Reports, May 2009: 7). Reasons cited for their behavior included too many steps in the process EO,—-X amount too small, missed deadline, fear of being placed on a mailing list, lost receipt, and doubts about r &—) receiving the money. Calculate an upper confidence bound at the 95% confidence level for the true proportion \ O-' — of such consumers.who never apply for a rebate. Based on this bound, is there compelling evidence that the true proportion of such consumers is smaller than 1/3? Explain your reasoning,. ",‘ = 139 . 0. e lowkY CL of WO(L™ %)°o fer Wfibrflw (9 1009 [ o, Ft =24 J33/n « 2/ int Qs . coreidence fewef | " ' T~ . = 0(-56.05 ; ([ + ‘l‘n/n 5 > 4S Y T Ry-#iReob 7 |0H. ~ P+ B/ an 015 + 015 » . A = 1900 _ g 950, A== 0.3 | L+ & n il ' 9y |+ ‘.V“g n = o0 Q |00V = ~ 1.'(" . ‘ o~ — Qgpes esnfidone Soued L g4 {?Vfl* 2z 5% =2 i o 48] covfidene (6 et The ot Q5] bt R V\ . st . =B ""“S‘jflfih (L1008 hee g 1 wehueer o and 0.2333 e Lo 4( 1000 . A2 < 0.33%2}3, Rew O -~ 0.0+ ‘.(oqs" Sin 0. b+ |0OD M"‘M auid R Nt Pae e v Y o GF
21. The desired percentage of SiO2 in a certain type of aluminous cement is To test whether the true average percentage is 5.5 for a particular production facility, @ndependently obtained samples are analyzed. Suppose that the percentage of SiO2 in a sample is normally distributed with o = ( and that X = 5.25. a. Does this indicate conclusively that the true average percentage differs from(5.5based on a = 0.01? b. If the true average percentage is # = 5.6 and a level a = 0.01 test based on 1= 16 is used, what is the probability of detecting this departure from H;? Moo= 5.5 o c. What value of 7715 tequired to satisfy a = 0.01 and A(5.6) = 0.01? . v Lo 4 bR ¥ drue ANtIUQR. pexiRntmug e of $0g . G =03 o i Q. o v M = 5.5 )":5'9‘5 > 2 -lest. : ) X = 0.0 oo a5 Z - Mo 5.6 ~§.5 . %% t = = > ?-V&W\-‘L: 0.9u & < v.of =) r(ydr Ho. S e e R S e e e K S AU L3 Ao I e L Gt S G A s T SIS R A ] N R el a2 A e, S B R SR A U BN T e A O R S ) A Wrosp F20ile g s C ey eaf 2 sl ) . SRSt DR
38. A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specimen, resulting in the accompanying data (from “Engineering Properties of Soil,” Soil Science, 1998: 93-102). 110 509 097 159 4.60 032 055 145 The values of the sample mean, sample standard deviation, 0.14 447 120 350 502 467 522 269 N=20 and (estimated) standard error of the mean are 2.481, 1.616, 398 3.17 303 221 069 447 331 I1.17 ~ and .295, respectively. Does this data suggest that the true 0}? L A e e D @fi_ge,percentage\of organic matter in such soil is something other than 3%? Carry out a test of the appropriate hypotheses at significance level 0. 101 Would your conclusion be different if a = .05 had been used? [Note: A normal probability plot of the data indicates the sample is normally distributed] O~ uncrowq, e Loy = Yoo orere? TR avXURR pexckNYaye of cryanc moktec in ruen sol LR LTOMQ A ?j -’E" )_.M\ s " 3 s = (blb . ¥ 45 Y (7)== = 05 X 3 ZAN-3 L -i3m B L= = 3 0245 t—tesT- 1 =% /r‘: T- vaWm €@ = 7—(006'“> o(:gg;\.o@ o= 2T\ - pe (Twy \’0\3) = 0. 08_. = o —% o Pc(Taq 7 |.‘+S‘l)] S et Ho - o =0 .02 7 0.05 = 2 aC be (Tua? L ) MEe08 petind T M, L B s oo e de e e B e e S R D R R o e L e W o A W A A T N