Question

Physioans depend on laboratory test results when managing medical problems such as diabetes or epilepsy. In a uniformity test for glucose tolerance, three different laboratories were each sent n-5 identical blood samples from a person who had drunk 50 milligrams (mg) of glucose dissolved in water. The laboratory results (in mg/d) are listed here. Lab 1 Lab 2 Lab 3 120.0 98.3 102.9 110.7 112.1 108.5 108.8 107.7 101.1 104.2 107.9 110.0 00.499.2 105.5 (a) Do the data indicate a difference in the average readings for the three laboratones? (Use α-0.05.) State the null and alternative hypotheses. O Hg: One or more pairs of population means differ : One or more pairs of population means differ H: None of the population means are the same. Find the test statistic. (Round your answer to two decimal places.) Find the rejection region, (Round your answer to two decimal places.]

(b) find the test statistic (t =?)

(c) Find the reject region (t > ?)

(d) estimate the difference in the mean discharge of effluents between plants A and D, using a 95% confidence interval. Round to three decimals

Answer 1

The okfect of the ud ts to test and delernsne cohetbey there is dileene tn tha average oood, foi the tree laborator e defned The null ard alatp as olloos to:At least ene af fla means ts different the others obai tha ANOVA oulput for rven daia 3 Columns s、 Speay lab,.lale s tabs &ba in the ses box ae comparis uten and select Turkey's click ok twice

ANovA : Lob , lab 2, Lab3, P MS F Focto a Cea.u 1,2 0,60 0564 2 u23 4 as. 3 Total u6S.7 Pooled. St DeV too.o tou-о 108.0 (12.0 : 5.94 "The test &tdi slotistesos o.60 and the covwesponderg Pvlu o56u g reater than tha guen level of ha stgncanre. ajected corres 步Here R can be obserued ttat tta p-value Ps Hence, ta nu

●,R can be conceded that ther {s ?asRfcient eufdence to ave tha Lhrec laboratoess dffer ALterYafle hypotheses. +4 : 어 le uk or e mean defferet n othess eves of stgnificance , «:o.os Ente tl data rto xcel sheet a to otepu. click ok The obtaind output One -sacfor ANOOVA as -klean Mean_.- l 576 l. 178 I. 5T2- 5 01379 A- 5 0.2194 0

ANoVA Talle Source Treat ment ous P-value Ms 01512500 O.0123 0.q838 16 0.03024 P.tolee: 0.012.3 test statishc Ps F-Soo conclusfon: The P.volue P, less than the ge Snificance (euel o.os, So oe vejec ard conclede that ehno Êsa & 棨cent evidence to value S the nul e书lents disclarged by the uplants at o.osled of sranicante Calclote mean słandard devialion of taplant Mean of plant A TA S- o.oI902 37 266

S calculato the test statistc value S76-sO Sa/ n 0-13วุโ5 ニ,, 2.3 reskam- ouo oo5 O 00

onclusion h statistic (t1 is no dantain the Yegion Sooe lail to ejeet tha null hypothes Conclude that tha dila psvuide insufbictent eviden te EDtndicata that tha tents eeedd at plant- A The follofirg talle repsents te of e uplants Count ean gample varance D9 0.0 .i-S46 78 1552-1.918 mple slandard 139.6 I38 46 manimum ramao o3 06 36 045 erence in ebluenks plants A o tterene Pant Mean s 16 1./8 O 342

rt flterndu hypothesis Hr: Ha Moo Calalt th 45. con-fidence eerual s lest: 1 Plamt A plant- I 576 0 138 18 mean 6 sld dev S N 7 d 0-342000 dffece plantA-plant- Ohypothesized ditlerence 0.00 P.value two laed) 73 confidencc interva . o. 23 half-odth Fsom the aboe o CamScanner

Conclusion The alove onderterl des not Corda%, null value above mulr hypothess and conclude thal thene deifference fn mean dis chaeoeflents erente tn mean ds charge bahusren plants AD d dke o-os level Cs Scanned with CamScanner