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In each of Exercises 9.107-9.112, we have provided a sample mean, sample standard deviation, and sample size. In each case, use the one-mean t-test to perform the required hypothesis test at the 5% significance level.

x=21,s=4,n=32,H0:=22,Ha:<22

Short Answer

Expert verified

Since 0.05<P=0.084<0.10, the evidence against null hypothesis is moderate

Step by step solution

01

Step 1. Given information is: 

x=21,s=4,n=32,H0:=22,Ha:<22

02

Step 2. Calculating P-value

Teststatic,t=x-0sn...(*)UndertheassumptionthatH0istrue,tfollowstdistributionwithdf=32-1=31Observedvalueofteststatic,t0=21-22432=-1.414Sincethegivenhypothesisisalefttailedtest,Pvalueisgivenby:P-value=P(tt0),wheret~t31=P(t-1.414)=0.084

03

Step 3. Calculating P using MINITAB

Theprobability,P(t-1.414)iscalculatedusingMINITABinthefollwingway:Step1:PresstheCalcmenu;Highlightthe'ProbabilityDistributions'.Step2:Presst...;Step3:TickCumulativeProbabilityandenterthedfDegreesoffreedom:31Step4:TickInputconstantandenterthevalue-1.414Inputconstant:-1.414Step5:PressOkNow,P=0.084>=0.05Therefore,at5%levelofsignificancewerejectthenullhypothesis,H0:=22

04

Step 4. Result

Since0.05<P=0.084<0.10,theevidenceagainstnullhypothesisismoderate

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