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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=20,s=4,n=24,H0:=22,Ha:22

Short Answer

Expert verified

Since 0.01<P=0.022<0.05, the evidence against null hypothesis is strong.

Step by step solution

01

Step 1. Given information is: 

x=20,s=4,n=24,H0:=22,Ha:22

02

Step 2. Calculating P-value 

Teststatic,t=x-0sn...(*)UndertheassumptionthatH0istrue,tfollowstdistributionwithdf=24-1=23Observedvalueofteststatic,t0=20-22424=-2.45Sincethegivenhypothesisisalefttailedtest,Pvalueisgivenby:P-value=P(t-t0)+P(tt0),wheret~t23=P(t-2.45)+P(t2.45)=2xP(t-2.45)=0.022

03

Step 3. Calculating P using MINITAB 

Theprobability,P(t-2.45)iscalculatedusingMINITABinthefollwingway:Step1:PresstheCalcmenu;Highlightthe'ProbabilityDistributions'.Step2:Presst...;Step3:TickCumulativeProbabilityandenterthedfDegreesoffreedom:23Step4:TickInputconstantandenterthevalue-2.45Inputconstant:-2.45Step5:PressOkNow,P=0.022<=0.05Therefore,at5%levelofsignificancewerejectthenullhypothesis,H0:=22

04

Step 4. Result

Since0.01<P=0.022<0.05,theevidenceagainstnullhypothesisisstrong

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