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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=23,s=4,n=15,H0:=22,Ha:>22

Short Answer

Expert verified

Since P=0.174<0.10, the evidence against null hypothesis is weak or none.

Step by step solution

01

Step 1. Given information is: 

x=23,s=4,n=15,H0:=22,Ha:>22

02

Step 2. Calculating P-value 

Teststatic,t=x-0sn...(*)UndertheassumptionthatH0istrue,tfollowstdistributionwithdf=15-1=14Observedvalueofteststatic,t0=23-22415=0.97Sincethegivenhypothesisisalefttailedtest,Pvalueisgivenby:P-value=P(tt0),wheret~t14=P(t0.97)=P(t-0.97)=0.174

03

Step 3. Calculating P using MINITAB 

Theprobability,P(t-0.97)iscalculatedusingMINITABinthefollwingway:Step1:PresstheCalcmenu;Highlightthe'ProbabilityDistributions'.Step2:Presst...;Step3:TickCumulativeProbabilityandenterthedfDegreesoffreedom:14Step4:TickInputconstantandenterthevalue-0.97Inputconstant:-0.97Step5:PressOkNow,P=0.174<=0.05Therefore,at5%levelofsignificancewerejectthenullhypothesis,H0:=22

04

Step 4. Result

SinceP=0.174<0.10,theevidenceagainstnullhypothesisisweakornone

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Most popular questions from this chapter

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