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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=24,s=4,n=15,H0:=22,Ha:>22

Short Answer

Expert verified

If thep-value<reject the null hypothesis.

Here, 0.036<0.05reject the null hypothesis.

Step by step solution

01

Step 1. Given information is: 

x=24,s=4,n=15,H0:=22,Ha:>22

02

Step 2. Calculating P-value 

Teststatic,t=x-0sn...(*)UndertheassumptionthatH0istrue,tfollowstdistributionwithdf=15-1=14Observedvalueofteststatic,t0=24-22415=1.94Sincethegivenhypothesisisalefttailedtest,Pvalueisgivenby:P-value=P(tt0),wheret~t14=P(t1.94)=P(t-1.94)=0.036

03

Step 3. Result

Comparep-valuewithlevelofsignificance:If,thep-value<rejectthenullhypothesis.Here,0.036<0.05rejectthenullhypothesis.Hence,concludethatHa:>22

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