document.write( "Question 1121207: The manager of the Marua Shopping Mall in Windhoek claims that visitors to this mall spend
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document.write( "always more than 90 minutes in the mall on any occasion. To test this claim, the chairman of the
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document.write( "Namibian Chamber of Commerce and Industry (NCCI) commissioned a study which found that, from
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document.write( "a random sample of 25 recent visitors to this mall, the average visiting time was 95.5 minutes, with a
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document.write( "standard deviation of 15 minutes.
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document.write( "2.2.1) Formulate a suitable null and alternative hypothesis for this situation. (2)
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document.write( "2.2.2) Which test statistic ( z or t ) is most appropriate to use? & Why? (3)
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document.write( "2.2.3) Compute the sample statistic (3)
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document.write( "2.2.4) Formulate the decision rule by using α = 10 % (3)
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document.write( "2.2.5) What conclusion can the chairman of the Namibian Chamber of Commerce and Industry
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document.write( "(NCCI) draw from the findings? \n" );
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Algebra.Com's Answer #737017 by Boreal(15235) You can put this solution on YOUR website! Ho: average visiting time <=90 minutes \n" ); document.write( "Ha: average visiting time >90 minutes. Want to reject Ho if there is a true difference. One way test, since we know which direction the difference will be if significant. \n" ); document.write( "alpha=0.10 \n" ); document.write( "One sample t-test 0.90, df=24 since using sd of sample as unbiased estimator of sd of population. \n" ); document.write( "critical value t>1.711 \n" ); document.write( "t=(x bar-mean)/s/sqrt(n)=(95.5-90)/15/sqrt (25) \n" ); document.write( "=5.5*5/15=1.83 \n" ); document.write( "Can reject Ho at the 10% level and say that the average visiting time is > 90 minutes.\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |