Question 1121207: The manager of the Marua Shopping Mall in Windhoek claims that visitors to this mall spend
always more than 90 minutes in the mall on any occasion. To test this claim, the chairman of the
Namibian Chamber of Commerce and Industry (NCCI) commissioned a study which found that, from
a random sample of 25 recent visitors to this mall, the average visiting time was 95.5 minutes, with a
standard deviation of 15 minutes.
2.2.1) Formulate a suitable null and alternative hypothesis for this situation. (2)
2.2.2) Which test statistic ( z or t ) is most appropriate to use? & Why? (3)
2.2.3) Compute the sample statistic (3)
2.2.4) Formulate the decision rule by using α = 10 % (3)
2.2.5) What conclusion can the chairman of the Namibian Chamber of Commerce and Industry
(NCCI) draw from the findings?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Ho: average visiting time <=90 minutes
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.
alpha=0.10
One sample t-test 0.90, df=24 since using sd of sample as unbiased estimator of sd of population.
critical value t>1.711
t=(x bar-mean)/s/sqrt(n)=(95.5-90)/15/sqrt (25)
=5.5*5/15=1.83
Can reject Ho at the 10% level and say that the average visiting time is > 90 minutes.
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