case 1.docx - PART I Plan Find the department of...

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PART I: Plan: Find the department of transportation's annual net profit at the 95%confidence. Do: Independence assumption: All the travelers are random and there are not travelers that would influence each other. 10% Condition: There are 500 random sample train travelers, which should be certainly less than 10% of the population which is 1 million. Success/Failure Condition: n = 500 p p × 0.82= 410 10 n(1- ) = 500 p p × 0.18 = 90 10 So, we can use a normal distribution model to find one proportion and z-interval. 95% confidence which is critical Z value =1.96 P(Yes)= 410/500=0.82 P(No)=90/500=0.18= p p A random sample of 500, which is n =500 1
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Margin of Error ME = (Critical Number) × (SE) ME = (1.96) × ( ^ p ( 1 ^ p ) n ) = 1.96 × ( 0.82 × ( 1 o .82 ) 500 ) =0.033712 Confidence Interval: Confidence interval = Estimate ± ME =0.18 ± 0.033712 (0.146288, 0.213712) It means use 95% of confidence, between 0.146288 and 0.213712, there are travelers who don’t buy a tickets. We know there are 1 million of traveler population per year, lost revenue from single ticket is $7.5. Then: Lower revenue in Department of Transportation:
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