In: Statistics and Probability
Answer :- Ships arriving in U.S. ports are inspected by Customs officials for contaminated cargo. Assume, for a certain port, 20% of the ships arriving in the previous year contained cargo that was contaminated. A random selection of 50 ships in the current year included 5 that had contaminated cargo.
Given that :-
• Sample Size (n) = 50
• X = 5
• Proportion (P) = 20℅ = [ 0.2 ]
• Null Hypothesis and Alternative Hypothesis are :-
=> Null Hypothesis : Ho = 0.2
=> Alternative Hypothesis : HA = P < 0.2
• Sample Proportion :- P^ = X/n
=> P^ = 5/50
=> [ P^ = 0.1 ]
• Test -Statistic :- Z = (P^ - P)/√P(1-P)/n
= (0.1 - 0.2)/√0.2(1-0.2)/50
= -0.1/√0.2(0.8)/50
= -01/√0.16/50
= -0.1/0.0 566
= -1.77
= Test-Statistic => [ Z = -1.77 ]
• P-VALUE :-
=> Significance level = (α = 0.01)
=> P-VALUE = P(z < -1.77)
(using Z table)
=> [ P-VALUE = 0.0384 ]
=> Here, P-value > α
[ P-value(0.0384) > α(0.01) ]
• We fail to reject Null Hypothesis Ho.
Conclusion :- There is no enough evidence to support the claim that the proportion of ships arriving in the port with the contaminated cargo has decreased in the current year.
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