Question

11) (7 pts) An airline's public relations department says that the airline rarely loses passengers' luggage. It further claims that on those occasions when luggage is lost, 88% is recovered and delivered to its owner within 24 hours. A consumer group who surveyed a large number of air travelers found that only 132 out of 160 people who lost luggage on that airline were reunited with the missing items by the next day. At the 5% level of significance, do the data provide sufficient evidence to conclude that the proportion of times that luggage is returned within 24 hours is less than 0.887 Test hypothesis using the confidence interval. HO: TEST USED: C-level Confidence Interval Decision: because

Answer 1

To Test :-

H0 :- P = 0.88

H1 :- P < 0.88

P0 = 0.88
q0 = 1 - P0 = 0.12
n = 160
P = X / n = 132/160 = 0.825

Test Statistic :-
Z = ( P - P0) / √(P0 * q0 / n)
Z = ( 0.825 - 0.88 ) / √(( 0.88 * 0.12) /160)
Z = -2.1409

Test Criteria :-
Reject null hypothesis if Z < -Z(α)
Z(α) = Z(0.05) = 1.6449
Z < -Z(α) = -2.1409 < -1.6449, hence we reject the null hypothesis
Conclusion :- We Reject H0

Decision based on P value
P value = P ( Z < -2.1409 )
P value = 0.0161
Reject null hypothesis if P value < α = 0.05
Since P value = 0.0161 < 0.05, hence we reject the null hypothesis
Conclusion :- We Reject H0

There is sufficient evidence to support the claim that the proportion of times that luggage is returned within 24 hours is less than 0.88.

p̂ = 132 / 160 = 0.825
p̂ = 1 - p̂ = 0.175
n = 160
p̂ ± Z(α/2) √( (p * q) / n)
0.825 ± Z(0.05/2) √( (0.825 * 0.175) / 160)
Z(α/2) = Z(0.05/2) = 1.96
Lower Limit = 0.825 - Z(0.05) √( (0.825 * 0.175) / 160) = 0.7661
upper Limit = 0.825 + Z(0.05) √( (0.825 * 0.175) / 160) = 0.8839
95% Confidence interval is ( 0.7661 , 0.8839 )
( 0.7661 < P < 0.8839 )

Since the values in the interval is positive, hence we can conclude to reject null hypothesis.