Question

Most major airlines allow passengers to carry two pieces of luggage (of a certain maximum size) onto the plane. However, their studies show that the more carry-on baggage passengers have, the longer it takes the plane to unload and load passengers. One regional airline is considering changing its policy to allow only one carry-on per passenger. Before doing so, the airline decided to collect some data. Specifically, a random sample of 1000 passengers was selected. Researchers observed the passengers and noted the number of bags each person carried on the plane. Out of the 1000 passengers, 340 had more than one bag.

1. Based on this sample, develop and interpret a 99% confidence interval estimate for the proportion of the traveling population that would have been impacted had the “one-bag” limit been in effect. Discuss your result.

2. The domestic version of Boeing’s 757 has a capacity for 220 passengers. Determine an interval estimate of the number of passengers you would expect to board the plane with more than one carry-on. Assume the plane is at its passenger capacity.

Answer 1

Level of Significance,   α =    0.01          
Number of Items of Interest,   x =   340          
Sample Size,   n =    1000          
                  
Sample Proportion ,    p̂ = x/n =    0.340          
z -value =   Zα/2 =    2.576   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0150          
margin of error , E = Z*SE =    2.576   *   0.0150   =   0.0386
                  
99%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.340   -   0.0386   =   0.3014
Interval Upper Limit = p̂ + E =   0.340   +   0.0386   =   0.3786
                  
99%   confidence interval is (   0.301   < p <    0.379   )

mulitiplying with capacity of 220,we get

confidence interval is ( 66.311 < p <    83.289 )