Hawk & Hummingbird Airline wants to determine the proportion of passengers that

Question

Hawk & Hummingbird Airline wants to determine the proportion of passengers that bring only carry-on luggage to the flight from New York to Boston. In a random sample of 195 passengers, 43 passengers had only carry-on luggage.

Is there enough evidence to conclude that more than 20% of all passengers have only carry-on luggage? Perform the appropriate test at a = 0.10.Enter the test statistic in the first box and your decision in the second box ( 0 if you reject and 1 if you don't (i.e if you accept)

Find the p-value of the test in part (b). (Note: Instead of usual 5% tolerance the tolerance on p value is set at 0.005 to take care of small values. Please be careful to enter up to 3 decimal places in case your p value is large)

Find the minimum sample size required in order to estimate the proportion of passengers who have only carry-on luggage to within 2% with 90% confidence, if it is known that this proportion is at most 0.34.

Explanation / Answer

let P= proportion of passengers that bring only carry-on luggage to the flight from New York to Boston

here P=43/193=0.2228, n=193

H0:p=0.2 and H1:p>0.2

SE(P)=sqrt(p(1-p)/n)=sqrt(0.2*(1-0.2)/193)=0.0288

we use z-test and z=(P-p)/SE(P)=(0.2228-0.2)/0.0288=0.7917

one tailed P-value=0.2143

since P-value is more than alpha=0.1 so we fail to reject H0 and conclude that P=0.2=20%

second part p=0.34

SE(p)=sqrt(0.34*(1-0.34)/n)

with (1-alpha)*100% confidence margin of error of p=z(alpha/2)*SE(p)

with 90% confidence margin of error=z(0.1/2)*SE(P)=1.645*sqrt(0.34*(1-0.34)/n)

or,0.02=1.645*sqrt(0.34*(1-0.34)/n)

or,sqrt(0.34*(1-0.34)/n)=0.02/1.645

or,sqrt(0.34*(1-0.34)/n)=0.0122

or,0.2244/n=0.0122*0.0122

or,n=0.2244/(0.0122*0.0122)=1507.66 (next whole number is 1508)

answer is 1508

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