An airline claims that the no-show rate for passengers is less than 5%. In a sample...

An airline claims that the no-show rate for passengers is less than 5%. In a sample of 420 randomly selected reservations, the rate of no-show is 4.5%. Set up the null and alternative hypothesis to test this claim. Describe what type 1 error and type 2 error are in this case. Find the test statistic and make a decision at 5% level of significance.

Expert Solution

Here claim is that p<0.05

So hypothesis is vs

Type 1 error is that p is not less than 0.05, but we conclude it that it is p<0.05

Type 2 error is that p is less than 0.05, but we conclude that it is not less than 0.05

So test statistics is

P value is

As P value is greater than alpha we fail to reject the null hypothesis

Hence we do not have sufficient evidence to support the claim that p<0.05

orchestra answered 2 years ago

An airline claims that the no-show rate for passengers is less than 7%. In a sample...

An airline claims that the no-show rate for passengers is less than 7%. In a sample of 420 randomly selected reservations, 18 w ere no-shows. At α = 0.01, test the airline's claim. A. P-value = 0.002 < 0.01; reject H0; There is not enough evidence to support the airline's claim, that is less than 7% B. P-value = 0.003 < 0.01; reject H0; There is not enough evidence to support the airline's claim, that is less than 7% C....

An airline claims that the no-show rate for passengers is 5%. In a sample of 420...

An airline claims that the no-show rate for passengers is 5%. In a sample of 420 randomly selected reservations, 19 were no-shows. At α = 0.05, test the airline’s claim. Also construct a 95% confidence interval for the proportions of no-show in the airline reservations to see the connection between confidence interval and hypothesis testing.

Because not all airline passengers show up for their reserved seat, an airline sells 125 tickets...

Because not all airline passengers show up for their reserved seat, an airline sells 125 tickets for a flight that holds only 124 passengers. The probability that a passenger does not show up is 0.10, and the passengers behave independently. Round your answers to two decimal places (e.g. 98.76). (a) What is the probability that every passenger who shows up gets a seat? (b) What is the probability that the flight departs with empty seats? (c) What are the mean...

Because not all airline passengers show up for their reserved seats, an airline sells 125 tickets...

Because not all airline passengers show up for their reserved seats, an airline sells 125 tickets for a flight that holds only 120 passengers. The proportion that a passenger does not show up is 10%, and the passengers behave independently. [Think Binomial Dist.] a. What is the proportion that every passenger who shows up gets a seat? b. What is the proportion that the flight departs with empty seats? c. What are the mean and standard deviation of the number...

A random sample of 400 passengers of an airline is polled after their flights. Of those...

A random sample of 400 passengers of an airline is polled after their flights. Of those passengers, 300 say they will fly again with the same airline. Which of the following is the 99% confidence interval for the proportion of passengers that will fly again with the same airline?

An airline estimates that 80% of passengers who reserve the tickets actually show up for the...

An airline estimates that 80% of passengers who reserve the tickets actually show up for the flights. Based on this information, it has to decide how many tickets it will sell for each flight, which is typically more than the number of seats actually available. In the economy section of a particular aircraft, 200 seats are available. The airline sells 225 seats. What is the probability that more passengers will show up than there are seats for?

An airline's public relations department says that the airline rarely loses passengers' luggage. It further claims...

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, 92 % 92% is recovered and delivered to its owner within 24 hours. A consumer group who surveyed a large number of air travelers found that only 137 of 169 people who lost luggage on that airline were reunited with the missing items by the next day. Does this cast doubt on the airline's claim? Explain

McJilton Airlines’ public relations department says that the airline rarely loses passengers’ luggage. It also claims...

McJilton Airlines’ public relations department says that the airline rarely loses passengers’ luggage. It also claims that when it does lose luggage, 90% is recovered and delivered to its owner within 24 hours. A consumer group who surveyed travelers with lost luggage found that 103 out of 122 people had their luggage returned to them within 24 hours. Can the consumer group dispute the claim, with 95% confidence?

Airline passengers arrive randomly and independently at the passenger check-in counter. The average arrival rate is 10 passengers per minute.

POISSON PROBABILITY DISTRIBUTION Airline passengers arrive randomly and independently at the passenger check-in counter. The average arrival rate is 10 passengers per minute. to. Calculate the probability that no passenger will arrive within one minute. μ = 10 in 1 minute x = 0 in 1 minute b. Find the probability that three or fewer passengers will arrive within one minute. μ = 10 in 1 minute x ≤ 3 in 1 minute >>> P (x = 0) + P...

A random sample of 19 airline passengers at Hartsfield Jackson International airport showed that the mean...

A random sample of 19 airline passengers at Hartsfield Jackson International airport showed that the mean time spent waiting in line to pass though the TSA checkpoint was 31 minutes with a standard deviation of 7 minutes. Construct a 95 % confidence interval for the mean waiting time by all passengers in the TSA checkpoint at this airport. Assume that such waiting time for all passengers is normally distributed. A) 31+1.65(7/P19) B) 31+1.746(7/P19) C) 31+1.96(7/P19) D ) 31+2.101(7/P19)

Question