A national air traffic control system handled an average of 47941 flights during

Question

A national air traffic control system handled an average of 47941 flights during 28 randomly selected days in a recent year. The standard deviation for this sample is 6204 flights per day. Complete parts a through c below.

a. Construct aa 99% confidence interval to estimate the average number of flights per day handled by the system. The 99% confidence interval to estimate the average number of flights per day handled by the system is from a lower limit to an upper limit of . (Round to the nearest whole numbers.)

b. Suppose an airline company claimed that the national air traffic control system handles an average of 50,000 flights per day. Do the results from this sample validate the airline company's claim?

C. Since the 99% confidence interval contains 50,000, it cannot be said with 99% confidence that the sample validates the airline company's claim. C. Since the 99% confidence interval contains 50,000, it can be said with 99% confidence that the sample validates the airline company's claim. This is the correct answer.D. Since the 9999% confidence interval does not contain 50,000, it cannot be said with 9999% confidence that the sample validates the airline company's claim.

What assumptions need to be made about this population? A. Since the sample size is not greater than or equal to 30, one needs to assume that the population distribution is not very skewed to one side. B. Since the sample size is not greater than or equal to 30, one needs to assume that the population follows the Student's t-distribution. C. Since the sample size is not greater than or equal to 30, one needs to assume that the population distribution is skewed to one side. D. Since the sample size is not greater than or equal to 30, one needs to assume that the population follows the normal probability distribution

Explanation / Answer

a. From information given, xbar=47941, n=28, s=6204. Sample size is small (n<30) and population standard deviation is unknown. Therefore, use Student's t distribution to compute the 99% confidence level.

The 99% c.i=xbar+-talpha/2, n-1(s/sqrt n), where, xbar is sample mean, t is t critical at alpha/2 (alpha=0.01, alpha/2=0.005) and n-1 degrees of freedom, s is sample standard deviation, n is sample size.

=47941+-2.77(6204/sqrt 28)

=(44693,51189)

b. From information given, H0:mu=50000 (average number of flights handled per day is 50000) versus H1:mu=/=50000 (average number of flisghts handled per day is different from 50000)

Per rejection rule based on confidence interval if the confidence interval doesnot contain population mean, reject null hypothesis. The confidence interval does contain 50000, therefore, fail to reject null hypothesis. There is insufficient sample evidence to warrant the rejection of the claim made by airline company. In other words, since the confidence interval does contain 50000, it can be said with 99% confidence sample validates the airline company's claim.

Since sample size is not greater than equal to 30 (n=28), therefore, one needs to assume population follows the Student's t-distribution. Option B. For large sample, assumption of normal population is suitable. Options A, C and D are wrong.

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