Your engineering position requires frequent travel to the west coast, typically

Question

Your engineering position requires frequent travel to the west coast, typically through Chicago. In reviewing historical airline data, you have determined that over the last year flights into Chicago on Southwest have been normally distributed, arriving an average of 75 minutes late with a standard deviation of 9.8 minutes. Yesterday, a travel agent informed you that Southwest recently changed their scheduling into Chicago and has improved arrival times. To confirm this claim, you collect the arrival times for a random sample of 40 Southwest flights arriving into Chicago, and find a sample mean of 71.6 minutes and a sample standard deviation of 7.2 minutes. (a) Has there been a change in the mean arrival times on Southwest? (b) Determine if there been a change in the variability of arrival times on Southwest. (c) Estimate your arrival time on your next Southwest flight into Chicago with 90% confidence (i.e., give an interval).

Explanation / Answer

a. To test whether there has been change in the mean arrival times on Southwest, states the hypotheses.

H0:mu=75 (mean arrival time is 75 minutes)

H1:mu=/=75 (mean arrival time is different from 75 minutes)

For normally distrubuted arrival times, sample size, n>30 and known population standard deviation, use 1-sample z test.

Z=(xbar-mu)/(sigma/sqrt N), where, xbar is sample mean, mu is population mean, sigma is population standard deviation, and N is sample size.

=(71.6-75)/(9.8/sqrt 40)

=-2.19

p value is: 0.0285.

Conclusion: Per rule, reject H0, if p value is less than 0.05. Here, p value is less than 0.05, therefore, reject H0 and conclude that mean arrival time is significantly different from 75 minutes.

b. Hypotheses are as follows:

H0:sigma=9.8 (population standard deviation is 9.8 minutes)

H1:sigma=/=9.8 (population standard deviation is different from 9.8 minutes)

alpha=0.05

Test statistic:

X^2={(n-1)s^2}/sigma^2

={(40-1)*7.2^2}/9.8^2

=21.05

p value at 39 degrees of freedom is: 0.9916.

The p value is not less than alpha=0.05, therefore, fail to reject H0. There is insufficient sample evidence to warrant the rejection of null hypothesis.

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