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Concluding Mean Waiting Time and Premium Rate

The Internal Revenue Service (IRS) provides a toll-free help line for taxpayers to call in and get answers to questions as they prepare their tax returns. In recent years, the IRS has been inundated with taxpayer calls and has redesigned its phone service as well as posted answers to frequently asked questions on its website (The Cincinnati Enquirer, January 7, 2010). According to a report by a taxpayer advocate, callers using the new system can expect to wait on hold for an unreasonably long time of 12 minutes before being able to talk to an IRS employee. Suppose you select a sample of 50 callers after the new phone service has been implemented; the sample results show a mean waiting time of 10 minutes before an IRS employee comes on the line. Based upon data from past years, you decide it is reasonable to assume that the standard deviation of waiting time is 8 minutes. Using your sample results, can you conclude that the actual mean waiting time turned out to be significantly less than the 12-minute claim made by the taxpayer advocate? Use ex = .05.


Fowle Marketing Research, Inc., bases charges to a client on the assumption that telephone surveys can be completed in a mean time of 15 minutes or less. If a longer mean survey time is necessary, a premium rate is charged. A sample of 35 surveys provided the survey times shown in the file named Fowle. Based upon past studies, the population standard deviation is assumed known with a = 4 minutes. Is the premium rate justified?

a. Formulate the null and alternative hypotheses for this application.
b. Compute the value of the test statistic.
c. What is the p-value?
d. At a = .01, what is your conclusion?

Solution Preview

The null hypothesis tested is
H0: Mean waiting time using the new phone service ≥ 12 minutes. (µ ≥ 12)
The alternative hypothesis is
H1: Mean waiting time using the new phone service < 12 minutes. (µ < 12)
Significance level = 0.05
The test statistic used is Z = x̅ - µ / σ/ √n. Given that x̅ = 10, n= 50, σ = 8
Therefore, Z = 10 - 12 / 8/ √50 = -1.767766953
Rejection criteria: Reject the null hypothesis, if the calculated value of the test statistic is less than the critical value at the 0.05 significance ...

Solution Summary

Concluding Mean Waiting Time and Premium Rate