Question

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 posting 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 line. Based upon data from past years, you decide it is reasonable to assume that the standard deviation of waiting times is 8 minutes. Use α-0.05.
a. State the hypotheses.
b. What is the p-value (to 4 decimals)
c. Using a-0.05, can you conclude that the actual mean waiting time is significantly less than the claim of 12 minutes made by the taxpayer advocate -
Select your answer- Yes / No

Answer 1

Answer:

(a) H₀: μ ≥ 12 vs. Hₐ: μ < 12.

(b) The p-value of the test is 0.0384.

(c) Yes.

Step-by-step explanation:

A hypothesis test for single mean can be used to determine whether the actual mean waiting time is significantly less than the claim of 12 minutes made by the taxpayer advocate.

(a)

The hypothesis can be defined as:

H₀: The actual mean waiting time is less than 12 minutes, i.e. μ ≥ 12.

Hₐ: The actual mean waiting time is not less than 12 minutes, i.e. μ < 12.

The information provided is:

$n=50\\\bar x=10\\\sigma=8\\\alpha =0.05$

Since the population standard deviation is provided use a z-test for single mean.

The test statistic is:

$z=\frac{\bar x-\mu}{\sigma/\sqrt{n}}=\frac{10-12}{8/\sqrt{50}}=-1.77$

(b)

The p value is computed as follows:

$P(Z<-1.77)=1-P(Z<1.77)\\=1-0.9616\\=0.0384$

*Use a z-table for the probability.

Thus, the p-value of the test is 0.0384.

(c)

The decision rule for the hypothesis test is:

If the p-value of the test is less than the significance level, α then the null hypothesis will be rejected and vice-versa.

The p-value = 0.0384 < α = 0.05

The p-value is less than α thus, the null hypothesis will be rejected at 5% level of significance.

Conclusion:

The actual mean waiting time is significantly less than the claim of 12 minutes made by the taxpayer advocate.