contestada

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 15 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 13 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 11 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.

Respuesta :

Answer:

From the result of the one-tailed z-test, we can conclude that the mean waiting time is not significantly less than the the 15 minute claim by the taxpayer advocate.

Step-by-step explanation:

Sample size = 50

Mean waiting time = 13 minutes

Standard deviation = 11 minutes

We need to perform one hypothesis test that the actual mean waiting time turned out to be significantly less than the 15-minute claim made by the taxpayer advocate.

The null hypothesis would be that there is no significant difference.

H₀: μ₀ = 15 minutes

The alternative hypothesis would be that the mean waiting time is indeed significantly less than the 15-minute claim by the taxpayer advocate.

Hₐ: μ₀ < 15 minutes.

This is evidently a one tail hypothesis test (we're investigating only in one direction; less than the claim). Hence, we can use the z-test

z = (x - μ)/σₓ

σₓ = (σ/√n) = (11/√50) = 1.556

z = (13 - 15)/1.556 = - 1.29

Using the z-table at significance level of 0.05.

p-value = 1 - 0.9115 (from the z-tables)

p-value = 0.0885

Since the p-value is more than the significance level (0.0885 > 0.05), we do not reject the null hypothesis.

Hence, we accept the null hypothesis and can conclude that the mean waiting time is not significantly less than the the 15 minute claim by the taxpayer advocate.

Hope this Helps!!!

The conclusion of the hypothesis is that; the mean waiting time is not significantly less than the the 15 minute claim by the taxpayer advocate.

What is the P-value of the hypothesis?

We are given;

Sample size; n = 50

Mean waiting time; x' = 13 minutes

Standard deviation; σ = 11 minutes

The null hypothesis is;

H₀: μ₀ = 15 minutes

The alternative hypothesis is;

Hₐ: μ₀ < 15 minutes.

Let us find the test statistic from the z-score formula;

z = (x - μ)/σₓ

Where;

σₓ = (σ/√n)

σₓ = (11/√50)

σₓ = 1.556

Thus;

z = (13 - 15)/1.556

z = - 1.29

From online p-value from z-score calculator with a significance value of 0.05, we have;

p-value = 0.0885

Since the p-value is more than the significance level we fail to reject the null hypothesis and conclude that the mean waiting time is not significantly less than the the 15 minute claim by the taxpayer advocate.

Read more about hypothesis testing at; https://brainly.com/question/15980493

Otras preguntas