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 minutes before being able to talk to an IRS employee. Suppose you select a sample of callers after the new phone service has been implemented; the sample results show a mean waiting time of minutes before an IRS employee comes on line. Based upon data from past years, you decide is it reasonable to assume that the standard deviation of waiting times is minutes 11?

The calculated p-value is greater than the significance level at which the test was performed, hence, we fail to reject the null hypothesis & conclude that there is no significant evidence to say that the actual mean waiting time turned out to be significantly less than the 15-minute claim made by the taxpayer advocate.

That is, the true mean waiting time is equal to or greater than the 15-minute claim by the taxpayer advocate.

Step-by-step explanation:

For hypothesis testing, we first clearly state our null and alternative hypothesis.

For hypothesis testing, the first thing to define is the null and alternative hypothesis.

In hypothesis testing, especially one comparing two sets of data, the null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the direction of the test.

The alternative hypothesis usually confirms the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the direction of the test.

For this question, we are to investigate 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 evidence to say that the actual mean waiting time turned out to be significantly less than the 15-minute claim made by the taxpayer advocate. That is, the true mean waiting time is equal to or greater than the 15-minute claim by the taxpayer advocate.

The alternative hypothesis is that there is significant evidence to suggest that the actual mean waiting time turned out to be significantly less than the 15-minute claim made by the taxpayer advocate.

This is evidently a one tail hypothesis test (we're investigating only in one direction; less than the claim

Mathematically, the null hypothesis is

H₀: μ ≥ 15

The alternative hypothesis is

Hₐ: μ < 15 minutes

To do this test, we will use the z-distribution because the population standard deviation is known.

So, we compute the z-test statistic

z = (x - μ₀)/σₓ

x = sample mean = 13 minutes

μ₀ = the advocate's claim = 15 minutes

σₓ = standard error of the poll proportion = (σ/√n)

where n = Sample size = 50

σ = population standard deviation = 11 minutes.

σₓ = (σ/√n) = (11/√50) = 1.556

z = (13 - 15) ÷ 1.556 = -1.29

checking the tables for the p-value of this z-statistic

p-value (for z = -1.29, at 0.05 significance level, with a one tailed condition) = 0.098525

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 5% = 0.05

p-value = 0.098525

0.098525 > 0.05

Hence,

p-value > significance level

This means that we fail to reject the null hypothesis & conclude that there is no significant evidence to say that the actual mean waiting time turned out to be significantly less than the 15-minute claim made by the taxpayer advocate.

That is, the true mean waiting time is equal to or greater than the 15-minute claim by the taxpayer advocate.

Hope this Helps!!!

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step-by-step explanation:

this is answer of 144 cubic meters