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.
No, you cannot conclude that the actual mean waiting time is significantly less than the 12-minute claim made by the taxpayer advocate. To do this, you would need to calculate the z-score for the sample mean of 10 minutes and compare it to the z-score for the 12-minute claim. Since the z-score for the sample mean is not greater than the z-score for the 12-minute claim, you cannot reject the null hypothesis that the mean waiting time is 12 minutes.
To determine if the actual mean waiting time is significantly less than the 12-minute claim made by the taxpayer advocate, we can perform a hypothesis test using the given sample data.
Step 1: Set up the hypotheses:
- Null hypothesis (H0): The actual mean waiting time is equal to or greater than 12 minutes. μ ≥ 12
- Alternative hypothesis (Ha): The actual mean waiting time is less than 12 minutes. μ < 12
Step 2: Determine the level of significance:
The significance level (α) is given as ex = 0.05. This means that we are willing to accept a 5% chance of making a Type I error (rejecting the null hypothesis when it is true).
Step 3: Compute the test statistic:
The test statistic for this case is the z-score, which can be calculated using the formula:
z = (x̄ - μ) / (σ / √n),
where x̄ is the sample mean, μ is the claimed mean waiting time, σ is the known standard deviation, and n is the sample size.
In this case:
x̄ = 10 minutes
μ = 12 minutes
σ = 8 minutes
n = 50
Calculating the z-score:
z = (10 - 12) / (8 / √50) ≈ -1.77
Step 4: Determine the critical value:
Since the alternative hypothesis is one-tailed (less than), we need to find the critical value corresponding to a left-tailed test with a 5% significance level.
Using a standard normal distribution table or a z-table, the critical value for a 5% level of significance is approximately -1.645.
Step 5: Make a decision:
If the test statistic falls below the critical value, we can reject the null hypothesis.
In this case, the z-score (-1.77) is smaller than the critical value (-1.645), which means it falls in the rejection region. Therefore, we reject the null hypothesis.
Step 6: Draw a conclusion:
Based on the sample results, we have sufficient evidence to conclude that the actual mean waiting time is significantly less than the 12-minute claim made by the taxpayer advocate at a 5% level of significance.
Note: It's important to remember that this conclusion is based on the assumption that the sample is representative and the data follows a normal distribution.
To determine if the actual mean waiting time is significantly less than the 12-minute claim made by the taxpayer advocate, we can perform a hypothesis test.
Step 1: State the hypotheses:
- Null hypothesis (H0): The actual mean waiting time is equal to 12 minutes.
- Alternative hypothesis (Ha): The actual mean waiting time is less than 12 minutes.
Step 2: Set the significance level:
The significance level (alpha) is given as 0.05, which means we want to be 95% confident in our conclusion.
Step 3: Compute the test statistic:
We will use a one-sample t-test since we have a sample mean and the population standard deviation is known.
The formula for the test statistic (t) is:
t = (sample mean - hypothesized mean) / (standard deviation / √sample size)
In this case:
Sample mean (x̄) = 10 minutes
Hypothesized mean (μ) = 12 minutes
Standard deviation (σ) = 8 minutes
Sample size (n) = 50
Plugging these values into the formula, we get:
t = (10 - 12) / (8 / √50) ≈ -1.118
Step 4: Determine the critical value:
We need to determine the critical value for a one-tailed t-test at a 95% confidence level (since we have a less than alternative hypothesis).
Using a t-table or a t-distribution calculator with 49 degrees of freedom (n-1), we find the critical value to be approximately -1.676.
Step 5: Make a decision:
Compare the test statistic (t) to the critical value. If the test statistic is less than the critical value, we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.
Since -1.118 > -1.676, we fail to reject the null hypothesis.
Step 6: Draw a conclusion:
Based on the sample results, there is not enough evidence to conclude that the actual mean waiting time is significantly less than the 12-minute claim made by the taxpayer advocate. The sample mean of 10 minutes does not provide sufficient evidence to support the alternative hypothesis that the mean waiting time is less than 12 minutes.
Please note that this conclusion is specific to the given sample and assumes the sample is representative of the population.