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IRS Audits The IRS examined approximately $$1 \%$$ of individual tax returns for a specific year, and the average recommended additional tax per return was $$\$ 19,150 .$$ Based on a random sample of 50 returns, the mean additional tax was $$\$ 17,020 .$$ If the population standard deviation is $$\$ 4080,$$ is there sufficient evidence to conclude that the mean differs from $$\$ 19,150$$ at $$\alpha=0.05 ?\( Does a \)95 \%$$ confidence interval support this result?

Short Answer

Expert verified
There is sufficient evidence at \(\alpha = 0.05\) to conclude the mean differs from \$19,150. The 95% confidence interval also supports this result.

Step by step solution

01

Identify the Hypotheses

Start by identifying the null and alternative hypotheses. The null hypothesis \( H_0 \) states that the mean additional tax is \(19,150, whereas the alternative hypothesis \( H_1 \) is that the mean additional tax is not equal to \)19,150. These can be formulated as: \( H_0: \mu = 19,150 \) and \( H_1: \mu eq 19,150 \).
02

Calculate the Test Statistic

We need to calculate the test statistic using the formula \( z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \), where \( \bar{x} = 17,020 \), \( \mu = 19,150 \), \( \sigma = 4,080 \), and \( n = 50 \). Substituting these values, we find \( z = \frac{17,020 - 19,150}{4,080/\sqrt{50}} \).
03

Compute Standard Error and Z-Value

First, calculate the standard error by \( SE = \frac{4,080}{\sqrt{50}} \approx 577.26 \). Then the z-value is \( z = \frac{-2,130}{577.26} \approx -3.69 \).
04

Determine the Critical Z-Value

Given that the significance level \( \alpha = 0.05 \) and it's a two-tailed test, the critical z-values are approximately \( \pm 1.96 \).
05

Decision Making

Since the calculated z-value of \(-3.69\) is outside the critical z-values range of \(-1.96\) to \(1.96\), we reject the null hypothesis \( H_0 \). There is sufficient evidence to conclude the mean differs from \$19,150.
06

Construct the 95% Confidence Interval

Calculate the 95% confidence interval for the mean using the formula: \( \bar{x} \pm z_{\alpha/2} \times SE \), where \( z_{\alpha/2} = 1.96 \). Thus, the confidence interval becomes \( 17,020 \pm 1.96 \times 577.26 \).
07

Compute the Confidence Interval

Calculate the interval bounds: lower bound is \( 17,020 - 1.96 \times 577.26 \approx 15,888.78 \), and upper bound is \( 17,020 + 1.96 \times 577.26 \approx 18,151.22 \). The 95% confidence interval is \([15,889, 18,151]\).
08

Conclusion on Confidence Interval

Since \$19,150 does not lie within the confidence interval \([15,889, 18,151]\), this supports the rejection of the null hypothesis \( H_0 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Null Hypothesis
Understanding the Null Hypothesis is a fundamental part of hypothesis testing. It's essentially a statement that there is no effect or no difference, and it is assumed true until evidence suggests otherwise. In statistical tests, such as the one detailed in this exercise about IRS audits, the null hypothesis (denoted as\( H_0 \)) sets the benchmark.

Here, the null hypothesis is that the mean additional tax owed is \( \\(19,150 \). This suggests that any observed difference is simply due to random variation. To determine if this assumption holds, we need to check if our calculated test statistic falls into a rare zone (based on our chosen significance level). If it does, then we reject the null hypothesis, indicating our observed data provides enough evidence that the true mean likely differs from \( \\)19,150 \).
Alternative Hypothesis
The Alternative Hypothesis is a counterpart to the null hypothesis. When you conduct a hypothesis test, you are essentially choosing between these two possibilities. Unlike the null, the alternative hypothesis (denoted as \( H_1 \) or \( H_a \)) suggests that there is an effect or a difference.

For our IRS audit exercise, the alternative hypothesis is that the mean additional tax is not \( \\(19,150 \). This is a two-tailed statement because we are interested in deviations both above and below the specified mean. Reaching a conclusion to support the alternative hypothesis entails showing that the null hypothesis is unlikely given the data. In practical terms, rejecting the null in favor of the alternative suggests there is strong evidence that the mean tax amount differs from \( \\)19,150 \), pointing towards the accuracy of our sample calculations.
Confidence Interval
Confidence Intervals are ranges calculated from sample data that likely contain the true population parameter. In hypothesis testing, confidence intervals can provide a visual measure to cross-check findings from a hypothesis test.

In our example, a 95% confidence interval for the mean tax was calculated. It ranges from \( \\(15,889 \) to \( \\)18,151 \). This interval does not include the null hypothesis mean of \( \$19,150 \), lending support to our test result that the true mean is likely different from the null value. Essentially, confidence intervals offer a range estimate that includes the expected parameter 95% of the time in repeated sampling, thereby reinforcing the hypothesis test's conclusion.
Z-Value
The Z-Value is a crucial part of hypothesis testing in statistics. It tells us how far away our sample statistic is from the population mean under the null hypothesis, measured in standard deviations.

Here, we calculated a z-value of \(-3.69\). This was derived by finding how many standard errors our sample mean is from the null mean of \( \$19,150 \). The critical z-values at a 5% significance level in a two-tailed test are approximately -1.96 and 1.96. Our calculated z-value lies outside this range, which means it's in the rejection zone. This informs us to reject the null hypothesis, offering robust evidence that our observations significantly deviate from what we would expect under the null hypothesis. The z-value serves as a pivotal indicator in assessing whether our observed sample results are statistically significant.

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Most popular questions from this chapter

In what ways is the \(t\) distribution similar to the standard normal distribution? In what ways is the \(t\) distribution different from the standard normal distribution?

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For Exercises 5 through \(20,\) perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Female Physicians The percentage of physicians who are women is \(27.9 \%\). In a survey of physicians employed by a large university health system, 45 of 120 randomly selected physicians were women. Is there sufficient evidence at the 0.05 level of significance to conclude that the proportion of women physicians at the university health system exceeds \(27.9 \% ?\)

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