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Who wins disputed cases whenever there is a change made to the tax laws, the taxpayer or the Internal Revenue Service (IRS)? The latest trend indicates that the burden of proof in all court cases has shifted from taxpayers to the IRS, which tax experts predict could set off more intrusive questioning. Of the accountants, lawyers, and other tax professionals surveyed by RIA Group, a tax-information publisher, \(55 \%\) expect at least a slight increase in taxpayer wins. Suppose samples of 175 accountants and 165 lawyers were asked, "Do you expect taxpayers to win more court cases because of the new burden of proof rules?" Of those surveyed, 101 accountants replied "yes" and 84 lawyers said "yes." Do the two expert groups differ in their opinions? Use a 0.01 level of significance to answer the question. a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

Short Answer

Expert verified
If the p-value obtained in the p-value approach is less than the 0.01 level of significance, it indicates that the proportion of accountants who believe taxpayers will win more court cases is significantly different from the proportion of lawyers who believe the same. The same conclusion would follow from the classical approach if the test statistic is greater than the critical value. This would imply that the two expert groups differ in their opinions.

Step by step solution

01

Set up the Hypotheses

Null hypothesis, \(H_0: p_A = p_L\), where \(p_A\) is the population proportion of accountants who believe taxpayers will win more court cases and \(p_L\) is the population proportion of lawyers who believe the same.\n\nAlternative hypothesis, \(H_1: p_A ≠ p_L\). Based on this, any result that rejects the null hypothesis will mean there is a difference in belief between accountants and lawyers.
02

Solve using the p-value approach

Calculate the pooled sample proportion (\(p\)), which is the total number of 'yes' responses divided by the total number of responses: \((101+84) / (175 + 165)\).\n\nCalculate the standard error using: \(\sqrt{ p(1 − p) [(1/175) + (1/165)]}\).\n\nCalculate the test statistic (\(z\)) using: \((p_A - p_L) / SE\), where \(SE\) is the standard error, \(p_A\) is the proportion of accountants who responded 'yes', and \(p_L\) is the proportion of lawyers who responded 'yes'. The \(z\) value will then be compared to a critical value from the z-table corresponding to the 0.01 significance level.\n\nThe \(p\)-value is then calculated using the test statistic. If the \(p\)-value is less than the level of significance (0.01), the null hypothesis is rejected, indicating a significant difference between the two expert groups.
03

Solve using the Classical Approach

Under the classical approach, instead of computing a \(p\)-value, we compare our test statistic directly to our critical value. We reject the null hypothesis if our test statistic is greater than our critical value.\n\nUsing a significance level of 0.01, and since this is a two-tailed test, the critical values are the z-scores below which 0.005 (0.01/2) of the distribution lies. These critical values can be found from a z-table. If the test statistic computed in Step 2 lies beyond either of these critical values, the null hypothesis is rejected.

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Key Concepts

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

P-Value Approach
In the realm of hypothesis testing, the p-value approach plays a critical role. This method involves calculating the probability of obtaining test results at least as extreme as the results actually observed, under the assumption that the null hypothesis is true.

For example, in the exercise provided, where accountants and lawyers were surveyed about their expectations of taxpayer wins in court due to tax law changes, the p-value calculated represents the probability of observing the difference in 'yes' responses between the two groups, given that there is no true difference in their opinions. If this p-value is smaller than our predetermined significance level (in this case, 0.01), we conclude that the observed difference is statistically significant, hence rejecting the null hypothesis.

The calculation involves a few steps, starting with finding the pooled proportion of 'yes' responses and then determining the standard error of the difference between the two sample proportions. With these values, the test statistic (often a z-score) is computed. By comparing the p-value to the significance level, we can make a decision: a p-value less than 0.01 indicates a significant difference in opinions between accountants and lawyers.
Null and Alternative Hypotheses
In hypothesis testing, the null hypothesis, denoted as is a statement of no effect or no difference. It's the starting point of any hypothesis test and is presumed true until evidence suggests otherwise. The alternative hypothesis, denoted as , reflects the outcome that the experiment aims to demonstrate or support — it is considered only when we have sufficient evidence against the null hypothesis.

In the exercise about the accountants and lawyers, the null hypothesis asserts that there is no difference in opinion regarding the tax law changes (). The alternative hypothesis argues the contrary, suggesting a difference exists in the proportion of 'yes' responses between the two groups (). The selected approach, whether it's the p-value or the classical method, will examine whether observed data provide enough proof to reject the null hypothesis in favor of the alternate one.
Statistical Significance
The concept of statistical significance is deeply tied to hypothesis testing as it helps determine whether the results observed in a study or experiment are unlikely to have occurred by chance. A result is statistically significant if it is unlikely to have occurred if the null hypothesis were true. The level of statistical significance is determined by the significance level, or alpha (), a threshold chosen by the researcher prior to examining the data.

In practical terms, if the p-value is less than the chosen significance level, like 0.01 in our tax law opinion exercise, we can reject the null hypothesis and conclude that the results are statistically significant. This threshold helps to control the probability of making a Type I error, which is the incorrect rejection of a true null hypothesis. A lower alpha level means that you require more substantial evidence to reject the null hypothesis, reducing the chance of a false positive result.
Two-Sample Proportion Test
The two-sample proportion test is a statistical procedure used to compare the proportions of success between two independent groups. In the case of our tax law challenge, we compare the proportion of accountants who expect taxpayer wins to the proportion of lawyers who have the same expectation.

To execute this test, we start by assuming that the population proportions are equal (the null hypothesis). We then calculate a pooled sample proportion, which is a combined estimate of the two success rates assumed to be the same under the null hypothesis. From there, we determine the standard error of the difference in proportions and use it to find the z-score, which measures how far, in standard error units, our sample statistic lies from the null hypothesis value.

If the absolute value of this z-score is higher than the critical value associated with our chosen significance level, or if the p-value is lower than this significance level, we reject the null hypothesis. The test thus helps to assess if the observed discrepancy between the accountant and lawyer groups' opinions is statistically meaningful or simply due to random chance.

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