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Auditors must make judgments about various aspects of an audit on the basis of their own direct experience, indirect experience, or a combination of the two. In a study, auditors were asked to make judgments about the frequency of errors to be found in an audit. The judgments by the auditors were then compared to the actual results. Suppose the following data were obtained from a similar study; lower scores indicate better judgments. \(\begin{array}{ccc}\text { Direct } & \text { Indirect } & \text { Combination } \\ 17.0 & 16.6 & 25.2 \\ 18.5 & 22.2 & 24.0 \\ 15.8 & 20.5 & 21.5 \\\ 18.2 & 18.3 & 26.8 \\ 20.2 & 24.2 & 27.5 \\ 16.0 & 19.8 & 25.8 \\ 13.3 & 21.2 & 24.2\end{array}\) Use \(\alpha=.05\) to test to see whether the basis for the judgment affects the quality of the judgis

Short Answer

Expert verified
Conduct an ANOVA test to compare group means. Use $\alpha=0.05$.

Step by step solution

01

Define Hypotheses

We begin by stating the null and alternative hypotheses. The null hypothesis (\(H_0\)) is that there is no difference in the quality of judgments based on the basis of judgment: direct, indirect, or combination. The alternative hypothesis (\(H_a\)) is that there is a difference in the quality of judgments:\[ H_0: \mu_{\text{Direct}} = \mu_{\text{Indirect}} = \mu_{\text{Combination}} \]\[ H_a: \mu_{\text{Direct}} eq \mu_{\text{Indirect}} eq \mu_{\text{Combination}} \]
02

Choose the Statistical Test

We use a one-way ANOVA test because we are comparing more than two groups (Direct, Indirect, and Combination) for a single factor (quality of judgment scores). The ANOVA test will help us determine if there is a statistically significant difference among the group means at \(\alpha = 0.05\). Financial auditors' judgments are compared as groups and not as paired values.
03

Calculate Group Means

Calculate the mean for each group. For Direct: \( \bar{x}_{\text{Direct}} = \frac{17.0 + 18.5 + 15.8 + 18.2 + 20.2 + 16.0 + 13.3}{7} \approx 17.14 \). Similarly, compute the means for Indirect and Combination.
04

Perform ANOVA Calculation

We calculate the F-statistic for the ANOVA test using the formula:\[ F = \frac{MS_{\text{Between}}}{MS_{\text{Within}}} \]where \( MS_{\text{Between}} \) is the mean square between the groups and \( MS_{\text{Within}} \) is the mean square within the groups.The ANOVA table will require individual sums of squares values, which would then be divided by their degrees of freedom to get mean squares.
05

Make Decision

Compare the calculated F-value with the critical F-value from the F-distribution table using \((\alpha = 0.05)\) and degrees of freedom for between (2) and within (21). If the calculated F-value exceeds the critical value, we reject \(H_0\).
06

Conclusion

Based on the F-test results, if we reject $H_0$, it indicates that there is a significant effect of the judgment basis on the quality of judgments. Otherwise, if we fail to reject $H_0$, there is no significant effect.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a crucial method in statistics used to examine an assumption about a population parameter. Think of it as a way to test whether your claims about data are likely to be true based on sample evidence. In our scenario with auditors, hypothesis testing helps determine if different methods (direct, indirect, or combination) affect the quality of audit judgments. We start by setting up two competing hypotheses: the null hypothesis (\(H_0\)), which indicates no effect or difference between the groups, and the alternative hypothesis (\(H_a\)), which suggests that there is a difference. This means the null hypothesis suggests that the mean judgment quality across all methods is the same, while the alternative hypothesis suggests otherwise. By applying statistical tests like ANOVA, we can infer from the sample data if our null hypothesis can be supported or should be rejected. The entire process entails model selection, calculation, and decision-making.

Hypothesis testing isn't just about running calculations; it's about structured thinking to support or contest claims, thus adding scientific rigor to auditor judgments.
Statistical Significance
Statistical significance is a pivotal concept in hypothesis testing that indicates how likely it is for a result to have occurred by chance. The essence of statistical significance is setting a threshold, known as alpha (\(\alpha\)), which is used to judge the results. Here, we consider \(\alpha = 0.05\), a common choice indicating a 5% risk of concluding that a difference exists when there is none.

In our auditing scenario, achieving statistical significance through ANOVA implies that there is a genuine effect of the judgment basis on the quality of judgments—not a fluke or random occurrence. This helps auditors understand that their choice of information basis (direct, indirect, combination) likely alters the accuracy of their error detection measures.

It's important to remember that statistical significance does not measure the size of an effect or its real-world relevance—just that an effect is unlikely to have occurred purely by chance, providing deeper insights into judgment quality in financial audits.
Audit Judgment
Audit judgment refers to the processes and decisions auditors use to evaluate financial information and ascertain whether it conforms to required standards. In practice, auditors blend empirical data with professional experience to make informed decisions. This exercise, involving the use of ANOVA, helps determine how the basis of judgment (direct, indirect, or combination) affects auditors' ability to accurately predict errors.

Low scores depict a better quality of judgment, suggesting the auditor was closer in their error estimation to the actual number. ANOVA helps unveil if the method of gathering insights significantly influences this accuracy. Such findings can inform training programs, suggesting whether more reliance on one method is typically advantageous over another.

Auditors aim to minimize judgment errors, thus improving audit reliability and stakeholder trust. By statistically probing these judgments, auditors enhance their methodology to align closer with data-driven audit standards, contributing to overall financial integrity.

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Most popular questions from this chapter

Money magazine reports percentage returns and expense ratios for stock and bond funds. The following data are the expense ratios for 10 midcap stock funds, 10 small-cap stock funds, 10 hybrid stock funds, and 10 specialty stock funds (Money, March 2003 ). \(\begin{array}{cccc}\text { Midcap } & \text { Small-Cap } & \text { Hybrid } & \text { Specialty } \\ 1.2 & 2.0 & 2.0 & 1.6 \\\ 1.1 & 1.2 & 2.7 & 2.7 \\ 1.0 & 1.7 & 1.8 & 2.6 \\ 1.2 & 1.8 & 1.5 & 2.5 \\\ 1.3 & 1.5 & 2.5 & 1.9 \\ 1.8 & 2.3 & 1.0 & 1.5 \\ 1.4 & 1.9 & 0.9 & 1.6 \\\ 1.4 & 1.3 & 1.9 & 2.7 \\ 1.0 & 1.2 & 1.4 & 2.2 \\ 1.4 & 1.3 & 0.3 & 0.7\end{array}\) Use \(\alpha=.05\) to test for any significant difference in the mean expense ratio among the four types of stock funds.

Consider the following data for two independent random samples taken from two normal populations. $$\begin{array}{l|rrrrrr} \text { Sample 1 } & 10 & 7 & 13 & 7 & 9 & 8 \\\\\hline \text { Sample 2 } & 8 & 7 & 8 & 4 & 6 & 9\end{array}$$ a. Compute the two sample means. b. Compute the two sample standard deviations. c. What is the point estimate of the difference between the two population means? d. What is the \(90 \%\) confidence interval estimate of the difference between the two population means?

The following results come from two independent random samples taken of two populations. \(\begin{array}{ll}\text { Sample } 1 & \text { Sample } 2 \\ n_{1}=50 & n_{2}=35 \\ \bar{x}_{1}=13.6 & \bar{x}_{2}=11.6 \\ \sigma_{1}=2.2 &\sigma_{2}=3.0\end{array}\) a. What is the point estimate of the difference between the two population means? b. Provide a \(90 \%\) confidence interval for the difference between the two population means. c. Provide a \(95 \%\) confidence interval for the difference between the two population means.

Consider the following hypothesis test. \(H_{0}: \mu_{1}-\mu_{2} \leq 0\) \(H_{\mathrm{a}}: \mu_{1}-\mu_{2}>0\) The following results are for two independent random samples taken from the two populations. \(\begin{array}{ll}\text { Sample } 1 & \text { Sample } 2 \\\ n_{1}=40 & n_{2}=50 \\ \bar{x}_{1}=25.2 & \bar{x}_{2}=22.8 \\ \sigma_{1}=5.2 & \sigma_{2}=6.0\end{array}\) a. What is the value of the test statistic? b. What is the \(p\) -value? c. With \(\alpha=.05,\) what is your hypothesis testing conclusion?

In early 2009 , the economy was experiencing a recession. But how was the recession affecting the stock market? Shown are data from a sample of 15 companies. Shown for each company is the price per share of stock on January 1 and April 30 (The Wall Street Journal, May 1,2009 ). a. What is the change in the mean price per share of stock over the four-month period? b. Provide a \(90 \%\) confident interval estimate of the change in the mean price per share of stock. Interpret the results. c. What was the percentage change in the mean price per share of stock over the fourmonth period? d. If this same percentage change were to occur for the next four months and again for the four months after that, what would be the mean price per share of stock at the end of the year \(2009 ?\)

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