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A random sample of \(n=44\) individuals with a B.S. degree in accounting who started with a Big Eight accounting firm and subsequently changed jobs resulted in a sample mean time to change of \(35.02\) months and a sample standard deviation of \(18.94\) months ("The Debate over Post-Baccalaureate Education: One University's Experience," Issues in Accounting Education \([1992]: 18-36)\). Can it be concluded that the true average time to change exceeds 2 years? Test the appropriate hypotheses using a significance level of \(.01\).

Short Answer

Expert verified
The conclusion depends on comparison of computed t-value with the critical t-value. Should the computed t-value be greater, this suggests that the true average time for accountants with a B.S. degree to change jobs significantly exceeds 2 years.

Step by step solution

01

State the Hypotheses

The null hypothesis \(H_0\) assumes the true average time to change jobs is equal to or less than 2 years (which is equivalent to 24 months). In math notation,\(H_0: \mu \leq 24\)The alternative hypothesis \(H_1\) assumes the true average time to change jobs is more than 2 years or 24 months. In math notation,\(H_1: \mu > 24\)
02

Compute the Test Statistic

We use the formula for the test statistic in a t-test for the population mean which is \(t = \frac{{\bar{x} - \mu}}{{s / \sqrt{n}}}\)Where:\(\bar{x}\) is the sample mean (35.02),\(\mu\) is the population mean under the null hypothesis (24),s is the sample standard deviation (18.94), andn is the sample size (44).On substituting these values, we calculate the t-score.
03

Find the Critical Value

Firstly, notice that we are conducting a right-tailed test because the alternative hypothesis has a 'greater than' symbol. The degrees of freedom for this test are \(n-1=44-1=43\), and the significance level is 0.01. To find the critical value, we refer to the t-distribution table.
04

Make A Decision

If the computed test statistic is greater than critical value, we reject the null hypothesis. Else, we fail to reject the null hypothesis.
05

Interpret the Results

If we reject the null hypothesis, we can conclude that the true average time for accountants with a B.S. degree to change jobs is significantly more than 2 years. If we fail to reject the null hypothesis, there isn't enough evidence to suggest that the true average time exceeds 2 years.

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Key Concepts

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

t-Test
The t-Test is a powerful statistical tool used for comparing the mean of a sample to a known value, typically a hypothesized population mean. In the context of our example calculation, this t-test helps determine if accountants with a B.S. degree in accounting take, on average, more than two years to change jobs. What makes the t-test valuable in this scenario is its ability to handle small sample sizes (like our 44 individuals) and unknown population standard deviations.
  • **Components**: The test statistic, derived from the sample mean, sample standard deviation, and the sample size.
  • **Application**: In our example, a right-tailed test determines if the average job-changing time exceeds 24 months.
The formula for the t-test statistic is given by:\[t = \frac{ \bar{x} - \mu }{ s / \sqrt{n} }\]where \( \bar{x} \) represents the sample mean, \( \mu \) is the population mean specified in the null hypothesis, \( s \) is the standard deviation, and \( n \) is the sample size. This structure allows the test to be adaptable, convenient, and informative for comparative statistical analysis.
Null Hypothesis
The Null Hypothesis, or \( H_0 \), is a starting assumption in hypothesis testing. It represents the status quo or a statement to be tested for validity. In any statistical test, we usually begin by assuming that the effect or relationship we are testing for does not exist. In our exercise, the null hypothesis states that the true average time for accountants to change jobs is equal to or less than two years (24 months):\[H_0: \mu \leq 24\]The null is essentially a hypothesis of no effect or no change. It provides a baseline that the t-test can compare against to determine any statistically significant results. By default, until evidence suggests otherwise, the null hypothesis is assumed true.
Alternative Hypothesis
The Alternative Hypothesis, or \( H_1 \), presents the opposite condition to the null hypothesis and indicates the presence of an effect or difference. This hypothesis is what the researcher aims to support. In the given problem, the alternative hypothesis suggests that accountants, on average, change jobs after more than two years. This can be expressed mathematically as:\[H_1: \mu > 24\]Here, the symbol ">" signifies a right-tailed test, meaning we're interested in findings showing a mean greater than 24 months. The alternative hypothesis is crucial since it determines how the data is tested and interpreted. If enough evidence exists against the null hypothesis, we may accept the alternative hypothesis, indicating the significant effect or difference we are testing for.
Significance Level
The Significance Level, denoted by \( \alpha \), is a threshold used in hypothesis testing to determine the cutoff for statistically significant results. It's the probability of rejecting the null hypothesis when it is actually true, a Type I error. For our problem, the significance level is set at 0.01 or 1%. This means we are willing to accept a 1% chance of mistakenly concluding that the average job switch time is greater than two years when it actually is not.
  • **Common Levels**: Commonly used significance levels include 0.05 (5%), 0.01 (1%), and 0.10 (10%).
  • **Impact on Decision**: The lower the significance level, the stronger the evidence needs to be to reject the null hypothesis.
Choosing a significance level is vital as it reflects the researcher’s tolerance for error and impacts the result's reliability. In practical terms, it helps balance between being cautious against false positives and being open to evidence of real effects.

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Most popular questions from this chapter

Optical fibers are used in telecommunications to transmit light. Current technology allows production of fibers that transmit light about \(50 \mathrm{~km}\) (Research at Rensselaer, 1984 ). Researchers are trying to develop a new type of glass fiber that will increase this distance. In evaluating a new fiber, it is of interest to test \(H_{0}: \mu=50\) versus \(H_{a}: \mu>50\), with \(\mu\) denoting the true average transmission distance for the new optical fiber. a. Assuming \(\sigma=10\) and \(n=10\), use Appendix Table 5 to find \(\beta\), the probability of a Type II error, for each of the given alternative values of \(\mu\) when a level \(.05\) test is employed: \(\begin{array}{llll}\text { i. } 52 & \text { ii. } 55 & \text { iii. } 60 & \text { iv. } 70\end{array}\) b. What happens to \(\beta\) in each of the cases in Part (a) if \(\sigma\) is actually larger than \(10 ?\) Explain your reasoning.

Medical personnel are required to report suspected cases of child abuse. Because some diseases have symptoms that mimic those of child abuse, doctors who see a child with these symptoms must decide between two competing hypotheses: \(H_{0}\) : symptoms are due to child abuse \(H_{a^{*}}\) symptoms are due to disease (Although these are not hypotheses about a population characteristic, this exercise illustrates the definitions of Type I and Type II errors.) The article "Blurred Line Between Illness, Abuse Creates Problem for Authorities" (Macon Telegraph, February 28,2000 ) included the following quote from a doctor in Atlanta regarding the consequences of making an incorrect decision: "If it's disease, the worst you have is an angry family. If it is abuse, the other kids (in the family) are in deadly danger." a. For the given hypotheses, describe Type I and Type II errors. b. Based on the quote regarding consequences of the two kinds of error, which type of error does the doctor quoted consider more serious? Explain.

Explain why the statement \(\bar{x}=50\) is not a legitimate hypothesis.

Do state laws that allow private citizens to carry concealed weapons result in a reduced crime rate? The author of a study carried out by the Brookings Institution is reported as saying, 'The strongest thing I could say is that \(\bar{I}\) don't see any strong evidence that they are reducing crime" (San Luis Obispo Tribune, January 23, 2003). a. Is this conclusion consistent with testing \(H_{0}:\) concealed weapons laws reduce crime versus \(H_{a}\) : concealed weapons laws do not reduce crime or with testing \(H_{0}\) : concealed weapons laws do not reduce crime versus \(H_{e}:\) concealed weapons laws reduce crime Explain. b. Does the stated conclusion indicate that the null hypothesis was rejected or not rejected? Explain.

Medical research has shown that repeated wrist extension beyond 20 degrees increases the risk of wrist and hand injuries. Each of 24 students at Cornell University used a proposed new mouse design, and while using the mouse their wrist extension was recorded for each one. Data consistent with summary values given in the paper "Comparative Study of Two Computer Mouse Designs" (Cornell Human Factors Laboratory Technical Report RP7992) are given. Use these data to test the hypothesis that the mean wrist extension for people using this new mouse design is greater than 20 degrees. Are any assumptions required in order for it to be appropriate to generalize the results of your test to the population of Cornell students? To the population of all university students? (data on next page) $$ \begin{array}{llllllllllll} 27 & 28 & 24 & 26 & 27 & 25 & 25 & 24 & 24 & 24 & 25 & 28 \\ 22 & 25 & 24 & 28 & 27 & 26 & 31 & 25 & 28 & 27 & 27 & 25 \end{array} $$

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