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In hypothesis testing, the null hypothesis is a statement that suggests there is no effect or no difference. It is denoted as \(H_0\). In this exercise, the null hypothesis is that the average time to change accounting jobs is equal to or less than 2 years. Mathematically, this is expressed as \(μ \leq 24\), where \(μ\) represents the average number of months it takes for individuals with a B.S. degree in accounting to change jobs.

This hypothesis acts as a starting point for statistical testing. The goal is to challenge the null hypothesis by collecting evidence to see if enough support exists to reject it in favor of the alternative hypothesis. By rejecting the null hypothesis, we infer that the observed sample provides significant evidence to propose a different hypothesis.

The alternative hypothesis, denoted as \(H_1\), represents what we aim to support through statistical testing. It suggests that there is an effect or difference. In the context of this problem, the alternative hypothesis claims that the average time to change accounting jobs is more than 2 years. This is mathematically represented as \(μ > 24\).

The alternative hypothesis is the opposite of the null hypothesis and is what we conclude if there is strong evidence against \(H_0\). Here, the purpose is to see if the data support this claim that individuals, on average, require more than 24 months to change jobs in the given field. If the test statistic falls into the critical region determined by the significance level, the alternative hypothesis is favored.

In hypothesis testing, especially when dealing with sample sizes larger than 30, the z-distribution is used due to the central limit theorem. The z-distribution is a standardized normal distribution and is key for determining how far and in what direction our sample statistic is from the mean posited by the null hypothesis.

The z-score is calculated to see how many standard deviations our sample mean (\(\bar{x}\)) is from the population mean (\(μ_0\)). The formula for this calculation is given by:

• \( z = \frac{\bar{x} - μ_0}{s/\sqrt{n}}, \)
  where \(\bar{x} = 35.02\), \(μ_0 = 24\), \(s = 18.94\), and \(n = 44\).
  
  The calculated z-value of 4.60 indicates how far away the sample mean is from the mean stated in the null hypothesis. If this value is beyond the critical value at our stated significance level, we have enough evidence to question \(H_0\).

The significance level, denoted by \(\alpha\), helps determine the threshold for rejecting the null hypothesis. It represents the probability of making a Type I error, which is rejecting a true null hypothesis. In this exercise, the significance level is set to 0.01 (or 1%).

This means that we have allowed for a 1% chance of incorrectly rejecting the null hypothesis when it is, in fact, true. At a significance level of 0.01, the critical z-value is approximately 2.33.

• We reject the null hypothesis if our calculated z-value exceeds this critical z-value, indicating that our sample provides substantial evidence against \(H_0\).

In our case, since the computed z-value of 4.60 is greater than 2.33, it provides strong evidence against the null hypothesis. Consequently, we conclude that individuals with a B.S. degree in accounting, on average, take more than 2 years to change jobs, at a 1% significance level.