Question

The report "Highest Paying Jobs for 2009-10 Bachelor's Degree Graduates" (National Association of Colleges and Employers, February 2010) states that the mean yearly salary offer for students graduating with accounting degrees in 2010 was \(\$ 48,722\). Suppose that a random sample of 50 accounting graduates at a large university who received job offers resulted in a mean offer of \(\$ 49,850\) and a standard deviation of \(\$ 3,300\). Do the sample data provide strong support for the claim that the mean salary offer for accounting graduates of this university is higher than the 2010 national average of \(\$ 48,722 ?\) Test the relevant hypotheses using \(\alpha=0.05 .\)

Short answer

Based on hypothesis testing, it can be concluded that there is strong support that the mean salary offer for accounting graduates of this university is higher than the 2010 national average of $48,722.

Step-by-step solution

Formulate Hypotheses

First, formulate the null and alternative hypotheses. The null hypothesis (H0) is that the university’s mean salary, denoted by µ1, is not higher than the 2010 national average of $48,722. H0: µ1 ≤ 48722. The alternative hypothesis (H1), counter to H0, is that the university’s mean salary is greater: H1: µ1 > 48722.

Compute Test Statistic

To apply the formula, fill in the known information like this: Sample mean x¯ = 49850, Population Mean µ = 48722, Standard Deviation σ= 3300, and Sample Size n = 50. The test statistic (Z) is calculated by the formula: Z = (x¯ - µ) / (σ/√n), which gives Z = (49850 - 48722) / (3300/√50) = 2.12.

Determine Critical Value

Given level of significance (α) is 0.05; since this is a right-tailed test, the critical value corresponding to 0.05 is 1.645.

Decision Making

With Z = 2.12 and critical value = 1.645, Z > critical value, reject H0 in favor of H1.

Key concepts

Null and Alternative Hypotheses

Understanding the null (\( H_0 \)) and alternative (\( H_1 \text{ or }H_a\text{ depending on notation }\)) hypotheses is foundational in hypothesis testing. The null hypothesis represents a statement of no effect or no difference and serves as the starting assumption for the test. For example, in the context of the exercise, it states that the mean salary offer from the university isn't higher than the national average (\( \$48,722 \text{ in this case }\)).

The alternative hypothesis is the claim we aim to support with evidence, it suggests that there is an effect or a difference. In our exercise, the alternative hypothesis is that the mean salary offer for the university graduates is indeed higher than the national average. Formulating these hypotheses correctly is a critical step because they directly influence the computation of the test statistic and the eventual conclusion of the test.

Test Statistic Computation

The computation of the test statistic is a procedure that converts the sample data into a single, standardized value that can be compared against a critical value to determine whether or not the null hypothesis should be rejected.

In the provided example, we use the following formula to compute the Z-test statistic:
\[ Z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}} \]
where \(\bar{x}\) is the sample mean, \(\mu_0\) is the population mean under the null hypothesis, \(\sigma\) is the standard deviation of the sample, and \(n\) is the sample size. This calculated Z-value helps us to understand if the observed sample mean is likely under the null hypothesis or if it's significantly different, leading us toward considering the alternative hypothesis.

Significance Level

The significance level, denoted as \( \alpha \text{ (alpha) }\), is the probability threshold at which we decide whether the evidence against the null hypothesis is strong enough to declare a statistically significant result. It's the risk we are willing to take of rejecting the null hypothesis when it is actually true, known as a Type I error.

Commonly used significance levels include 0.01, 0.05, and 0.10. In our exercise, we use \( \alpha = 0.05 \text{ or }5\%\) as our significance level. This means we accept a 5% probability of incorrectly rejecting the null hypothesis. The choice of the significance level affects the strictness of the hypothesis test and should be determined before any data analysis is performed.

Critical Value

The critical value is a point on the scale of the test statistic that is compared to the calculated test statistic to decide whether to reject the null hypothesis. It represents the threshold beyond which we consider the results as statistically significant, assuming the null hypothesis is true.

For a given significance level, the critical value can be found from statistical tables or computational tools. In a right-tailed test like in our example, where we are testing if the mean is greater than a certain value, the critical value is positive. Here, with an \( \alpha = 0.05 \) for a one-tailed test, our critical value is 1.645. If our test statistic exceeds this critical value, it indicates that such a result would be very unlikely if the null hypothesis were true.

Decision Making in Hypothesis Testing

The decision-making process in hypothesis testing is a logical conclusion based on the test statistic and critical value. If the test statistic falls beyond the critical value in the direction of the alternative hypothesis, we reject the null hypothesis. If not, we do not have enough evidence to reject it and thus fail to reject it.

In our exercise, the calculated Z-value is 2.12 which is greater than the critical value of 1.645. This tells us that the sample mean salary is statistically significantly higher than the national mean at the 0.05 significance level. Hence, we reject the null hypothesis in favor of the alternative hypothesis. This decision asserts that, based on our sample, the group of university graduates is earning more on average than the national mean salary for graduates in 2010.