Question

A sample of 105 sanitation workers for the city of Euonymus, Texas, earns an average of \(\$ 24,375\) per year. The average salary for all Euonymus city workers is \(\$ 24,230\), with a standard deviation of \(\$ 523\). Are the sanitation workers overpaid? Conduct both one- and two-tailed tests.

Short answer

The sanitation workers are overpaid as their salaries are significantly higher than the average city worker salary.

Step-by-step solution

- State the Hypotheses

Define the null hypothesis (H_0) and the alternative hypothesis (H_A) for both the one-tailed and two-tailed tests. For one-tailed test (to check if sanitation workers are overpaid): \[H_0: \mu \leq \ 24,230\] \[H_A: \mu > \ 24,230\] For two-tailed test (to check if sanitation workers' pay is simply different from the average city worker): \[H_0: \mu \ = \ 24,230\] \[H_A: \mu \ eq \ 24,230\]

- Calculate the Test Statistic

Calculate the Z statistic using the formula: \[Z = \frac{(\bar{X} - \mu)}{(\sigma / \sqrt{n})}\]where \bar{X} = 24,375, \mu = 24,230, \sigma = 523, and n = 105.Plug the values into the formula: \[Z = \frac{(24,375 - 24,230)}{(523 / \sqrt{105})} \approx 2.56 \]

- Determine the Critical Value

For the one-tailed test at a significance level of 0.05, the critical value is approximately 1.645.For the two-tailed test at a significance level of 0.05, the critical values are approximately ±1.96.

- Compare Test Statistic and Critical Values

One-tailed test: Compare the calculated Z value (2.56) to the critical value (1.645). Since 2.56 > 1.645, reject the null hypothesis. Two-tailed test: Compare the calculated Z value (2.56) to the critical values (±1.96). Since 2.56 > 1.96, reject the null hypothesis.

- Interpret the Results

Since the null hypothesis is rejected in both tests, it can be concluded that the average salary of sanitation workers is significantly higher than that of all city workers, indicating they are overpaid.

Key concepts

one-tailed test

In hypothesis testing, a one-tailed test is used when the research question is focused on determining whether a population parameter is either greater than or less than a certain value. For example, in our exercise, we tested if the sanitation workers' average salary is greater than \( \$24,230 \).

To perform a one-tailed test:

• State your null hypothesis (H\textsubscript{0}): \( \mu \leq 24,230 \). This suggests that the average salary is not greater than the city workers' average.
• State your alternative hypothesis (H\textsubscript{A}): \( \mu > 24,230 \). This indicates that the average salary is actually higher.
• Calculate the test statistic.
• Determine the critical value from the standard normal distribution for your chosen significance level (typically 0.05).
• Compare your test statistic to the critical value to decide whether to reject your null hypothesis.

In our example, the test statistic z was calculated to be approximately 2.56, which is greater than the critical value of 1.645, leading us to reject the null hypothesis and infer that sanitation workers are indeed overpaid.

two-tailed test

A two-tailed test is used in hypothesis testing when we are interested in detecting any significant difference from a specific value, without concern for direction. For instance, checking if the sanitation workers' average salary is simply different (either higher or lower) from \( \$24,230 \).

To perform a two-tailed test:

• State your null hypothesis (H\textsubscript{0}): \( \mu = 24,230 \). This suggests that the average salary is equal to the city workers' average.
• State your alternative hypothesis (H\textsubscript{A}): \( \mu eq 24,230 \). This indicates that the salary is different than \( \$24,230 \).
• Calculate the test statistic.
• Determine the critical values from the standard normal distribution for your significance level (e.g., ±1.96 for 0.05 significance level).
• Compare your test statistic to the critical values to decide whether to reject your null hypothesis.

In this case, the test statistic (2.56) was outside the range of −1.96 to 1.96, so we rejected the null hypothesis and concluded that the average salary is indeed different from the average city worker salary.

test statistic

The test statistic is a standardized value that is calculated from sample data during a hypothesis test. It is used to decide whether to reject the null hypothesis.

In our example, we computed the Z statistic using the formula:
\[ Z = \frac{(\bar{X} - \mu)}{(\sigma / \sqrt{n})} \]
Where:

• \( \bar{X} \) is the sample mean (24,375)
• \( \mu \) is the population mean (24,230)
• \( \sigma \) is the standard deviation (523)
• \( n \) is the sample size (105)

Plugging in the values, we found:
\[ Z = \frac{(24,375 - 24,230)}{(523 / \sqrt{105})} \approx 2.56 \]
The test statistic serves as a measure of how far our sample mean is from the population mean, in terms of standard deviations.

critical value

A critical value is a point on the test distribution that is compared to the test statistic to determine whether to reject the null hypothesis. The critical value depends on the significance level (\alpha) of the test.

For example, common significance levels are 0.05, 0.01, and 0.10.

• For a one-tailed test at \( \alpha = 0.05 \), the critical value is 1.645.
• For a two-tailed test at \( \alpha = 0.05 \), the critical values are ±1.96.

In the given exercise:

• For the one-tailed test, our Z statistic (2.56) was greater than the critical value (1.645), leading us to reject the null hypothesis.
• For the two-tailed test, our Z statistic (2.56) was beyond the critical values of ±1.96, again leading us to reject the null hypothesis.

Critical values help in making the final decision in hypothesis testing by setting a threshold beyond which the null hypothesis is unlikely to be true.

null hypothesis

In hypothesis testing, the null hypothesis (H\textsubscript{0}) is the assumption that there is no effect or no difference, and it serves as the starting point for testing statistical significance.

In our exercise, the null hypotheses were:

• For the one-tailed test: \( H\textsubscript{0}: \mu \leq 24,230 \). This suggests that sanitation workers are not overpaid.
• For the two-tailed test: \( H\textsubscript{0}: \mu = 24,230 \). This states that there is no difference in salaries.

The purpose of the hypothesis test is to determine if there is enough evidence to reject the null hypothesis.

It's essential to note that rejecting the null hypothesis doesn't prove it false with absolute certainty; rather, it indicates that there's strong enough evidence to support the alternative hypothesis.

alternative hypothesis

The alternative hypothesis (H\textsubscript{A}) is what you aim to support in a hypothesis test. It suggests that there is an effect or a difference. When the null hypothesis is rejected, it provides support for the alternative hypothesis.

In our example, the alternative hypotheses were:

• For the one-tailed test: \( H\textsubscript{A}: \mu > 24,230 \). This suggests sanitation workers are overpaid.
• For the two-tailed test: \( H\textsubscript{A}: \mu eq 24,230 \). This suggests there is a difference in salaries either way.

Supporting the alternative hypothesis indicates there is significant evidence to show an effect or difference, based on the data.

In both cases in the exercise, the results led to rejecting the null hypotheses which provided evidence to support that sanitation workers' salaries are indeed significantly different and higher than the average salary of city workers.