Question

For Exercises 5 through \(20,\) perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. According to the Women's Bureau of the U.S. Department of Labor, the occupation with the highest median weekly earnings among women is pharmacist with median weekly earnings of \(\$ 1603 .\) Based on the weekly earnings listed from a random sample of female pharmacists, can it be concluded that the median is less than \(\$ 1603 ?\) Use \(\alpha=0.05 .\) $$ \begin{array}{lll} 1550 & 1355 & 1777 \\ 1430 & 1570 & 1701 \\ 2465 & 1655 & 1484 \\ 1429 & 1829 & 1812 \\ 1217 & 1501 & 1449 \end{array} $$

Short answer

There is insufficient evidence to conclude that the median is less than \(\$1603\).

Step-by-step solution

State the Hypotheses

We set up our null and alternative hypotheses as follows. The null hypothesis, \( H_0 \), states that the median weekly earnings of female pharmacists is at least \( \\(1603 \). Hence, \( H_0: \text{median} \geq \\)1603 \). The alternative hypothesis, \( H_1 \), claims that the median is less than \( \\(1603 \): \( H_1: \text{median} < \\)1603 \).

Identify the Claim

The claim in this problem is that the median weekly earnings is less than \( \$1603 \). Thus, our claim corresponds to the alternative hypothesis \( H_1 \).

Determine the Critical Value(s)

Since we are testing whether the median is less than a value, we will conduct a left-tailed test. For a significance level of \( \alpha = 0.05 \), the critical value for a normal distribution can be found using a Z-table. The critical value is approximately \(Z = -1.645\).

Compute the Test Value

### Calculation of the Test StatisticFirstly, arrange the data in ascending order and identify the number of values below \( \\(1603 \): \(1217, 1355, 1429, 1430, 1449, 1484, 1501, 1550, 1570, 1655, 1701, 1777, 1812, 1829, 2465 \)Count the number of data points below \( \\)1603 \). There are 8 values below \( \$1603 \) out of a total of 15 data points.To conduct the sign test for median, we calculate the test statistic,\[ Z = \frac{X - np}{\sqrt{npq}} \]Here, \( n = 15 \), \( p = 0.5 \), \( q = 1-p = 0.5 \), and \( X = 8 \).\[ Z = \frac{8 - 15(0.5)}{\sqrt{15(0.5)(0.5)}} \approx \frac{0.5}{1.936} \approx 0.258 \]

Make the Decision

Compare the computed test value \( Z = 0.258 \) with the critical value \( Z = -1.645 \). Since \( 0.258 \) is greater than \( -1.645 \), we do not reject the null hypothesis \( H_0 \).

Summarize the Results

The results suggest that there is not enough statistical evidence to support the claim that the median weekly earnings of female pharmacists is less than \( \$1603 \) at the \( 0.05 \) significance level.

Key concepts

Significance Level

In hypothesis testing, the significance level, often denoted as \( \alpha \), plays a crucial role. It is a predefined threshold used to decide whether to reject the null hypothesis. A common significance level is \( \alpha = 0.05 \), meaning there is a 5% risk of concluding that a difference exists when there is no actual difference. In other words, it represents the probability of making a Type I error, which is rejecting a true null hypothesis.
Choosing a smaller significance level, such as \( \alpha = 0.01 \), means you're being more cautious and demanding more evidence before rejecting the null hypothesis. Conversely, a larger \( \alpha \) increases the risk of making a Type I error. Always choose your significance level before conducting the test to avoid bias in data interpretation.

Null Hypothesis

The null hypothesis, represented as \( H_0 \), is a statement of no effect or no difference. In the context of the given exercise, the null hypothesis asserts that the median weekly earnings of female pharmacists is at least \( \$1603 \). The null hypothesis is typically what you aim to test, rather than prove.
The null hypothesis provides a baseline or a point of comparison. Like a status quo, it assumes that any deviation in the sample results is due to chance. If the null hypothesis is not rejected, it does not prove the hypothesis but suggests insufficient evidence to support the alternative hypothesis. It's important to craft a clear and testable null hypothesis to properly set up your hypothesis test.

Alternative Hypothesis

Contrary to the null hypothesis, the alternative hypothesis, denoted as \( H_1 \) or \( H_a \), suggests that there is an effect or a difference. For the exercise under discussion, the alternative hypothesis claims that the median is less than \( \$1603 \). This hypothesis is what researchers typically want to prove.
When you conduct a hypothesis test, you’re really testing whether there’s enough evidence to accept the alternative hypothesis over the null. If the test results lead to rejecting \( H_0 \), you accept the alternative hypothesis. The alternative hypothesis is directional in this case, indicating that you're conducting a one-tailed test. It is crucial to define your alternative hypothesis clearly to ensure that your test measures the correct effect.

Critical Value

The critical value is a key concept in hypothesis testing that dictates the boundary or cutoff point for deciding whether to reject the null hypothesis. When you perform a test, you calculate a test statistic from your sample data, which is then compared to the critical value. If the test statistic falls beyond the critical value, the null hypothesis is rejected.
In the exercise, since a left-tailed test is used with a significance level of \( \alpha = 0.05 \), the critical value is approximately \( Z = -1.645 \). Using critical values from statistical tables, like the Z-table for normal distribution, helps set the rejection region of a hypothesis test. It serves as a vital point of reference for interpreting your test results.

Test Statistic

The test statistic is a standardized value that results from the test conducted, providing a basis for making a decision about the hypotheses. It measures how far your sample statistic is from the null hypothesis. In this scenario, you calculate the Z value using the formula:

• \( n = 15 \)
• \( p = 0.5 \)
• \( q = 1-p = 0.5 \)
• \( X = 8 \)
The computed Z value of approximately 0.258 indicates how your sample statistic compares against theoretical distribution.
Once calculated, this Z value is compared against the determined critical value. By comparing the test statistic to the critical value, you determine whether to reject the null hypothesis. In the exercise, since 0.258 is greater than -1.645, the null hypothesis was not rejected, concluding that there isn't enough evidence to claim the median earnings is less than \( \$1603 \).