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School Workers In Exercise \(8.109,\) the average hourly wage for public school cafeteria workers was given as \(\$ 10.33 .^{26}\) If \(n=40\) randomly selected public school cafeteria workers within one school district are found to have an average hourly wage of \(\bar{x}=\) \(\$ 9.75\) with a standard deviation of \(s=\$ 1.65,\) would this information contradict the reported average of \(\$ 10.33 ?\) a. What are the null and alternative hypotheses to be tested? c. Use the Large-Sample Test of a Population Mean applet to find the \(p\) -value of this test. d. Based on your results from part c, what conclusions can you draw about the average hourly wage b. Use the Large-Sample Test of a Population Mean applet to find the observed value of the test statistic.

Step by step solution

01

State the Null and Alternative Hypotheses

In order to perform a hypothesis test, we need to state the null hypothesis (H0) and alternative hypothesis (Ha). The null hypothesis states that there is no difference between the observed average hourly wage and the reported average hourly wage, while the alternative hypothesis states that there is a significant difference. Null Hypothesis (H0): \(\mu = \$10.33\) Alternative Hypothesis (Ha): \(\mu \ne \$10.33\)

02

Calculate the Test Statistic

We will now calculate the test statistic using a large-sample test, as the sample size is 40. The formula for the test statistic is: \(z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}\) where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size. Plugging in the given values, we get: \(z = \frac{\$9.75 - \$10.33}{\frac{\$1.65}{\sqrt{40}}}\) After calculating, we find the observed value of the test statistic: \(z \approx -2.10\)

03

Find the p-value Using the Large-Sample Test of a Population Mean Applet

Using the Large-Sample Test of a Population Mean applet, input the observed value of the test statistic (\(z=-2.10\)) and the type of test (two-tailed, as the alternative hypothesis is \(\mu \ne \$10.33\)). The applet will provide the p-value for this test. The p-value is found to be approximately 0.036.

04

Draw Conclusions Based on the p-value

Now we need to compare the p-value to our chosen significance level, usually denoted as \(\alpha\). If the p-value is less than the significance level, we reject the null hypothesis. In this case, let's choose a common significance level of \(\alpha=0.05\). Since our p-value (0.036) is less than the chosen significance level (0.05), we reject the null hypothesis. This means that there is evidence to suggest that the average hourly wage of the randomly selected cafeteria workers in the school district is significantly different from the reported average hourly wage of \(\$10.33\).

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Key Concepts

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

Null Hypothesis

The null hypothesis is a fundamental part of hypothesis testing. It represents a default or initial claim that there is no effect or difference between the involved variables. In this context, the null hypothesis (\( H_0 \)) assumes that the true average hourly wage of the cafeteria workers in the school district is equal to the reported average. It provides a baseline measure against which the alternative hypothesis is tested.

• Statement: \( \mu = \$10.33 \)
• Purpose: To validate or refute the claim that there's no deviation in the observed mean wage from the expected value.

Establishing the null hypothesis lays the groundwork for statistical analysis, where we'll either find evidence to reject it or not.

Alternative Hypothesis

In direct contrast to the null hypothesis, the alternative hypothesis proposes that there is indeed a significant difference or effect. In our case, the alternative hypothesis (\( H_a \)) suggests that the average hourly wage of cafeteria workers within the district differs from the reported average.

• Statement: \( \mu eq \$10.33 \)
• Role: To propose a different possibility from the null, indicating a potential discrepancy in average wages.

Testing against the alternative hypothesis determines if findings are statistically significant enough to suggest that the null hypothesis might not be true.

p-value

The p-value is a critical component in hypothesis testing. It measures the probability of observing a test statistic at least as extreme as the one observed if the null hypothesis is true. In simpler terms, it helps us understand the "unlikeliness" of obtaining our sample results given that the null hypothesis holds true.

• Lower p-values indicate stronger evidence against the null hypothesis.
• In our scenario, the p-value is approximately 0.036.

A p-value less than or equal to a chosen significance level usually leads us to reject the null hypothesis, signaling that our results are statistically significant.

Test Statistic

The test statistic quantifies the difference between the sample data and the null hypothesis. For large samples, a z-test is often used, which results in a z-value that we can compare against known statistical distributions.

• Formula: \( z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \)
• Calculation: Substituting the values gives \( z \approx -2.10 \)

This z-value indicates how many standard deviations our sample mean is from the population mean under the null hypothesis, crucially aiding in determining the skew of sample results.

Significance Level

The significance level, symbolized as \( \alpha \), defines the threshold for determining statistical significance. It represents the probability of wrongly rejecting the null hypothesis when it is true (Type I error).

• Commonly set at 0.05, but may vary depending on the field of study.
• Our analysis used \( \alpha = 0.05 \).

Decision-making leverages this level by comparing it to the p-value. If the p-value is less than or equal to \( \alpha \), the difference is considered significant, warranting a rejection of the null hypothesis and supporting the findings in terms of statistical significance.