91Ó°ÊÓ

The Employment and Training Administration reported the U.S. mean unemployment insurance benefit of \(\$ 238\) per week (The World Almanac, 2003 ). A researcher in the state of Virginia anticipated that sample data would show evidence that the mean weekly unemployment insurance benefit in Virginia was below the national level. a. Develop appropriate hypotheses such that rejection of \(H_{0}\) will support the researcher's contention. b. For a sample of 100 individuals, the sample mean weekly unemployment insurance benefit was \(\$ 231\) with a sample standard deviation of \(\$ 80 .\) What is the \(p\) -value? c. At \(\alpha=.05,\) what is your conclusion? d. Repeat the preceding hypothesis test using the critical value approach.

Step by step solution

01

Define the Hypotheses

To test if the mean weekly unemployment insurance benefit in Virginia is below the national average, set up the hypotheses. The null hypothesis (\(H_0\)) is that the mean benefit in Virginia equals the national level: \(H_0: \mu = 238\). The alternative hypothesis (\(H_a\)) is that the mean benefit in Virginia is less than the national level, \(H_a: \mu < 238\).

02

Determine the Test Statistic

The test statistic for the sample mean can be calculated using the formula for the z-test: \[ z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \] where \( \bar{x} = 231\), \( \mu = 238\), \( s = 80\), and \( n = 100\). Substitute these values into the formula to get the test statistic.

03

Calculate the Test Statistic

Substitute into the formula: \[ z = \frac{231 - 238}{\frac{80}{\sqrt{100}}} = \frac{-7}{8} = -0.875 \]. Thus, the calculated \(z\)-value is -0.875.

04

Find the p-value

Since this is a left-tailed test, find the p-value corresponding to the \(z\)-value of -0.875 using the standard normal distribution table or a software. The p-value is approximately 0.1908.

05

Make a Decision at \(\alpha = 0.05\)

Compare the \(p\)-value (0.1908) with \(\alpha = 0.05\). Since the \(p\)-value is greater than \(\alpha\), we fail to reject the null hypothesis \(H_0\). Therefore, there is not enough evidence to support that the mean weekly benefit in Virginia is below the national level.

06

Repeat Using the Critical Value Approach

For \( \alpha = 0.05\), the critical value for a left-tailed test is approximately -1.645. Since the calculated \(z\)-value of -0.875 is greater than -1.645, we fail to reject \(H_0\). This confirms the decision made using the \(p\)-value approach: no sufficient evidence that the mean benefit is below \(\$238\).

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Key Concepts

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

Null Hypothesis

In hypothesis testing, the null hypothesis (\( H_0 \)) is a statement that there is no effect or difference, and it serves as a baseline for comparison.
It’s the assumption that any observed effect is due to random chance. Here, we are asked to investigate the unemployment benefits in Virginia relative to the national average.

• The null hypothesis is set up to show that the mean benefit in Virginia is the same as the national level: \( H_0: \mu = 238 \).
• By default, this hypothesis assumes there is no difference in the average benefits.

When conducting a test, the goal is to determine if there is enough evidence to reject this null hypothesis. If rejected, it suggests that an alternative explanation might be true.
In hypothesis testing, rejecting \( H_0 \)means we have found statistical evidence against the default assumption. However, failing to reject it simply indicates insufficient evidence, not proof that \( H_0 \) is true.

Alternative Hypothesis

The alternative hypothesis (\( H_a \)) contrasts with the null hypothesis. It proposes a specific effect or difference exists. In this exercise, the researcher suspects the mean benefit in Virginia is less than the national mean of \( \$238 \).
The statement of the alternative hypothesis is:

• \( H_a: \mu < 238 \).

In statistical testing, the goal is to find sufficient evidence in the data to support the alternative hypothesis.
If evidence shows that the mean benefit in Virginia is significantly lower, it can lead to supporting \( H_a \).This gives insights into differences that aren't attributed to just random variations but might point to underlying causes. A well-formulated \( H_a \) helps to ensure that the test is objective and scientifically valid.

p-value

The \(p\)-value is a key concept in hypothesis testing. It's a probability measure that helps us determine the significance of the test results.
The \(p\)-value tells us how likely we are to observe the results of our test—assuming the null hypothesis is true.

• A lower \(p\)-value indicates stronger evidence against the null hypothesis.
• In our scenario, a \(p\)-value of 0.1908 means the data presents insufficient evidence to conclude that Virginia's benefits are lower than the national average.

The comparison of the \(p\)-value to a predetermined significance level (\(\alpha=0.05\) in this case) helps in decision making.

• If \(p\) < \(\alpha\), we reject the null hypothesis, supporting the alternative hypothesis.
• If \(p\) >\(\alpha\), we fail to reject the null hypothesis.

In our exercise, since 0.1908 is greater than 0.05, we are not able to rejectthe null hypothesis.

z-test

The \(z\)-test is a statistical method used to determine if there is any significant difference between the sample and the population mean.
This is applicable when the sample size is large enough (typically \(n > 30\)), and the population standard deviation is known or the sample standard deviation is a good approximation.

• The \(z\)-test uses the formula:\( z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \).
• Here, this formula compares the sample mean (\(\bar{x}\)) to the population mean (\(\mu\)), adjusting for the number of data points (\(n\)) and the sample standard deviation (\(s\)).

In this problem, we calculate\(z = -0.875\)based on a sample mean of\(\\(231\),compared to the population mean of \(\\)238\),with a sample standard deviation of\(\$80\),across 100 individuals.

• The \(z\)-value indicates how far, in terms of standard deviations, the sample mean is from the population mean.
• Since the calculated \(z\)-value does not exceed the critical value of -1.645 for a left-tailed test, we fail to reject \(H_0\).
  This decision confirms the \(p\)-value approach.