91Ó°ÊÓ

The Employment and Training Administration reported that the U.S. mean unemployment insurance benefit was \(\$ 238\) 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 average. 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. \(\quad\) At \(a=.05,\) what is your conclusion? d. Repeat the preceding hypothesis test using the critical value approach.

Step by step solution

01

Formulate Hypotheses

We are testing if the mean weekly unemployment insurance benefit in Virginia is less than the national average of $238. Thus, the null and alternative hypotheses are:- Null Hypothesis (H_0): \( \mu = 238 \)- Alternative Hypothesis (H_a): \( \mu < 238 \)

02

Calculate Test Statistic

Use the following formula to calculate the test statistic:\[z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \]where:- \(\bar{x}\) = 231 (sample mean)- \(\mu\) = 238 (population mean)- \(s\) = 80 (sample standard deviation)- \(n\) = 100 (sample size)Substitute the values into the formula:\[z = \frac{231 - 238}{\frac{80}{\sqrt{100}}} = \frac{-7}{8} = -0.875 \]

03

Determine P-value

Since this is a left-tailed test, the \(p\)-value is the probability that the test statistic is less than or equal to -0.875. Using the standard normal distribution table or calculator, find:\[p\text{-value} \approx 0.1915 \]

04

Decision Based on P-value

Compare the \(p\)-value with the significance level \(\alpha = 0.05\). If \(p\)-value \( < \alpha\), reject \(H_0\).Here, \(p\)-value (0.1915) > \(\alpha (0.05)\), so we fail to reject \(H_0\). There is not enough evidence to support the claim that the mean is below $238.

05

Critical Value Approach

Determine the critical value for \(\alpha = 0.05\) in a left-tailed test. The critical z\-value at \(\alpha = 0.05\) is -1.645. Compare z\ from Step 2 (-0.875) with the critical z\-value.Since -0.875 > -1.645, we fail to reject \(H_0\) with the critical value approach.

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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, often denoted as \( H_0 \), represents a statement of no effect or no difference. It's the default or status quo assumption in hypothesis testing. In the context of our unemployment insurance benefit example, the null hypothesis posits that the mean benefit in Virginia is the same as the national average. Formally, this is expressed as:

• \( H_0: \mu = \$238 \)

Here, \( \mu \) represents the average weekly unemployment benefit. The null hypothesis is what you set out to test, hoping to find evidence against it in favor of an alternative explanation.
Rejecting \( H_0 \) suggests something noteworthy happened, whereas failing to reject it means there lacks sufficient evidence to conclude a difference or change from the specified claim.

Alternative Hypothesis

The alternative hypothesis, denoted as \( H_a \), is the claim you are trying to find support for. It presents a statement indicating a difference or effect, suggesting that something new is happening, contrary to the status quo. In this particular test, the researcher wants to prove that the mean unemployment benefit is less in Virginia. Thus, the alternative hypothesis is expressed as:

• \( H_a: \mu < \\(238 \)

Here, \( \mu \) still denotes the average weekly unemployment benefit, but the alternative hypothesis implies it's less than the national average of \\)238. Adopting \( H_a \) means that evidence is strong enough to suggest the mean benefit actually falls below the assumed national average.

Statistical Significance

Statistical significance is a threshold set by the researcher to decide whether the evidence is strong enough to reject the null hypothesis. It's denoted by \( \alpha \), which is the probability of rejecting \( H_0 \) when it is true, also known as the Type I error rate. In most public studies, a common choice for \( \alpha \) is 0.05, indicating a 5% risk of making a Type I error.
In this test:

• The significance level \( \alpha \) was set to 0.05.

If the probability of observing the test results, under the assumption that \( H_0 \) is true, is less than \( \alpha \), then the result is statistically significant, and \( H_0 \) can be rejected. Otherwise, \( H_0 \) cannot be rejected due to lack of sufficient evidence.

p-value

The p-value is a key concept in hypothesis testing that quantifies the probability of obtaining a test result at least as extreme as the observed one, under the assumption that the null hypothesis is true. It helps in determining the strength of the evidence against \( H_0 \).
For the unemployment benefit example, the p-value was calculated to be approximately 0.1915. This is the probability of observing a sample mean as low as \\(231, assuming the true mean is \\)238.

• If \( \text{p-value} < \alpha \), then reject \( H_0 \).
• If \( \text{p-value} > \alpha \), then fail to reject \( H_0 \).

In this case, since 0.1915 > 0.05, there is insufficient evidence to reject \( H_0 \).

Critical Value

The critical value is a point on the test statistic's distribution that serves as the threshold for significance. In hypothesis testing, it helps determine the boundary beyond which we consider the results to be statistically significant.
For the left-tailed test of the unemployment benefit, the critical value at \( \alpha = 0.05 \) is found to be -1.645 using a standard normal distribution.

• If the test statistic falls below this critical value, reject \( H_0 \).
• If it's above, fail to reject \( H_0 \).

In the exercise example, the calculated test statistic of -0.875 doesn't fall beyond this critical threshold, indicating insufficient evidence to reject the null hypothesis.