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Question

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. At$\alpha$ =.05, what is your conclusion?

d. Repeat the preceding hypothesis test using the critical value approach.

Solution

VerifiedAnswered 1 year ago

Answered 1 year ago

Step 1

1 of 8**a)** The task is to formulate the null and alternative hypotheses. This can be done by looking at the goal of the study.
The researcher anticipates that the mean weekly unemployment insurance benefit in Virginia is below the national average of $\$238$ per week.
Since the U.S. mean unemployment insurance benefit is $\$238$ per week, then we will make the tentative assumption that the mean weekly unemployment insurance benefit in Virginia is less than this. Therefore, the research hypothesis $\mu < 238$ is the alternative hypothesis.
Conversely, the null hypothesis is the assumption that the mean weekly unemployment insurance benefit in Virginia is above or equal to the national average of $\$238$ per week, which is written as $\mu \ge 238$.
Therefore, the null and alternative hypotheses are shown below which gives an **lower tail test**:
$$\begin{aligned}
\color{#4257b2} H_0: & \color{#4257b2} \> \mu \ge 238\\
\color{#4257b2} H_a: & \color{#4257b2} \> \mu < 238\\
\end{aligned}$$

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