Question

Duck hunting in populated areas faces opposition on the basis of safety and environmental issues. In a survey to assess public opinion regarding duck hunting on Morro Bay (located along the central coast of California), a random sample of 750 local residents included 560 who strongly opposed hunting on the bay. Does this sample provide convincing evidence that a majority of local residents oppose hunting on Morro Bay? Test the relevant hypotheses using α = 0.01.

Answer 1

Using the z-distribution, as we are working with a proportion, it is found that since the test statistic is greater than the critical value for the right-tailed test, the sample provides convincing evidence that a majority of local residents oppose hunting on Morro Bay.

What are the hypothesis tested?

At the null hypothesis, we test if no more than a majority of local residents oppose hunting on Morro Bay, that is:

$H_0: p \leq 0.5$

At the alternative hypothesis, we test if more than a majority of local residents oppose hunting on Morro Bay, that is:

$H_1: p > 0.5$

What is the test statistic?

The test statistic is given by:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

In which:

• $\overline{p}$ is the sample proportion.

• p is the proportion tested at the null hypothesis.

• n is the sample size.

In this problem, the parameters are:

$p = 0.5, n = 750, \overline{p} = \frac{560}{750} = 0.7467$

Hence, the value of the test statistic is:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

$z = \frac{0.7467 - 0.5}{\sqrt{\frac{0.5(0.5)}{750}}}$

$z = 13.5$

What is the decision?

The critical value for a right-tailed test, as we are testing if the proportion is greater than a value, with a significance level of 0.01, is of $z^{\ast} = 2.327$.

Since the test statistic is greater than the critical value for the right-tailed test, the sample provides convincing evidence that a majority of local residents oppose hunting on Morro Bay.

More can be learned about the z-distribution at brainly.com/question/16313918