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Chapter 10: Problem 83

Duck hunting in populated areas faces opposition on the basis of safety and environmental issues. The San Luis Obispo Telegram-Tribune (June 18,1991 ) reported the results of 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 sufficient evidence to conclude that the majority of local residents oppose hunting on Morro Bay? Test the relevant hypotheses using \(\alpha=.01\).

Short Answer

Expert verified
If the p-value will be smaller than \(\alpha = .01 \), we will conclude that there is sufficient evidence to say the majority of local residents oppose hunting on Morro Bay. If the p-value will bigger than \(\alpha = .01 \), we will conclude that there is not enough evidence to say the majority of residents oppose hunting on Morro Bay. However, without calculating the exact p-value based on the test statistic Z, it cannot be determined that whether we reject or fail to reject the null hypothesis.

Step by step solution

01

Setting up the Hypotheses

Since the problem is asking if the majority of residents oppose this means more than half the residents need to be opposing. Therefore, we set our Null Hypothesis ( \(H_0\) ) to state 'less than or equal to half the residents oppose hunting', and our Alternative Hypothesis ( \(H_1\) ) to state 'more than half the residents oppose hunting'. Therefore: \(H_0: p \leq 0.5\), \(H_1: p > 0.5\). Here, \(p\) is the population proportion of residents who oppose hunting on Morro Bay.
02

Computing the Test Statistic

Calculate the sample proportion, \( \hat{p} \), and use this to compute the test statistic. The sample proportion is: \( \hat{p} = \frac{560}{750} = 0.747\). The test statistic is computed as: \( Z = \frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\). Here, \(p_0\) is the hypothesized population proportion in the Null Hypothesis, and \(n\) is the sample size. Plugging in the values, we get: \( Z = \frac{0.747-0.5}{\sqrt{\frac{0.5(1-0.5)}{750}}}\).
03

Calculating the p-value

The calculation in step 2 has given us our test statistic Z. Now, we can use the typical normal distribution to calculate our p-value, which is the probability of getting a result as extreme, or more so, than our test statistic given that the null hypothesis is true. In most statistical calculators or packages, this value can be computed using a function often named as 'pnorm' or 'qnorm'.
04

Making the Decision

We compare the p-value with the significance level, \( \alpha = .01 \). If the p-value is less than \( \alpha \), we reject the null hypothesis. If the p-value is greater than \( \alpha \), we fail to reject the null hypothesis.

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