Answer:
To determine whether we believe the national director's claim that more than 110 guests participated in the conservation program, we can use statistical hypothesis testing. Let's set up the hypotheses:
- **Null Hypothesis (\(H_0\)):** The proportion of guests participating is 110 or fewer out of 200 (\(p \leq 0.55\)).
- **Alternative Hypothesis (\(H_1\)):** The proportion of guests participating is more than 110 out of 200 (\(p > 0.55\)).
Given that the probability of a guest participating is \(0.45\), we can calculate the expected number of guests participating (\(np\)) under the null hypothesis.
\[ np = 200 \times 0.45 = 90 \]
Now, we can use the normal approximation to the binomial distribution (since \(np\) and \(n(1-p)\) are both greater than 5) and calculate the standard deviation (\(\sigma\)).
\[ \sigma = \sqrt{np(1-p)} = \sqrt{90 \times 0.55} \]
Finally, we can calculate the z-score for observing more than 110 guests participating.
\[ z = \frac{X - np}{\sigma} \]
where \(X\) is the observed number of guests participating. If \(z > Z_{\alpha}\), we reject the null hypothesis.
After calculating the z-score, compare it to the critical value for your chosen level of significance (\(\alpha\)). If the z-score is greater than the critical value, you would reject the null hypothesis, indicating evidence to support the claim that more than 110 guests participated in the conservation program. If not, you would fail to reject the null hypothesis.
