A local park claims that less than 15% of visitors litter. A random sample of 120 visitors finds that 25 litter. At the 0.05 significance level, test if the proportion of visitors who litter is greater than 15%.

• A. Reject $H_0$, we have enough evidence to suggest p > $.15$
• B. Fail to reject $H_0$, we do NOT have enough evidence to suggest p > $.15$

Answer: (A)

Verified step by step guidance

1. Step 1: Define the null hypothesis (H₀) and the alternative hypothesis (H₁). The null hypothesis is H₀: p ≤ 0.15, which states that the proportion of visitors who litter is less than or equal to 15%. The alternative hypothesis is H₁: p > 0.15, which states that the proportion of visitors who litter is greater than 15%.
2. Step 2: Calculate the sample proportion (p̂). The sample proportion is calculated as p̂ = x / n, where x is the number of visitors who litter (25) and n is the total sample size (120).
3. Step 3: Compute the test statistic (z). The formula for the z-test statistic for proportions is z = (p̂ - p₀) / √(p₀(1 - p₀) / n), where p₀ is the hypothesized population proportion (0.15), p̂ is the sample proportion, and n is the sample size.
4. Step 4: Determine the critical value for the z-test at the 0.05 significance level. Since this is a one-tailed test (greater than), find the z-value corresponding to a cumulative probability of 0.95 in the standard normal distribution.
5. Step 5: Compare the test statistic (z) to the critical value. If the test statistic is greater than the critical value, reject the null hypothesis (H₀). Otherwise, fail to reject the null hypothesis. Interpret the result in the context of the problem.