Questions: A pollster wants to know whether the percentage of residents of a city are satisfied with a new downtown revitalization project differs from 90%. Let p be the proportion of residents who are satisfied with the project. A test will be made of H0: p=0.9 versus H,: p ≠ 0.9. Let p-hat be the sample proportion of residents that are satisfied with the project. Which value of p-hat provides stronger evidence against H0: p-hat=0.86 or p-hat=0.98?

Solution Steps

The null hypothesis \( H_0 \) states that the proportion of residents satisfied with the project is \( p = 0.9 \). The alternative hypothesis \( H_a \) is that the proportion differs from 0.9, i.e., \( p \neq 0.9 \).

We are given two sample proportions, \( \widehat{p} = 0.86 \) and \( \widehat{p} = 0.98 \). We need to determine which of these provides stronger evidence against the null hypothesis \( H_0: p = 0.9 \).

To assess which sample proportion provides stronger evidence against \( H_0 \), we calculate the absolute deviation of each sample proportion from the null proportion \( p = 0.9 \):

• For \( \widehat{p} = 0.86 \), the deviation is \( |0.86 - 0.9| = 0.04 \).
• For \( \widehat{p} = 0.98 \), the deviation is \( |0.98 - 0.9| = 0.08 \).

The larger the deviation from the null proportion, the stronger the evidence against the null hypothesis. Since \( 0.08 > 0.04 \), the sample proportion \( \widehat{p} = 0.98 \) provides stronger evidence against \( H_0 \).

Final Answer