1. **State the problem:** We want to find the p-value for a one-sample z-test for proportions where the null hypothesis is $H_0: p = 0.75$ and the alternative hypothesis is $H_a: p \neq 0.75$. The sample size is $n=90$ and the sample proportion is $\hat{p} = 0.82$. 2. **Formula for the test statistic:** The z-test statistic for a proportion is given by $$ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} $$ where $p_0$ is the hypothesized population proportion. 3. **Calculate the standard error:** $$ SE = \sqrt{\frac{0.75 \times (1 - 0.75)}{90}} = \sqrt{\frac{0.75 \times 0.25}{90}} = \sqrt{\frac{0.1875}{90}} = \sqrt{0.0020833} \approx 0.0456 $$ 4. **Calculate the z statistic:** $$ z = \frac{0.82 - 0.75}{0.0456} = \frac{0.07}{0.0456} \approx 1.54 $$ 5. **Find the p-value:** Since this is a two-tailed test, the p-value is $$ p\text{-value} = 2 \times P(Z > 1.54) $$ Using standard normal tables or a calculator, $$ P(Z > 1.54) \approx 0.0618 $$ So, $$ p\text{-value} = 2 \times 0.0618 = 0.1236 $$ 6. **Interpretation:** The p-value is approximately 0.124, which means there is about a 12.4% chance of observing a sample proportion as extreme as 0.82 if the true proportion is 0.75. **Final answer:** $$ p\text{-value} \approx 0.124 $$