1. **Problem Statement:** We are testing if the proportion of American adults who eat salad at least once a week is 85% based on a sample of 200 adults where 178 eat salad weekly. 2. **Hypotheses:** The null hypothesis $H_0$ assumes the population proportion $p = 0.85$. The alternative hypothesis $H_a$ tests if $p$ is greater than 0.85 because the sample proportion is higher. So, $H_0: p = 0.85$ and $H_a: p > 0.85$ (option (a)). 3. **Test Statistic Formula:** The z-test statistic for a proportion is $$z = \frac{x - np_0}{\sqrt{np_0(1-p_0)}}$$ where $x$ is the number of successes, $n$ is the sample size, and $p_0$ is the hypothesized proportion. 4. **Calculate Test Statistic:** Given $x=178$, $n=200$, $p_0=0.85$, $$z = \frac{178 - 200 \times 0.85}{\sqrt{200 \times 0.85 \times 0.15}} = \frac{178 - 170}{\sqrt{25.5}} = \frac{8}{5.0496} = 1.58$$ So, the test statistic is $z = 1.58$ (option (d)). 5. **Critical Value at 10% Significance:** For a one-tailed test at the 10% level, the critical z-value is approximately 1.28 (option (c)). 6. **Decision Rule:** Since $z = 1.58 > 1.28$, we reject $H_0$ (option (b)). 7. **Conclusion:** We reject $H_0$ and conclude that the percentage of American adults who eat salad at least once a week is higher than 85% (option (c)). 8. **Type I Error:** This error occurs if we reject $H_0$ when it is actually true. So, concluding the percentage is higher when it is not (option (c)).