Let's break down the main question (a) into 3 clear steps with explanations and a final answer: --- ### **Step 1: State the Hypotheses** We are testing if the true average stopping distance exceeds 30 ft. - **Null Hypothesis** (\( H_0 \)): \( \mu = 30 \) - **Alternative Hypothesis** (\( H_a \)): \( \mu > 30 \) (since we are interested in whether it **exceeds** 30) --- ### **Step 2: Calculate the Test Statistic** First, list the data: - Observations: 32.1, 30.9, 31.5, 30.4, 31.0, 31.9 Calculate the sample mean (\( \bar{x} \)) and sample standard deviation (\( s \)): \[ \bar{x} = \frac{32.1 + 30.9 + 31.5 + 30.4 + 31.0 + 31.9}{6} = \frac{187.8}{6} \approx 31.3 \] To calculate \( s \): \[ s = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n-1}} \] - (32.1 - 31.3)^2 = 0.64 - (30.9 - 31.3)^2 = 0.16 - (31.5 - 31.3)^2 = 0.04 - (30.4 - 31.3)^2 = 0.81 - (31.0 - 31.3)^2 = 0.09 - (31.9 - 31.3)^2 = 0.36 Sum = 0.64 + 0.16 + 0.04 + 0.81 + 0.09 + 0.36 = **2.10** \[ s = \sqrt{\frac{2.10}{5}} = \sqrt{0.42} \approx 0.65 \] Now, calculate the t-statistic: \[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{31.3 - 30}{0.65 / \sqrt{6}} = \frac{1.3}{0.265} \approx 4.91 \] --- ### **Step 3: Find the P-value and Conclusion** Degrees of freedom: \( n - 1 = 5 \). Using a t-table or calculator for \( t = 4.91 \) and \( df = 5 \) (one-tailed test): - The P-value will be much **less** than 0.01 (critical value for \(\alpha = 0.01\) is about 3.365 for df = 5). **Conclusion:** Since the calculated t-value (4.91) is **greater** than the critical value, and the P-value is much less than 0.01: - **Reject the null hypothesis.** - **There is sufficient evidence to conclude the true average stopping distance exceeds 30 ft.** --- ## **Final Answer Summary** ### **Step 1:** - \( H_0: \mu = 30 \) - \( H_a: \mu > 30 \) ### **Step 2:** - \( \bar{x} = 31.3 \), \( s = 0.65 \), \( t = 4.91 \) ### **Step 3:** - P-value \( < 0.01 \) - **Reject \( H_0 \). There is sufficient evidence to conclude the true average stopping distance exceeds 30 ft.**