1. **State the problem:** We want to test the claim that the mean braking distance for compound 1 tires ($\mu_1$) is less than that for compound 2 tires ($\mu_2$) using a significance level of 0.05. 2. **Recall the hypotheses:** - Null hypothesis: $H_0: \mu_1 = \mu_2$ - Alternative hypothesis: $H_a: \mu_1 < \mu_2$ 3. **Given data:** - $\bar{x}_1 = 74$, $\sigma_1 = 13.4$, $n_1 = 41$ - $\bar{x}_2 = 77$, $\sigma_2 = 14.3$, $n_2 = 41$ 4. **Test statistic formula for difference of means (z-test):** $$ z = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} $$ Since under $H_0$, $\mu_1 - \mu_2 = 0$, the formula simplifies to: $$ z = \frac{74 - 77}{\sqrt{\frac{13.4^2}{41} + \frac{14.3^2}{41}}} $$ 5. **Calculate the denominator:** $$ \sqrt{\frac{179.56}{41} + \frac{204.49}{41}} = \sqrt{4.379 + 4.987} = \sqrt{9.366} \approx 3.061 $$ 6. **Calculate the z-value:** $$ z = \frac{-3}{3.061} \approx -0.980 \text{ (rounded to three decimals)} $$ 7. **Find the p-value:** Since the alternative hypothesis is $\mu_1 < \mu_2$, this is a left-tailed test. Using standard normal distribution tables or a calculator: $$ p = P(Z < -0.980) \approx 0.1635 $$ **Final answer:** The p-value is approximately **0.1635**. This p-value is greater than the significance level 0.05, so we do not reject the null hypothesis at this step.