1. **State the problem:** We want to test if the doors are too short, i.e., if the mean height is less than 2058.0 mm. 2. **Set up hypotheses:** - Null hypothesis $H_0$: $\mu = 2058.0$ (doors are not too short) - Alternative hypothesis $H_a$: $\mu < 2058.0$ (doors are too short) 3. **Given data:** - Sample size $n = 20$ - Sample mean $\bar{x} = 2044.0$ - Population standard deviation unknown, sample standard deviation $s = 17.0$ - Significance level $\alpha = 0.05$ 4. **Test statistic:** Since population standard deviation is unknown and sample size is small, use the t-test: $$ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} $$ where $\mu_0 = 2058.0$. 5. **Calculate the test statistic:** $$ t = \frac{2044.0 - 2058.0}{17.0 / \sqrt{20}} = \frac{-14.0}{17.0 / 4.4721} = \frac{-14.0}{3.799} \approx -3.685 $$ 6. **Degrees of freedom:** $df = n - 1 = 19$ 7. **Find the P-value:** The P-value is the probability of observing a t-value less than $-3.685$ with 19 degrees of freedom. Using a t-distribution table or calculator: $$ P = P(T < -3.685) \approx 0.0009 $$ 8. **Interpretation:** Since $P = 0.0009 < 0.05$, we reject the null hypothesis and conclude there is strong evidence that the doors are too short and unusable. **Final answer:** The P-value is approximately $0.0009$.