1. **State the problem:** We have a sample of 25 glass bottles with sample mean $\bar{x} = 4.05$ mm and sample standard deviation $s = 0.08$ mm. We want to find a 95% lower confidence bound for the population mean wall thickness, assuming normal distribution. 2. **Formula and explanation:** For a normally distributed population with unknown variance, the lower confidence bound for the mean is given by: $$\text{Lower bound} = \bar{x} - t_{\alpha, n-1} \times \frac{s}{\sqrt{n}}$$ where: - $\bar{x}$ is the sample mean - $s$ is the sample standard deviation - $n$ is the sample size - $t_{\alpha, n-1}$ is the t-score from the t-distribution with $n-1$ degrees of freedom corresponding to the confidence level 3. **Identify parameters:** - $n = 25$ - Degrees of freedom $df = 24$ - Confidence level = 95%, so $\alpha = 0.05$ - For a lower bound, we use the $t$ value for $1 - \alpha = 0.95$ quantile 4. **Find the t-score:** From t-distribution tables or calculator, $t_{0.05, 24} \approx 1.7109$ 5. **Calculate the margin of error:** $$\text{ME} = t_{0.05, 24} \times \frac{s}{\sqrt{n}} = 1.7109 \times \frac{0.08}{\sqrt{25}} = 1.7109 \times 0.016 = 0.02737$$ 6. **Calculate the lower confidence bound:** $$\text{Lower bound} = 4.05 - 0.02737 = 4.02263$$ 7. **Interpretation:** We are 95% confident that the true mean wall thickness is at least approximately 4.023 millimeters. **Final answer:** $$\boxed{4.02 \text{ millimeters (rounded to two decimals)}}$$