1. **State the problem:** We want to construct a 95% confidence interval for the mean length of computer chips produced, given a sample mean $\bar{x} = 1.15$ cm, sample standard deviation $s = 0.08$ cm, and sample size $n = 15$. 2. **Formula used:** For a normally distributed population with unknown variance and small sample size, the confidence interval for the mean is given by: $$\bar{x} \pm t_{\alpha/2, n-1} \times \frac{s}{\sqrt{n}}$$ where $t_{\alpha/2, n-1}$ is the critical value from the t-distribution with $n-1$ degrees of freedom. 3. **Find the critical t-value:** For a 95% confidence level and $n-1=14$ degrees of freedom, from t-tables or calculator: $$t_{0.025, 14} \approx 2.145$$ 4. **Calculate the margin of error:** $$E = t_{0.025, 14} \times \frac{s}{\sqrt{n}} = 2.145 \times \frac{0.08}{\sqrt{15}}$$ Calculate $\sqrt{15} \approx 3.873$: $$E = 2.145 \times \frac{0.08}{3.873} = 2.145 \times 0.02065 \approx 0.0443$$ 5. **Construct the confidence interval:** $$\text{Lower limit} = \bar{x} - E = 1.15 - 0.0443 = 1.1057$$ $$\text{Upper limit} = \bar{x} + E = 1.15 + 0.0443 = 1.1943$$ 6. **Interpretation:** We are 95% confident that the true mean length of the chips lies between $1.1057$ cm and $1.1943$ cm. **Final answer:** The 95% confidence interval for the mean chip length is $$\boxed{(1.106, 1.194)}$$ (rounded to three decimal places).