1. **State the problem:** We want to find a 95% confidence interval for the population proportion $p$ of spare parts that are not usable. 2. **Given data:** Sample size $n=80$, sample proportion $\hat{p}=0.25$ (25% not usable). 3. **Formula for confidence interval for a population proportion:** $$\hat{p} \pm z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$ where $z_{\alpha/2}$ is the critical value from the standard normal distribution for a 95% confidence level. 4. **Find $z_{\alpha/2}$:** For 95% confidence, $\alpha=0.05$, so $z_{0.025}=1.96$. 5. **Calculate the standard error (SE):** $$SE = \sqrt{\frac{0.25 \times (1-0.25)}{80}} = \sqrt{\frac{0.25 \times 0.75}{80}} = \sqrt{\frac{0.1875}{80}} = \sqrt{0.00234375} \approx 0.0484$$ 6. **Calculate margin of error (ME):** $$ME = 1.96 \times 0.0484 \approx 0.0949$$ 7. **Calculate confidence interval:** $$0.25 - 0.0949 = 0.1551$$ $$0.25 + 0.0949 = 0.3449$$ 8. **Interpretation:** We are 95% confident that the true proportion of unusable parts is between 0.1551 and 0.3449. **Final answer:** The 95% confidence interval for the population proportion is $$\boxed{(0.155, 0.345)}$$.