1. **Problem statement:** A researcher wants to estimate the mean motivation level of high school students. Given: - Population standard deviation $\sigma = 3.5$ - Sample size $n = 100$ - Sample mean $\bar{x} = 8.25$ - Motivation scale from 1 to 10 We need to find confidence intervals (CI) for the mean at different confidence levels and analyze sample size and CI changes. 2. **Formula for confidence interval when population standard deviation is known:** $$CI = \bar{x} \pm z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}$$ where $z_{\alpha/2}$ is the z-score for the desired confidence level. 3. **Find $z$-scores:** - For 95% confidence, $\alpha = 0.05$, so $z_{0.025} = 1.96$ - For 93% confidence, $\alpha = 0.07$, so $z_{0.035} \approx 1.81$ 4. **Calculate margin of error (ME) and CI for 95%:** $$ME = 1.96 \times \frac{3.5}{\sqrt{100}} = 1.96 \times 0.35 = 0.686$$ $$CI = 8.25 \pm 0.686 = (7.564, 8.936)$$ 5. **Calculate margin of error and CI for 93%:** $$ME = 1.81 \times 0.35 = 0.634$$ $$CI = 8.25 \pm 0.634 = (7.616, 8.884)$$ 6. **Minimum sample size to reduce CI from part (b) by 25%:** Reducing CI by 25% means new margin of error $ME_{new} = 0.75 \times 0.634 = 0.4755$ Using formula for margin of error: $$ME = z \times \frac{\sigma}{\sqrt{n}} \Rightarrow n = \left(\frac{z \times \sigma}{ME}\right)^2$$ For 93% confidence, $z=1.81$, $\sigma=3.5$, $ME=0.4755$ Calculate: $$n = \left(\frac{1.81 \times 3.5}{0.4755}\right)^2 = \left(13.635\right)^2 = 185.9$$ Minimum sample size is $n=186$ (round up). 7. **Effect of increasing sample size to 200 and reducing confidence level from 0.95 to 0.90:** - Increasing sample size decreases margin of error, making CI narrower. - Reducing confidence level decreases $z$-value, also narrowing CI. **Therefore, the confidence interval will become narrower due to both changes.**